tag:blogger.com,1999:blog-37414104180967168272019-06-25T14:46:24.627-07:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger369125tag:blogger.com,1999:blog-3741410418096716827.post-21346508164714122752019-06-25T14:41:00.000-07:002019-06-25T14:46:12.921-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Radiation<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ABTz7UVEuQo" width="480"></iframe><br /><br /><br/><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Heat Transfer - Radiation</u><br /><br /><i>Heat radiation</i> IS NOT the same as <i>radioactivity</i>. Though, under certain circumstances (like an explosion of an atomic bomb) the heat radiation and radioactivity are both present. When we discuss the <i>heat radiation</i> we talk about a process that occurs in any object with a temperature greater than absolute zero, while <i>radioactivity</i> occurs in extreme cases of very high energy output.<br/><br/>Heat transfer through <i>radiation</i> is totally different from <i>conduction</i> and <i>convection</i>. The most important property of heat transfer by <i>radiation</i> is that heat transfer occurs without any visible material conduit that carries the heat, like molecular movement in two other cases.<br/><br/>Let's start from an example.<br/>The brightest example is our Sun, as a source of heat energy. Between Sun and Earth there is no visible material conduit, yet the heat comes to Earth and is a source of life on our planet.<br/><br/>The fundamental concept that lies in the foundation of a process of <i>heat radiation</i> is a concept of a <i><b>field</b></i>.<br/>The <i>field</i> is a region of space, where certain forces act on certain objects without visible material medium.<br/>As an example of the <i>field</i>, consider <i>gravity</i>. The Sun keeps planets on their orbits, the Earth keeps the Moon circling around, people are walking on the ground without flying away to stars etc. We did not know much about WHY the <i>gravitational field</i> exist, yet we did study its behavior, the forces involved and the laws of motion in this field.<br/><br/>There are other fields.<br/><i>Magnetic field</i> around our planet, acting on a compass, forces the arrow to point North.<br/><i>Electric field</i> exists around electrically charged objects, so other electrically charged objects are attracted to or repulsed from it.<br/><br/>In Physics we successfully study these fields, but complete understanding of WHY they have the properties that we observed is not completely clear. So, we will concentrate on properties, answering the question HOW?, not on a more fundamental question WHY?.<br/><br/>Let's start with a particular field called <i>electro-magnetic</i>. Very simplified description of this field is as follows.<br/><br/>Any electron creates an <i>electric field</i> around itself. Moving electrons, which we call <i>electric current</i> or <i><b>electric field that changes in time</b></i>, also create a <i>magnetic field</i> around them. So, <i><b>changing in time electric field creates magnetic field that changes in space</b></i>. It's an experimental fact, and we have the whole theory about properties of these fields.<br/><br/>Consider an experiment, when you move a metal rod or any other electrical conductor in a <i>magnetic field</i> or change a <i>magnetic field</i> around any electrical conductor, thus creating a <i><b>magnetic field that changes in time</b></i>. You will observe that there is an electric current in the conductor, thus creating an <i><b>electric field that changes in space</b></i>. It's an experimental fact, and we also have the whole theory about properties of this process.<br/><br/>So, <i>electric field</i> creates <i>magnetic field</i>, which, in turn, creates <i>electric field</i> etc. This is a loop of energy conversion that propagates extremely fast, with a speed of light, about 3·10<sup>8</sup>m/sec.<br/><br/>The combination of electric and magnetic forces form <i>electro-magnetic field</i> that propagates much faster than its physical medium - electrons. So, the propagation of the <i>electro-magnetic field</i> seems to be a self-sufficient process, occurring without the medium. This is a very brief and unsatisfactory explanation of the nature of the <i>electro-magnetic field</i>. We will not go much further in this explanation, but rather concentrate on the properties of the <i>electro-magnetic field</i>.<br/><br/>Assuming that we accept the existence of the <i>electro-magnetic field</i> and, however uncomfortably we feel about it, but accept that there is no need for a medium to propagate this field, we can talk about frequency of electro-magnetic transformations, that can be considered similar to oscillation of molecules in a solid. Inasmuch as the oscillations of molecules in a metal are propagated, thus transferring heat energy from hot area to cold one, oscillations of the electric and magnetic components of the <i>electro-magnetic field</i> transfers energy.<br/>This energy transfer by <i>electro-magnetic field</i> is called <i><b>radiation</b></i>.<br/><br/>As in a case of oscillating molecules in a solid, carrying more heat energy when oscillation is more intense (higher frequency), the <i>electro-magnetic field</i> oscillation carries energy with higher frequencies being more "energetic" than lower.<br/><br/>Interestingly, receptors in our skin feel the temperature of a solid object, that is, we feel the intensity of oscillation of its molecules. Similarly, we feel the warm rays of Sun on our skin, that is, we feel the intensity of electro-magnetic oscillation of the electro-magnetic field.<br/>What's more remarkable, we see the light. Apparently, electro-magnetic oscillations in certain frequency range act upon censors in our eyes, thus we see the light. Moreover, in this visible range of frequencies different frequencies of electro-magnetic oscillation produce effect of different colors in our eyes.<br/><br/>As you see, the light and heat of radiation have the same source - the oscillation of electro-magnetic field, the only difference is the frequency. In other words, the <i>light</i> and <i>heat radiation</i> are manifestations of the same process of transferring energy by the oscillations of the components of the <i>electro-magnetic field</i>.<br/><br/>An object does not have to have a temperature of the Sun to emit <i>heat radiation</i>. All objects that have temperature higher than absolute zero emit thermal radiation of some frequencies. Usually, the whole spectrum of frequencies of electro-magnetic oscillations is emitted by objects. Lower frequencies (usually called infrared) are felt by skin receptors, higher frequencies are visible by an eye. Frequencies higher than those visible by a human eye are called ultraviolet. Even higher frequencies are called X-rays, which can be produced by special equipment and, depending on intensity and time of exposure, can represent a health hazard. Even higher intensity and high frequencies are called gamma rays, and they are produced in extreme cases like nuclear explosion or nuclear reactor meltdown, and they are extremely dangerous and are usually meant, when the term <i>radioactivity</i> is used. All frequencies can be observed using some scientific instruments.<br/><br/>Any object, placed in the outer space will emit its heat energy through radiation until its temperature will reach absolute zero. Our Sun emits huge amounts of energy in all spectrum of frequencies in all directions and, eventually, run out of heat energy and go dark.<br/><br/>The intensity of radiation, that is amount of heat radiated per unit of time per unit of area of an object depends, as in other cases of heat transfer, on the temperature of an object and temperature of surrounding environment.<br/>In the complete vacuum with no other source of energy the radiation intensity of an object is proportional to the fourth degree of its absolute temperature in °K:<br/><i><b>q = </b>σ<b>·T<sup>4</sup></b></i> where<br/><i>σ = 5.67·10<sup>−8</sup> W/(m<sup>2</sup>·°K<sup>4</sup>)</i><br/>is the Stefan-Boltzmann constant.<br/>This is the Law of Stefan-Boltzmann. Its derivation is complex and is outside of the scope of this course.<br/><br/>Radiation is not only emitted by objects with temperatures above absolute zero, but also can be absorbed by them and even reflected. While ability to absorb the heat is common for other heat transfer types (conduction and convection), reflection is a specific property of <i>heat radiation</i>. More precisely, it's a specific property of oscillations of the electro-magnetic field.<br/>Obvious application of this property is the usage of mirrors that reflect the oscillations of the electro-magnetic field in a very broad spectrum of frequencies, including the visible light.<br/>An example of absorbed radiation is a slice of bread toasted in the electric toaster. It absorbs the thermal radiation emitted by electric coils, that changes the bread's structure.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-9125641408687227262019-06-21T14:21:00.001-07:002019-06-21T14:21:58.347-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Convection<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/WfhCtIcRJTU" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat Transfer - Convection</u><br /><br />As in a case of conduction, we start with a statement: <b>heat</b> is a form of <b>internal energy</b> that is related to molecular movement.<br />However, while heat transfer during the process of <i>conduction</i> occurs between molecules oscillating around their relatively fixed positions and transferring their internal energy by "shaking" the neighboring molecules, <i>convection</i> occurs when molecules are free to travel in different directions and carry their internal energy with them.<br /><br />In other words, <i>conduction</i> is a pure transfer of energy on a micro level from one oscillating molecule in a relatively fixed position to another such molecule, while <i>convection</i>occurs when molecules freely fly away from their positions, carrying their internal energy with themselves, thus transferring energy on a macro level.<br /><br />It should be noted that, when dealing with solid objects, <i>conduction</i> is a prevailing way of heat transfer, while in liquids and gases the main way of heat transfer is <i>convection</i>. It does not mean that <i>conduction</i> does not occur in liquids or gases, it does, but it does not constitute the major way of heat transfer. Much more heat is transferred through the mechanism of <i>convection</i><br /><br />Here are a few examples of heat transfer through <i>convection</i>:<br />(a) heating up water in a pot; heat is carried from hot bottom of a pot up by hot (fast moving with high kinetic energy) molecules;<br />(b) circulation of air in the atmosphere from hot places to cold;<br />(c) circulation of water in oceans from hot places to cold.<br /><br />Describing <i>convection</i>mathematically is not a simple task.<br />While in case of <i>conduction</i> we can use a relatively simple Fourier's Law of Thermal Conduction<br /><i><b>q(x) = −k·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />that describes the heat flow as a function of how fast the temperature between the layers of conducting material changes (<i>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i>) and properties of the material itself (<i>conductivity coefficient <b>k</b></i>), the process of <i>convection</i> is significantly more complex, described by convection-diffusion differential equations that are beyond the scope of this course.<br /><br />However, for practical purposes we can use a similar formula that puts the amount of heat transferred by <i>convection</i>process in a liquid or gas during a unit of time through a unit of area as proportional to a difference of temperatures between the layers of liquid or gas and a <i>convective heat transfer coefficient <b>h</b></i> that depends on the physical properties of this liquid or gas:<br /><i><b>q = −h·(T<sub>2</sub>−T<sub>1</sub>)</b></i><br /><br />This formula puts amount of heat <i><b>q</b></i> going through a layer of a unit area of liquid or gas during a unit of time as proportional to a difference of temperatures between bounding surfaces of this layer <i><b>T<sub>2</sub>−T<sub>1</sub></b></i> and some physical properties of liquid or gas expressed in <i>convective heat transfer coefficient <b>h</b></i> that, in turn, depend on such properties as <i>viscosity</i>, <i>density</i>, the type of flow (turbulent or laminar) etc.<br /><br />Consider an example.<br />A round steam pipe of temperature 100°C goes through a room with air temperature 25°C. We have to calculate the amount of heat from the pipe to select an air conditioner required to neutralize the heat from a pipe and keep the room temperature at that level.<br />Assume that the pipe's length is 4m, diameter 0.2m and the <i>convective heat transfer coefficient</i> of air is 40J/(sec·m²·°C). As we know, J/sec is a unit called "watt", so we will use W instead of J/sec.<br /><br />The heat transfer per unit of time through a unit of area of a pipe is, therefore,<br /><i><b>q = 40·(100−25) = 3000(W/m²)</b></i><br />The pipe's area is<br /><i><b>A = π·0.2·4 = 2.512(m²)</b></i><br />Therefore, the pipe is producing the following amount of heat:<br /><i><b>Q = 3000·2.512 = 7536(W)</b></i><br /><br />So, we need an air conditioner that can extract 7536W of heat from the room to maintain stable temperature of 25°C.<br />Usually, the power of air conditioners is measured in BTU/hr (1 watt = 3.41 BTU/hr). So we need an air conditioner of approximately 2200 BTU/hr - a relatively small one.<br /><br />Another example.<br />Outside temperature is 40°C, inside a room we want temperature 25°C. The glass wall between a room and outside air has an area of 20m². What kind of air conditioner is needed to maintain the room temperature at 25°C, assuming the <i>convective heat transfer coefficient</i> of air is 40W/(m²·°C)?<br /><br /><i><b>Q = 40·(40−25)·20 = 12000(W)</b></i><br />This is equivalent to about 3500 BTU/hr.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-2963730674516771812019-06-20T12:36:00.001-07:002019-06-20T12:36:23.797-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Conduction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/0UaoqluO4OE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat Transfer - Conduction</u><br /><br />As we know, heat is a form of internal energy that is related to molecular movement.<br /><br />For solids the molecular movement is usually restricted to molecules' oscillation around some neutral positions.<br /><br />For liquids the freedom of molecular motion is greater, but still restricted by external forces, like gravity, and surface tension. The average distance between molecules of liquids is relatively constant.<br /><br />Gas molecules are usually taking all the space available for them. Such forces as gravity also restrict their movement (otherwise, the air molecules would fly away from our planet), but still allow substantial freedom. The average distance between molecules of gas mostly depends on a reservoir the gas is in, the larger the reservoir - the larger average distance between molecules.<br /><br />Transfer of heat is transfer of molecular movement from one object or part of an object to another object or part of an object, from an object or part of an object with more intense molecular movement (relatively warmer) to an object or part of an object with less intense movement (relatively cooler).<br /><br />There are three major ways to transfer heat from a hot object to a cold one:<br /><i>Conduction</i>,<br /><i>Convection</i>,<br /><i>Radiation</i>.<br />This lecture explains a concept of <i>conduction</i>.<br /><br /><i><b>Conduction</b></i><br /><br /><i>Conduction</i> of heat energy is a transfer of molecular movement mostly applicable to solid objects - between two solid objects that touch each other, having an area of a contact, or within one object, one part of it having different temperature than another.<br />The <i>conductivity</i> is present in heat transfer in liquids and gases, but there it's usually combined with another form of heat transfer - <i>convection</i>, while in solids it's not the case, and we can study <i>conductivity</i> by itself.<br /><br />An example is a building wall, one side of which towards the outside having temperature of the air outside, while inner surface of the wall having room temperature. The heat energy constantly flows from a warmer surface of the wall to the opposite cooler one with some <i>rate of flow</i> that depends on the <i>thermal conductivity</i> of the wall material. The wall material with higher level of <i>thermal conductivity</i> will transfer more heat energy to a cooler side within unit of time per unit of area, which is usually not a desirable property of the building walls.<br /><br />The mechanism of heat transfer through <i>conductivity</i> can be explained as follows.<br />Imagine two objects (or two parts of the same object), a hot one with higher intensity of molecular movement and a cold one with lower intensity level of molecular movement, that touch each other along some surface, while completely insulated from heat around them. For example, you put a cold metal spoon into a styrofoam cup with hot tea.<br /><br />Molecules of a hot object are hitting the molecules of a cold one, thus forcing the molecules of a cold object to move faster. These faster molecules of a cold object, in turn, hit their neighbors, forcing them to move faster. This process of transferring heat energy through contacting surfaces continues until the intensity of molecular movement gradually equalizes on average. A hot object will lose some energy of molecular movement, while a cold one will gain it. As a result, the temperatures of both will equalize.<br /><br />Since the heat energy is a kinetic energy of molecular movement, that is a sort of mechanical energy, we expect that the total amount of energy for an isolated system will remain constant, whatever a hot object loses in its kinetic energy of molecular movement will be gained by a cold object. The total amount of heat energy will remain the same.<br /><br />If you put a silver spoon into a cup of hot tea, it will heat up faster than a spoon made of steel, which, in turn, will heat up much faster then a spoon made of wood. The reason for this is that the <i>thermal conductivity</i> of different materials is different.<br /><br />We can experimentally measure the thermal conductivity of different solids by having a standard rod of any solid material at certain starting temperature and heating its one end by bringing it to contact with some hot object. Measuring the temperature on the other end after different time intervals will give us a picture of growing temperature.<br /><br />Some materials with higher thermal conductivity will have the temperature at the opposite end of a rod growing faster than in case of other materials.<br />Metals have much higher heat conductivity then plastic or wood, for example. That's why the handle of a tea kettle is usually made of plastic or wood. Diamonds have one of the highest thermal conductivity, even higher than silver.<br /><br />More precise definition of <i>thermal conductivity</i> is related to a concept of <i>heat flux</i>(sometimes, called <i>heat flow density</i> or <i>thermal flux</i>, or <i>thermal flow density</i>). <b>Heat flux is an amount of heat energy flowing through a unit of area during a unit of time</b>.<br /><br />Let's examine how heat flows through a building wall made of some uniform material from a warm room to cold air outside the building.<br />Assume, the room temperature is <i><b>T<sub>room</sub></b></i> and the cold air outside the building has temperature <i><b>T<sub>air </sub></b></i>. If the thickness of a wall is <i><b>L</b></i>, the temperature inside the wall <i><b>T(x)</b></i>, as a function of the distance <i><b>x</b></i> from the surface facing outside, gradually changes from <i><b>T(0)=T<sub>air</sub></b></i> to <i><b>T(L)=T<sub>room</sub></b></i>.<br /><br />It is intuitively understandable and experimentally confirmed that amount of heat energy flowing through a unit of area of such a wall during a unit of time (<i>heat conductivity</i> of a wall) is proportional to a difference between temperatures <i><b>T<sub>room</sub></b></i>and <i><b>T<sub>air</sub></b></i> and inversely proportional to a thickness of a wall <i><b>L</b></i>:<br /><i><b>q = −k·(T<sub>room</sub> − T<sub>air </sub>) <span style="font-size: medium;">/</span> L</b></i><br />(negative sign is used because the flow of heat is opposite to a direction of temperature growth).<br /><br />The situation with <i>heat flux</i>might be compared with a water flow down a river between two points A and B. The difference in levels above the sea level of these points is similar to a difference in temperature between the inside and outside walls of a building. The distance between points A and B is similar to a thickness of a wall. It's reasonable to assume that amount of water flowing through a unit of area in a unit of time will be proportional to a difference between the levels of points A and B above the sea level and inversely proportional to a distance between these two points.<br /><br />To make this definition of the <i>heat flux</i> more precise and independent of the way how the heat flows inside the wall, let's consider a thin slice of wall parallel to both sides from a point at distance <i><b>x</b></i> from the outdoor cold side to a point at distance <i><b>x+</b></i>Δ<i><b>x</b></i>.<br />The heat flow through this thin slice of a wall, as a function of distance <i><b>x</b></i>, can be expressed similarly to the above:<br /><i><b>q(x)=−k·</b></i>[<i><b>T(x+</b></i>Δ<i><b>x)−T(x)</b></i>]<i><b> <span style="font-size: medium;">/</span></b></i>Δ<i><b>x</b></i><br /><br />Next step is, obviously, to reduce the thickness of the slice by making Δ<i><b>x</b></i> infinitesimal, that is Δ<i><b>x→0</b></i>, which leads to the following definition of the <i>heat flux</i>:<br /><i><b>q(x) = −k·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />This definition was formulated by Fourier in 1822 and is called <i>Fourier's law of thermal conduction</i>.<br />The coefficient <i><b>k</b></i> is called <i>thermal conductivity</i>.<br /><br />To find the amount of heat <i><b>Q(x,A)</b></i> going through an area <i><b>A</b></i>at distance <i><b>x</b></i> from the outside wall during a unit of time, we have to multiply the <i>heat flux</i> by an area:<br /><i><b>Q(x,A) = −k·A·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />As in a case of water flow along the river, if the temperature is linearly dependent on the distance from the outside wall, the derivative is constant and the flow of heat is constant. But, if the wall material is uneven, like in case of a river bed not being a straight line down, the heat flow rate will change.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-23446821878253532262019-06-12T19:55:00.001-07:002019-06-12T20:09:50.697-07:00Unizor - Physics4Teens - Energy - Measuring Heat - Problems<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/d6iBVozDJ5w" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Measuring Heat - Problems</u><br /><br /><i>Problem 1</i><br/><br/>How much heat energy is required to raise the temperature of <b><i>1 kg</i></b> of water from <i><b>20°C</b></i> to a boiling point of <i><b>100°C</b></i>?<br/>Assume the specific heat capacity of water is<br/><i><b>C</b><sub>w</sub><b> = 4184 J/(kg·°C)</b></i>.<br/><br/><i>Answer</i><br/><b><i>Q = C·m·(T</b><sub>end</sub><b>−T</b><sub>beg</sub><b>) =<br/>= 4184·1·(100−20) =<br/>= 334,720 J</i></b><br/><br/><i>Problem 2</i><br/><br/>A piece of unknown metal of mass <i><b>M<sub>m</sub></b></i> and temperature <i><b>T<sub>m</sub></b></i> was put into an isolated reservoir filled with <i><b>M<sub>w</sub></b></i> mass of water at temperature <i><b>T<sub>w</sub></b></i>. After the system of water and metal came to thermal equilibrium, its temperature became <i><b>T</b></i>.<br/>Assume that the metal is not too hot (so, water will not vaporize) and not too cold (so, the water will not freeze).<br/>Assuming that the water's specific heat capacity is known and equals to <i><b>C<sub>w</sub></b></i>, what is the specific heat capacity <i><b>C<sub>m</sub></b></i> of the unknown metal?<br/><br/><i>Answer</i><br/><br/><i><b>C<sub>w</sub>·M<sub>w</sub>·(T−T<sub>w</sub>) =<br/>= C<sub>m</sub>·M<sub>m</sub>·(T<sub>m</sub>−T)</b></i><br/>from which <i><b>C<sub>m</sub></b></i> equals to<br/><i><b>C<sub>w</sub>·M<sub>w</sub>·(T−T<sub>w</sub>)<font size=4>/</font></b></i>[<i><b>M<sub>m</sub>·(T<sub>m</sub>−T)</b></i>]<i><b></b></i><br/><br/><i>Problem 3</i><br/><br/>A burger has about <i><b>300 kcal</b></i> of energy in it.<br/><b><i>1 kcal = 4184 J</i></b>.<br/>A person, who ate it, wants to spend this energy by climbing up the stairs. A person's mass is <i><b>75 kg</b></i>, the height between the floor is <i><b>3 m</b></i>.<br/>Assume that only 25% of energy in the food can be used for climbing, while the other 75% is needed to maintain our body's internal functions.<br/>Counting from the ground floor as floor #0, to what floor can a person climb using that energy from a burger?<br/><br/><i>Answer</i><br/><b><i>E</b><sub>bur</sub><b> = 300</b>kcal<b> · 4184</b>J/kcal<b> =<br/>= 1,255,200</b>J<b></i></b><br/><b><i>E</b><sub>climb</sub><b> = 0.25·E</b><sub>bur</sub><b> =<br/>= 313,800</b>J<b></i></b><br/><b><i>E</b><sub>floor</sub><b> = 75</b>kg<b> · 9.8</b>m/sec²<b> · 3</b>m<b> =<br/>= 2205</b>J<b></i></b><br/><b><i>N = E</b><sub>climb</sub><b> / E</b><sub>floor</sub><b> ≅ 142</b> floors</i><br/><br/><i>Problem 4</i><br/><br/>An ice of mass <i><b>0.1</b>kg</i> has temperature <i><b>−10</b>°C</i>.<br/>What's the minimum amount of water <i><b>M</b></i> at temperature <i><b>20</b>°C</i> needed to melt it?<br/>Assume, specific heat capacity of water is <i><b>4183</b>J/(kg·°C)</i> and that of ice is <i><b>2090</b>J/(kg·°C)</i>. Assume also that the amount needed to melt ice at <i><b>0</b>°C</i> is <i><b>333,000</b>J/kg</i>.<br/><br/><i>Answer</i><br/><i><b>E</b><sub>warm</sub><b> = 2090·0.1·10 = 2090</b>J</i><br/><i><b>E</b><sub>melt</sub><b> = 333000·0.1 = 33300</b>J</i><br/><i><b>E</b><sub>need</sub><b> = 2090 + 33300 = 35390</b>J</i><br/><i><b>E</b><sub>water</sub><b> = 4183·M·20 = 83660·M</b></i><br/><i><b>E</b><sub>water</sub><b> = E</b><sub>need</sub></i><br/><i><b>35390 = 83660·M</b></i><br/><i><b>M = 0.423</b>kg</i><br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-62844118903892419332019-06-10T16:28:00.001-07:002019-06-10T16:28:10.421-07:00Unizor - Physics4Teens - Energy - Heat - Heat & Temperature<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/QQOqz5qTRXE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat and Temperature</u><br /><br /><i>Specific Heat Capacity</i><br /><br />Let's discuss the relationship between <i>heat</i> and <i>temperature</i>.<br />From the unscientific standpoint these two concepts are almost identical. Heating an object results in increase of its temperature. Increasing a temperature of an object constitutes its heating.<br />Yet, from the strictly scientific viewpoint these two concepts are different.<br /><br /><i>Heat</i> is energy of some specific type, that can be transferred from one object to another, while <i>temperature</i> is an average kinetic energy of molecules of an object.<br />The same amount of <i>heat</i>, transferred into different objects, will result in different growth of their <i>temperatures</i>, depending on many factors, like mass, chemical composition, state etc. of these objects. Analogy of this is that same amount of fuel in different cars results in achieving very different speeds and, therefore, different kinetic energies in different cars, even if the gas pedal is pushed all the way down for all cars. It's just because cars are different and their internal structure converts fuel into movement differently.<br /><br />It has been experimentally observed that to increase a temperature of an isolated object of a unit mass by a unit of temperature is independent (within reasonable level of precision) of the initial temperature of an object, but depends only on the type of object's material, composition, state etc. In other words, amount of heat needed to increase a temperature of 1 kg of water from 20°C to 21°C is the same as from 50°C to 51°C. If, instead of water, we take copper, the amount of heat, needed to increase its temperature from 20°C to 21°C, will be the same as to increase it from 50°C to 51°C, but different than that for water.<br /><br />The above experimental fact allowed to establish a concept of <i><b>specific heat capacity</b></i> for each material as an <b>amount of heat required to increase the temperature of a unit of mass of this material (1 kg in SI) by a unit of temperature (1°C or 1°K in SI)</b>.<br /><br />Thus, <i>specific heat capacity</i> of water is, as we know, one kilocalorie per kilogram per degree - <i>1 kcal/(kg·°K</i>), that is about 4184 joules per kilogram per degree - <i>4183 J/(kg·°K</i>).<br />For copper the specific heat capacity is <i>385 J/(kg·°K</i>).<br />Gold has the specific heat capacity of <i>129 J/(kg·°K</i>).<br />Uranium's specific heat capacity is <i>116 J/(kg·°K</i>).<br />Cotton's specific heat capacity is <i>1400 J/(kg·°K</i>).<br />Hydrogen's specific heat capacity is <i>14304 J/(kg·°K</i>).<br />Generally speaking, but not always, more dense, more solid materials have less specific heat capacity than less dense or liquids, which have, in turn, less specific heat capacity than gases.<br /><br />Knowing <i>specific heat capacity</i><b><i>C</i></b> of material of an object and its mass <b><i>m</i></b>, we can easily determine amount of energy Δ<b><i>Q</i></b>needed to heat it up by Δ<b><i>T</i></b>degrees:<br />Δ<b><i>Q = C·m·</i></b>Δ<b><i>T</i></b><br />Inversely, knowing the amount of heat supplied, we can determine an increase in temperature:<br />Δ<b><i>T = </i></b>Δ<b><i>Q<span style="font-size: medium;">/</span>(C·m)</i></b><br />Notice, that increment of temperature Δ<b><i>T</i></b> and increment of heat energy Δ<b><i>Q</i></b> can be both positive or both negative, which means that an object, that has increased its temperature, has increased (gained, consumed) energy, and the object that decreased its temperature, has decreased (lost, released) energy.<br /><br /><i>Change of State</i><br /><br />Consider specific heat capacity of ice and water:<br />Ice: <i>2090 J/(kg·°K)</i><br />Water: <i>4183 J/(kg·°K)</i><br />Both these substances exist at temperature about 0°C=273°K. That means that we can heat <nobr>1 kg</nobr> of ice up to 0°C spending 2090 joules per each degree of temperature, but to increase the temperaature of the water around 0°C we have to spend 4183 joules per each degree. But, while ice is melting into water, which takes some time, both states, solid and liquid, exist side by side and the temperature of the water will not rise while the ice is not completely melted, we have to spend heat energy just on melting without actually changing the temperature of a substance, that remains around 0°C during the melting process.<br /><br />This experimental observation leads us to believe that, while graduate change of temperature for any specific state of matter linearly depends on the amount of heat supplied, change of state (like <b>melting</b> or <b>freezing</b>, or <b>evaporating</b> etc.) brings an element of non-linearity to this dependency.<br />More precisely, the temperature, as a function of amount of heat supply, in case of an object going through transformation of state from solid to liquid, looks like this (for better view, right click on the picture and open it in another tab of a browser):<br /><img src="http://www.unizor.com/Pictures/HeatTemperature.png" style="height: 150px; width: 200px;" /><br />As seen on this graph, supplying heat to ice will increase its temperature proportionally to amount of heat supplied.<br />Then, when the temperature reaches 0°C, ice will start melting and new heat will not change the temperature of ice/water mix, but will be used to change the state of matter from solid (ice) to liquid (water).<br /><br />This process of melting consumes heat without increasing the temperature until all ice is melted. Such a transformation of state that requires supply of heat energy is called <i><b>endothermic</b></i>.<br />Then, when the process of melting is complete, and all ice is transformed into water, the temperature of water will start increasing, as the new heat is supplied, but with a different coefficient of linearity relatively to the heat than in case of ice, since water's <i>specific heat capacity</i> is different than that of ice.<br />The graph above is characteristic to any heating process, where the transformation of the state of matter is involved. It is relatively the same for transforming liquid to gas (evaporation) or solid to gas (sublimation).<br /><br />In case of a reversed transformation from liquid to solid (freezing) or gas to liquid (condensation) or gas to solid (deposition) the heat energy must be taken away from a substance. It is the same amount in absolute value as was needed to supply to solid to melt it into liquid or to liquid to vaporise it into gas, or to solid to vaporize it into gas. So, in case of freezing, condensation or deposition we deal with the process of decreasing heat energy of a substance. Such a process is called <i><b>exothermic</b></i>.<br /><br />Amount of heat energy needed to transform a substance from one state to another is also experimentally determined and, obviously, depends on a substance, its mass and a kind of transformation it undergoes.<br />For example, melting of 1 kg of ice requires 333,000 joules of heat energy to be supplied. The same amount of heat energy should be extracted from the water at 0°C to freeze it into ice.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-89101864694300283532019-06-10T14:19:00.001-07:002019-06-10T14:19:39.219-07:00Unizor - Physics4Teens - Energy - Heat - Measuring<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/5ekDwNd0PCo" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Measuring Heat - Calorie</u><br /><br />Heat and temperature were known and researched by scientists long before the molecular movement was determined as their essence. As a result, attempts of measuring the amount of heat were unrelated to kinetic energy of the molecules.<br /><br />The unit of amount of heat was defined as amount of heat needed to increase the temperature of one gram of water by one degree Celsius (or Kelvin), and this unit of amount of heat was called a <i>calorie (cal)</i>.<br /><br />Obviously, this definition has its flaws. For example, the amount of heat needed to heat the same amount of water by the same temperature depends where exactly on Earth we are. The higher we are above the sea level - the less heat is needed. It also depends on the chemical purity of water. Also it's not obvious that the amount of heat needed to heat the water from 1°C to 2°C is the same as amount of heat needed to heat it from 88°C to 89°C, though within reasonable level of precision we do assume that this is true.<br /><br />With the development of molecular theory of heat and establishing relationship between heat and kinetic energy of molecules there was a need to put into correspondence existing units of heat (<i>calories</i>) and units of mechanical energy (<i>joules</i>). A simple experiment allowed to do just that.<br /><br />Imagine a standing on the ground large reservoir with known amount of water <i><b>M</b></i> of depth <i><b>H</b></i> from the surface to the bottom, kept at certain known temperature <i><b>T</b></i>°, and a relatively small stone of known mass <i><b>m</b></i>, having the same temperature <i><b>T</b></i>°, kept on the level of the surface of the water in a reservoir. This stone has certain known amount of potential energy <i><b>E=m·g·H</b></i>relatively to the bottom level of a reservoir.<br />Now we let the stone go down the reservoir to its bottom. Its potential energy relatively to the bottom of a reservoir decreases to zero. Where did the potential energy go? It's used to stir the water, thereby increasing the kinetic energy of its molecules.<br /><br />Because of this more intense movement of molecules, the water will increase its temperature. Potential energy of a stone will turn into kinetic energy of the molecular movement of water. If the temperature of the water has risen by Δ<i><b>T</b></i>°, potential energy of a stone <i><b>m·g·H</b></i> (in <i>joules</i>) equals to <i><b>M</b></i>·Δ<i><b>T</b></i> (in <i>calories</i>).<br /><br />Precise experiments like the above allowed to determine the correspondence between historical measure of the amount of heat in <i>calories</i> under different conditions and contemporary one in units of energy in SI - <i>joules</i>. This correspondence had been established approximately as<br /><i><b>1 calorie = 4.184 joules</b></i><br />but under different conditions (initial temperature, air pressure etc.) it might be equal to a slightly different value.<br /><br />Besides <i>calorie</i>, which is a relatively small amount of heat, the unit <i>kilocalorie (kcal)</i> had been introduced.<br />As is obvious from its name, <nobr><i><b>1 kilocalorie = 1000 calories</b></i>.</nobr><br /><br />The amount of energy contained in food and in some other practical cases very often is measured in <i>kilocalories</i>, which sometimes are called <i>large calories</i>, while a <i>calorie</i> is sometimes called <i>small calorie</i>. Unfortunately, the word "large" in many cases is omitted, which might cause misunderstanding.<br />From the experimental viewpoint, <i><b>1 kcal</b></i> is amount of heat needed to heat <i><b>1 kg</b></i> of water by one degree Celsius (or Kelvin).<br /><br />To avoid problems with the definition of <i>calorie</i> as amount of heat needed to warm up one gram of water by one degree Celsius, the contemporary <b>scientific definition</b> of <i>thermocalorie</i> is<br /><i><b>1 thermocalorie = 4.1833 joules</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-79690377419782082032019-06-03T14:31:00.003-07:002019-06-03T14:32:23.557-07:00Unizor - Physics4Teens - Energy - Ideal Gas Kinetics - Problems <a href="https://youtu.be/MYEgnMaBJ0w"></a> <i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Problem 1</u><br /><br />How much kinetic energy have all the molecules in the room?<br/>How fast should an average size car move to have this amount of kinetic energy?<br/><br/>Assume the following:<br/>(a) the room dimensions are <i>4</i>x<i>4</i>x<i>3 meters</i> (that is, <i><b>V=48m³</b></i>);<br/>(b) normal atmospheric pressure is <i>100,000 Pascals</i> (that is, <i><b>p=100,000N/m²</b></i>);<br/>(c) a mass of an average size car is <i>2,000 kg</i> (that is, <i><b>M=2,000kg</b></i>).<br/><br/><i>Solution</i><br/><br/>From the lecture on <i>Kinetics of Ideal Gas</i> we know the relationship between the pressure on the walls of a reservoir, volume of a reservoir and total kinetic energy of gas inside this reservoir:<br/><i><b>p = (2/3)E</b><sub>tot </sub><b><font size=5>/</font>V</b></i><br/><br/>From this we derive a formula for total kinetic energy:<br/><i><b>E</b><sub>tot</sub><b> = (3/2)·p·V</b></i><br/><br/>Substituting the values for pressure and volume, we obtain<br/><i><b>E</b><sub>tot</sub><b> = (3/2)·100,000·48 =<br/>= 7,200,000</b>(joules)</i><br/><br/>The kinetic energy of a car is <br/><i><b>E = M·v²<font size=5>/</font>2</b></i><br/>Therefore, given the kinetic energy and mass, we can determine the car's speed:<br/><i><b>v = √<span style='text-decoration:overline'>2·E/M</span></b></i><br/>Substituting calculated above <i><b>E</b><sub>tot</sub><b>=7,200,000</b>(J)</i> for <i><b>E</b></i> and the value for mass <i><b>M=2,000</b>(kg)</i>, we obtain<br/><i><b>v = √<span style='text-decoration:overline'>2·7,200,000/2,000</span> ≅<br/>≅ 85</b>(m/sec)<b> ≅<br/>≅ 306</b>(km/hour)<b> ≅<br/>≅ 190</b>(miles/hour)</i><br/><br/><br/><u>Problem 2</u><br /><br />Given the temperature, pressure and volume of the air in a room, determine the number of gas molecules in it.<br/><br/>Assume the following:<br/>(a) the room dimensions are <i>4</i>x<i>4</i>x<i>3 meters</i> (that is, <i><b>V=48m³</b></i>);<br/>(b) normal atmospheric pressure is <i>100,000 Pascals</i> (that is, <i><b>p=100,000=10<sup>5</sup>N/m²</b></i>);<br/>(c) temperature is <i>20°C</i> (that is, <i><b>T=20+273=293°K</b></i>).<br/><br/><i>Solution</i><br/>Recall the combined law of ideal gas<br/><i><b>p·V<font size=5>/</font>T = k</b><sub>B</sub><b>·N = const</b></i><br/>where<br/><i><b>k</b><sub>B</sub><b> = 1.381·10<sup>−23</sup> (J/°K)</b></i> is Boltzmann's constant and<br/><i><b>N</b></i> is the number of gas molecules in a reservoir.<br/>From this we derive the number of molecules<br/><i><b>N = p·V<font size=5>/</font>(k</b><sub>B</sub><b>·T)</b></i><br/>Substituting the values,<br/><i><b>N = 10<sup>5</sup>·48<font size=5>/</font>(1.381·10<sup>−23</sup>·293) = 0.12·10<sup>28</sup></b></i><br/>It's a lot!<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-36808442494385698552019-05-06T12:07:00.001-07:002019-05-06T12:07:05.904-07:00Unizor - Physics4Teens - Energy - Heat - Temperature, Pressure, Volume o...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/jPWoBQEEH4s" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Temperature, Pressure<br />and Volume of Ideal Gas</u><br /><br /><i>Temperature</i> is an observable macro property of an object. It's related to a particular instrument we use to measure this property.<br />Let's examine the mechanism of this measurement using a classic mercury-based or alcohol-based thermometer.<br /><br />Our first step in measurement is to make a physical contact between a thermometer and an object of measurement (for example, a human body or air in a room). When accomplished, we expect that the measurement of a thermometer would correspond to a state of average kinetic energy of molecules of an object. The reason it happens (and it takes some time to happen) is that on a micro level molecules at the surface of an object are colliding with molecules at the surface of a thermometer, and exchange the kinetic energy, eventually equalizing it. The molecules close to a surface, in turn, collide with surface molecules and also eventually equalize their average kinetic energies. This process continues until the average level of kinetic energy in all parts of an object and in a thermometer equalize.<br /><br />What is important in this case is that the total amount of kinetic energy of all molecules of an object and a thermometer remains the same. So, if an object has more intense movement of molecules and a thermometer's molecules are moving slower, the kinetic energy is transferred from an object to a thermometer. If the molecules of a thermometer are, on average, faster, then the exchange of kinetic energy will be from a thermometer to an object.<br .="" br="" /> In any case, the average kinetic energy of molecules of both an object and a thermometer equalize.<br /><br />An important consideration is that the contact between an object and a thermometer changes the average level of kinetic energy in both. The process of measuring, therefore, is not completely neutral towards an object. However, what happens in most cases is that the number of molecules inside an object we measure is usually significantly greater that the number of molecules in a thermometer. As a result, equalizing the average kinetic energy of all molecules does not significantly change the level of average kinetic energy of molecules of an object, and the level of average kinetic energy of the molecules of a thermometer is a good representation of this characteristic of an object.<br /><br />As explained in the <i>Heat and Energy</i> lecture of this course, the temperature in mercury or alcohol thermometers is an observable expansion of the volume of liquid inside a thermometer. We also indicated in that lecture that this thermo-expansion is proportional to an average of squares of velocities of molecules, that is proportional to average kinetic energy of the molecules of a thermometer, which, in turn, is equalized with average kinetic energy of molecules of an object.<br /><br />Thus, by observing the expansion of liquid in a thermometer we measure the average kinetic energy of molecules of an object, which allows us to write the following equation:<br /><i><b>T ≅ </b>AVE(<b>E</b><sub>kin</sub>)<b> = E</b><sub>ave</sub></i><br />where <i><b>T</b></i> is an observable level of liquid inside a thermometer in some units (that is, <i>temperature</i>) and<br /><i>AVE(<b>E</b><sub>kin</sub>)<b> = E</b><sub>ave</sub></i> is average kinetic energy of molecules of an object.<br />This looks more natural in a form<br /><i>AVE(<b>E</b><sub>kin</sub>)<b> = E</b><sub>ave</sub><b> ≅ T</b></i><br />since average kinetic energy of molecules <i><b>E</b><sub>ave</sub></i> (micro characteristic) cannot be easily observed, while the temperature <i><b>T</b></i> (macro characteristic) can.<br /><br />So, we have established that the temperature (in some units, starting from absolute zero) and average kinetic energy of molecules are proportional. The coefficient of proportionality between the temperature and average kinetic energy of molecules remains unknown and is different for different substances.<br /><br />Obviously, the measure of liquid in a thermometer should be calibrated and, for this equation to be true, we have to assign the zero temperature to a state of an object when all its molecules are at rest, which happens when there is no source of energy around, like in open space far from stars.<br />The convenient scale is the Kelvin scale with zero temperature on this scale corresponding to this state of molecules at complete rest and a unit of measurement of the temperature is a <i>degree</i> with the distance between the temperature of melting ice and boiling water assigned as 100 degrees on this scale.<br /><br />From the previous lecture about kinetics of <i>ideal gas</i> we know the relationship between the <i>pressure</i> of ideal gas on the walls of a reservoir, the <i>volume</i>of a reservoir and <i>average kinetic energy of the molecules</i><br /><i><b>p = (2/3)N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V</b></i><br />where<br /><i><b>p</b></i> is the gas pressure against the walls of a reservoir,<br /><i><b>N</b></i> is the number of gas molecules in a reservoir,<br /><i><b>V</b></i> volume of a reservoir,<br /><i><b>E</b><sub>ave</sub></i> is average kinetic energy of molecules,<br /><br />Alternatively, it can be written as<br /><i><b>E</b><sub>ave</sub><b> = (3/2)·p·V<span style="font-size: large;">/</span>N</b></i><br /><br />Comparing this with a derived above relationship between <i>temperature</i> (counting from the absolute zero) and <i>average kinetic energy of molecules</i>, we have established a relationship<br /><i><b>E</b><sub>ave</sub><b> = (3/2)·p·V<span style="font-size: large;">/</span>N ≅ T</b></i><br />and<br /><i><b>p·V<span style="font-size: large;">/</span>N = k·T</b></i><br />where <i><b>k</b></i> is an unknown coefficient of proportionality.<br /><br />The only thing that prevents us from determining average kinetic energy of molecules by its temperature is an unknown coefficient of proportionality <b><i>k</i></b>.<br /><br />Now we will concentrate attention on gases, as an object of measuring temperature and average kinetic energy of molecules.<br />Different gases have different molecules and different molecular mass. Using certain theory and using chemical and physical experiments, we can compare the masses of different molecules and even measure this mass in certain "units of mass" called <i>atomic mass units</i>.<br />For such a unit of molecular mass scientists used 1/12 of a mass of a single atom of carbon. So, in this system of units hydrogen molecule of 2 hydrogen atoms <i><b>H<sub>2</sub></b></i> has approximate mass of 2, oxygen molecule of two oxygen atoms <i><b>O<sub>2</sub></b></i> - approximately 32, carbon dioxide molecule <i><b>CO<sub>2</sub></b></i> had a measure of, approximately, 44 atomic mass units, etc.<br /><br />Using this measurement, we can always establish experiments with the same number of molecules of different gases. For example, if the mass of certain amount of oxygen (atomic mass of molecules <i><b>O<sub>2</sub></b></i> is 32) is 16 times greater than the mass of certain amount of hydrogen (atomic mass of molecules <i><b>H<sub>2</sub></b></i> is 2), we can assume that the number of molecules in both cases is the same.<br /><br />It has been experimentally established that, if the same number of different gas molecules are placed in reservoirs of the same volume and hold them at the same temperature, the pressure on the walls in both cases will be the same. Alternatively, if the pressure is the same, the temperature will be the same too.<br />In other words, the coefficient <b><i>k</i></b>in formula<br /><i><b>p·V<span style="font-size: large;">/</span>N = k·T</b></i><br />does not depend on the type of gas we deal with, it's a universal constant called the <i>Boltzman's constant</i>, which is equal to<br /><i><b>k</b><sub>B</sub><b> = 1.381·10<sup>−23</sup> (J/°K)</b></i><br />This was the reason to introduce a concept of <i>ideal gas</i>. All gases are, approximately, ideal to a certain degree of precision. This is related to the fact that molecules of the gas are flying with high speeds and on large distances from each other, much larger than their geometric sizes.<br /><br />Now we can write the equation between the temperature <i><b>T</b></i> (in degrees <b><i>°K</i></b> from absolute zero), average kinetic energy of molecules (in units of SI <i>joules J</i>), pressure <i><b>p</b></i> (in units of SI <i>newton/m²</i>), volume <i><b>V</b></i> (in <i>m³</i>) and number of molecules in a reservoir for <i>ideal gas</i>:<br /><i><b>E</b><sub>ave</sub><b> = (3/2)·p·V<span style="font-size: large;">/</span>N = (3/2)·k</b><sub>B</sub><b>·T</b></i><br /><br />Consequently, if we are dealing with certain fixed amount of gas (<i><b>N</b></i> molecules) then<br /><i><b>p·V<span style="font-size: large;">/</span>T = k</b><sub>B</sub><b>·N = const</b></i><br />That means that changing the pressure, volume and temperature of the same amount of gas preserves the expression <i><b>p·V<span style="font-size: large;">/</span>T</b></i><br />which is called the <b>Combined Ideal Gas Law</b>.<br /><br />For example, <u>if the absolute temperature remains the same</u>, but volume taken by certain amount of gas increases (decreases) by some factor, the pressure will decrease (increase) by the same factor, that is pressure and volume are inversely proportional to each other (<b>Boyle-Mariotte's Law</b>).<br /><br /><u>If the pressure remains the same</u>, but the volume taken by certain amount of gas increases (decreases) by some factor, the absolute temperature will increase (decrease) by the same factor, that is volume and absolute temperature are proportional to each other (<b>Charles' Law</b>).<br /><br /><u>If the volume taken by certain amount of gas remains the same</u>, but the absolute temperature increases (decreases) by some factor, the pressure will increase (decrease) by the same factor, that is absolute temperature and pressure are proportional to each other (<b>Gay-Lussac's Law</b>).<br /><br />Out of curiosity, let's use the formula<br /><i><b>E</b><sub>ave</sub><b> = (3/2)·k</b><sub>B</sub><b>·T</b></i><br />to calculate how fast the molecules of oxygen are flying in the room at some normal temperature.<br />Assume the pressure at the ground level is about <i>100,000N/m²</i> and the temperature in the room is about <i>20°C=293°K</i>. Then the average kinetic energy of a molecule of oxygen is<br /><i><b>E</b><sub>ave</sub><b> = (3/2)·1.381·10<sup>−23</sup>·293 =<br />= 6.06·10<sup>−21</sup> J</b></i><br />Mass of a molecule of oxygen <i><b>O<sub>2</sub></b></i> is <i><b>m=5.31·10<sup>−26</sup> kg</b></i><br />From the formula for kinetic energy <i><b>E=m·v²/2</b></i> we derive the average of squares of velocities of oxygen molecules as<br /><i>AVE(<b>v²</b>)<b> = 2·E</b><sub>ave </sub><b>/m =<br />= 2·6.06·10<sup>−21</sup>/5.31·10<sup>−26</sup> =<br />= 2.28·10<sup>5</sup></b></i><br />Therefore, the average speed of oxygen molecule will be equal to a square root of this number:<br /><i>AVE(<b>v</b>)<b> = 478 m/sec</b></i><br />Pretty fast moving! Take into consideration, however, that real oxygen molecules, as molecules of any real gas, are chaotically colliding with other and change the direction all the time.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-72352227384622028192019-05-01T20:26:00.001-07:002019-05-01T20:26:41.656-07:00Unizor - Physics4Teens - Energy - Heat - Kinetics of Ideal Gas<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/cIDMFCcne4E" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Kinetics of Ideal Gas</u><br /><br />In this lecture we will discuss a concept of <i>pressure</i> of gas, enclosed in some reservoir, against the walls of this reservoir.<br /><br />Usually, speaking about pressure, we have in mind an object of certain weight on a horizontal table, in which case the constant force of its weight exerts a pressure on a table equal to this weight divided by an area of the table occupied by an object or, as it's sometimes presented, the weight per unit of area.<br /><br />As an introduction to kinetics of gases, let's consider a reservoir shaped as a cube with sides aligned parallel to coordinate planes and with the length of each edge <i><b>L</b></i>.<br />Assume that there is a single molecule of gas of mass <i><b>m</b></i>flying between the opposite walls perpendicularly to them, elastically reflecting off these walls.<br />Keep in mind that, according to the Third Newton's Law, the force exerted by the wall towards the molecule, which causes the reflection of a molecule from the wall, is equal to the force exerted by a molecule towards the wall, which causes <i>pressure</i>.<br /><br />The area of each of these opposite walls equals to <i><b>L²</b></i>, which will be used to calculate the <i>pressure</i>.<br /><br />In this case of a single gas molecule, flying between the opposite walls of a reservoir, the force is variable. It exists during the time when a molecule hits the wall and then the force becomes zero until the next time this molecule hits the same wall.<br />A proper definition of <i>pressure</i>in this case is based on a concept of the average force during certain amount of time.<br /><br />Firstly, let's evaluate the average force during a single period of oscillation of a molecule between walls from a moment of time it's near one wall to a next moment it's at this position after flying to the opposite wall and returning back.<br /><br />As we know, a change of a momentum of an object equals to an impulse of the force that caused this change<br /><i><b>m·v</b><sub>end</sub><b> − m·v</b><sub>beg</sub><b> = F·τ</b></i><br />where<br /><i><b>m</b></i> is a mass of an object<br /><i><b>v</b><sub>beg</sub><b></b></i> is the velocity of an object in the beginning of the action of the force<br /><i><b>v</b><sub>end</sub><b></b></i> is the velocity of an object at the end of the action of the force<br /><i><b>F</b></i> is the force, acting on an object<br /><i><b>τ</b></i> is the time duration of the force, acting on an object.<br />Keep in mind that velocity and force are vectors in the above equation.<br /><br />In our case we consider one period of oscillation of a molecule as the time period of <i>average force</i> acted on it.<br />So, <i><b>τ</b></i> is the time between two consecutive events when a molecule is at the wall opposite to the one it collides with:<br /><i><b>τ = 2L/|v|</b></i>.<br />where <i><b>|v|</b></i> is now an absolute speed of a molecule (scalar).<br /><br />Since we are considering elastic reflection of a molecule in the opposite direction to its initial trajectory,<br /><i><b>v<sub>end</sub> = −v<sub>beg</sub></b></i><br />and the equation above can be written in scalar form<br /><i><b>2m·|v| = |F|·τ = |F|·2L/|v|</b></i><br />from which we determine the absolute average force<br /><i><b>|F| = m·|v|²<span style="font-size: medium;">/</span>L</b></i><br />Mass <i><b>m</b></i> of a molecule is, obviously, a constant during this process.<br />The average pressure of a molecule on a wall is the average force divided by the area of a wall:<br /><i><b>p = m·|v|²<span style="font-size: medium;">/</span>L³ = m·|v|²<span style="font-size: medium;">/</span>V =<br />= 2E</b><sub>kin </sub><b><span style="font-size: medium;">/</span>V</b></i><br />where <i><b>V</b></i> is the volume of a reservoir and <i><b>E</b><sub>kin</sub></i> is a <b>kinetic energy</b> of a molecule.<br /><br />Our first step towards a comprehensive theory is to consider a case of many molecules flying parallel to each other in the same reservoir as above, but with, generally speaking, different speeds. Obviously, the pressure against the wall will be greater because each molecule contributes its own force against the wall during a collision.<br />If <i><b>v<sub>i</sub></b></i> is a speed of the <i><b>i</b><sup>th</sup></i>molecule, the combined pressure will be<br /><i><b>p =</b></i> <span style="font-size: large;">Σ</span><i><b>p<sub>i</sub> =</b></i> <span style="font-size: large;">Σ</span><i><b>m·v<sub>i</sub>²<span style="font-size: large;">/</span>V =<br />= 2E</b><sub>tot </sub><b><span style="font-size: large;">/</span>V</b></i><br />where <i><b>E</b><sub>tot</sub></i> is a <b>total kinetic energy of all molecules</b> moving parallel to each other.<br /><br />Considering the number of molecules <i><b>N</b></i> remains the same, it's more convenient to express this formula in terms of an <b>average of squares of molecular speed</b> and <b>average kinetic energy</b> of molecules:<br /><i><b>v²</b><sub>ave</sub><b> = (1/N)</b></i><span style="font-size: large;">Σ</span><i><b>v<sub>i</sub>²</b></i><br /><i><b>E</b><sub>ave</sub><b> = (1/N)</b></i><span style="font-size: large;">Σ</span><i><b>E<sub>i</sub></b></i><br />Using this <b>average of squares of molecular speed</b>, the pressure is<br /><i><b>p = N·m·v²</b><sub>ave </sub><b><span style="font-size: large;">/</span>V = 2N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V</b></i><br />which brings us to a principle of <b>proportionality between pressure of the gas against the walls of a reservoir and an average of squares of the molecular speed or to an average of kinetic energy of the molecules</b>.<br /><br />Granted, we showed this only for a case of all molecules moving between two opposite walls parallel each other and perpendicularly to the walls they collide with.<br /><br />Let's make another step towards comprehensive theory and consider the chaotic movement of all gas molecules.<br />To quantitatively approach molecular movement, we will use a model of an <i>ideal gas</i>.<br />This model assumes that molecules of <i>ideal gas</i> are point-mass objects completely chaotically moving in all directions with equal probabilities and elastically colliding at random times among themselves or with all the walls of a reservoir that contains this gas.<br /><br />We further assume that no interacting forces (like gravity or electromagnetic forces) exist between them. So, only <i>kinetic energy</i> of these molecules plays the role in evaluation of the characteristics of the ideal gas.<br /><br />Every vector of velocity of any molecule can be represented as a sum of three vectors along the XYZ-coordinates<br /><i><b><span style="text-decoration-line: overline;">v</span> = <span style="text-decoration-line: overline;">v</span><sub>x</sub>+<span style="text-decoration-line: overline;">v</span><sub>y</sub>+<span style="text-decoration-line: overline;">v</span><sub>z</sub></b></i><br /><br />So, every movement of a molecule can be represented as simultaneous movement in three different directions. Depending on which wall of the reservoir is hit by a molecule, its <i>pressure</i>against the wall is determined only by one component - the one that is perpendicular to the surface of a wall.<br /><br />If molecules inside the reservoir are moving completely chaotically, they are hitting all walls with approximately the same frequency. So, for each of the six walls of a cube we can use the same logic as above for one side, except for the walls parallel to YZ coordinate plane the pressures <i><b>p<sub>x</sub></b></i> will be proportional to the average of <i><b>v<sub>x</sub>²</b></i>, for the walls parallel to XZ coordinate plane the pressures <i><b>p<sub>y</sub></b></i> will be proportional to the average of <i><b>v<sub>y</sub>²</b></i> and for the walls parallel to XY coordinate plane the pressures <i><b>p<sub>z</sub></b></i> will be proportional to the average of <i><b>v<sub>z</sub>²</b></i>.<br /><br />Considering completely chaotic character of the molecular motion in the <i>ideal gas</i>, the three pressures <i><b>p<sub>x</sub></b></i>, <i><b>p<sub>y</sub></b></i> and <i><b>p<sub>z</sub></b></i>must be the same. Similarly, averages of <i><b>v<sub>x</sub>²</b></i>, <i><b>v<sub>y</sub>²</b></i> and <i><b>v<sub>z</sub>²</b></i> must be also the same.<br /><br />Since <i><b><span style="text-decoration-line: overline;">v</span> = <span style="text-decoration-line: overline;">v</span><sub>x</sub>+<span style="text-decoration-line: overline;">v</span><sub>y</sub>+<span style="text-decoration-line: overline;">v</span><sub>z</sub></b></i>, according to three-dimensional equivalent of Pythagorean Theorem,<br /><i><b>v² = v<sub>x</sub>² + v<sub>z</sub>² + v<sub>z</sub>²</b></i><br />and, similarly, for averages<br /><i><b></b>AVE(<b>v²</b>)<b> = v²</b><sub>ave</sub><b> =<br />=</b>AVE(<b>v<sub>x</sub>²</b>)<b>+</b>AVE(<b>v<sub>y</sub>²</b>)<b>+</b>AVE(<b>v<sub>z</sub>²</b>)</i><br /><br />From the equality of averages for the ideal gas we come to the following equalities:<br /><i><b>v²</b><sub>ave</sub><b> = 3·</b>AVE(<b>v<sub>x</sub>²</b>)</i><br /><i><b></b>AVE(<b>v<sub>x</sub>²</b>)<b> = (1/3)·v²</b><sub>ave</sub><b></b></i><br /><br />Therefore,<br /><i><b>p<sub>x</sub> = N·m·</b>AVE(<b>v<sub>x</sub>²</b>)<b><span style="font-size: large;">/</span>V =<br />= (1/3)N·m·v²</b><sub>ave </sub><b><span style="font-size: large;">/</span>V =<br />= (2/3)N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V</b></i><br /><br />Similarly,<br /><i><b>p<sub>y</sub> = N·m·</b>AVE(<b>v<sub>y</sub>²</b>)<b><span style="font-size: large;">/</span>V =<br />= (1/3)N·m·v²</b><sub>ave </sub><b><span style="font-size: large;">/</span>V =<br />= (2/3)N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V</b></i><br />and<br /><i><b>p<sub>z</sub> = N·m·</b>AVE(<b>v<sub>z</sub>²</b>)<b><span style="font-size: large;">/</span>V =<br />= (1/3)N·m·v²</b><sub>ave </sub><b><span style="font-size: large;">/</span>V =<br />= (2/3)N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V</b></i><br /><br />And, since pressures on all walls are the same in the ideal gas,<br /><i><b>p = (1/3)N·m·v²</b><sub>ave </sub><b><span style="font-size: large;">/</span>V =<br />= (2/3)N·E</b><sub>ave </sub><b><span style="font-size: large;">/</span>V = (2/3)E</b><sub>tot </sub><b><span style="font-size: large;">/</span>V</b></i><br /><br />The expression of a <i>pressure</i> in terms of average kinetic energy and volume is more general than in terms of mass, average of squares of molecular speed and volume, as it encompasses a case with molecules of different masses.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-76393475022064104892019-04-17T06:40:00.001-07:002019-04-17T06:40:36.534-07:00Unizor - Physics4Teens - Energy - Temperature and Kinetic Energy<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/0wO3LmBTN7c" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat, Temperature &<br />Kinetic Energy of Molecules</u><br /><br />The goal of this lecture is to demonstrate that <i>temperature</i>, as a measure of <i>heat</i>, is proportional to <i>average of a square of molecular speed</i>.<br /><br /><i>Temperature</i> is what we see on a thermometer that measures the intensity of the molecular movement. Let's examine this process of measurement using the classical mercury-based thermometer.<br /><br />The thermometer shows different values of temperature based on physical property of mercury to expand, as its molecules are moving faster. The reason for this is that, as the molecules move faster, they cover longer distance before they are stopped by collision with other molecules or the walls or by a surface tension forces. The faster molecules push against obstacles with greater forces, and that is the reason why the volume of hot mercury is larger than the volume of the same mass of cold mercury.<br /><br />Let's quantify this increase in volume.<br />There is a force of resistance that acts against the movement of the molecule, as it pushes its way through, that stops this movement. In some way it's analogous to throwing a stone vertically up against the force of gravity. The force of gravity eventually stops the movement upward, and the stone falls back on the ground.<br /><br />The height of the trajectory of a stone depends on the initial speed it has at the beginning of movement on the ground level. Assume, the initial speed is <i><b>V</b></i>. Then the kinetic energy in the beginning of motion is <i><b>M·V²/2</b></i>, where <i><b>M</b></i> is a mass of a stone. The potential energy relative to the ground in the beginning of motion is zero.<br />At the highest point at height <i><b>H</b></i>the kinetic energy is zero but potential is <i><b>M·g·H</b></i>, where <i><b>g</b></i> is a free fall acceleration.<br /><br />As we know, the total mechanical energy is constant in this case, from which follows that kinetic energy in the beginning of motion (with potential energy zero) equals to potential energy at the highest point (with kinetic energy zero):<br /><i><b>E<sub>kin</sub> = M·V²/2 = M·g·H</b></i><br />From this follows<br /><i><b>H = V²/(2g)</b></i><br />So, the height of the rise and the square of initial velocity are proportional to each other, independently of mass.<br /><br />Obviously, one mercury molecule does not do much, but when the heat is relatively evenly distributed among all molecules, their combined efforts push the surface of the mercury in the tube sufficiently noticeably, and the height it rises is proportional to average of a square of molecular speed.<br /><br />The thermometers are different, their design and construction differ, they all must be calibrated, but one thing remains - a principle of <b>proportionality between geometrical expansion (that is, temperature, as we read it on a thermometer's scale) and an average of squares of the molecular speed</b>.<br /><br />It is important to point out that this proportionality should be understood in absolute terms. That is, if the speed is zero, the temperature is zero as well. So, we are talking about temperature scales with zero representing <i>absolute zero</i> on the Kelvin's scale, like in space far from any source of energy.<br />The coefficient of proportionality, of course, depends on the units of measurement.<br /><br />So, if the average speed of molecules doubles, the temperature on the Kelvin's scale rises by a factor of 4.<br /><br />Inversely, if an object is at temperature 0°K (absolute zero), the kinetic energy of its molecule is zero, that is no molecular movements.<br />If we observe that the temperature has risen by a factor of 4, we can safely assume that the average speed of molecules doubles.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-25128332571872825572019-04-15T10:52:00.001-07:002019-04-15T10:52:03.781-07:00Unizor - Physics4Teens - Energy - Heat - Molecular Movement<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/AO9b5Mn9Eao" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Molecular Movement</u><br /><br /><i><b>Molecules</b></i><br /><br />In order to understand the nature of <i>heat</i> and <i>temperature</i>, we have to go inside the objects we experiment with.<br />If we divide a drop of water into two smaller drops, each of these smaller drops will still be water. Let's continue dividing a smaller drop into even smaller and smaller. There will be a point in this process of division, when further division is not possible without changing a nature of the object, it will no longer be water.<br /><br />From what we know now, the water consists of hydrogen and oxygen - two gases, connected in some way. In our process of division of a drop of water, we will reach such a point that, if we divide this tiniest drop of water, the result will be certain amount of hydrogen and certain amount of oxygen, but no water.<br />That tiniest amount of water, that still preserves the quality of being water, is called a <b><i>molecule</i></b>.<br />Similarly, tiniest amount of any substance, that retains the qualities of this substance, is called a <b><i>molecule</i></b>.<br />Incidentally, if we divide a <i>molecule</i> of any substance, the result will be certain <n><i>atoms</i>, that possess completely different qualities.<br /><br />In our discussion about <i>heat</i> <u>we will not go deeper than the <i>molecules</i></u> because our purpose at this stage is to study the nature of <i>heat</i> as it relates to different objects and substances, so the preservation of the qualities of these objects and substances is important. That's why in this part of a course we will not cross the border between <i>molecular</i> and more elementary <i>atomic</i> level.<br /><br />Any object or substance we are dealing with consists of certain number of <i>molecules</i> - the smallest particles that retains the qualities of this object or substance. This number of <i>molecules</i>, by the way, for regular objects we see and use in practical life, is extremely large because the size of molecules is extremely small. We cannot see individual molecules with a naked eye. Only special equipment, different in different cases, can help us to see individual molecules. And, being so tiny, molecules of different substances are different in size among themselves.<br />And not only in size. Since the molecules contain different, more elementary particles called <i>atoms</i>, the configuration of these <i>atoms</i> that form a <i>molecule</i> is different for different <i>molecules</i>. Thus, a molecule of water contains two atoms of hydrogen and one atom of oxygen that connects hydrogen atoms into some three-dimensional construction. A molecule of protein consists of many different types of atoms and its structure and size are quite different from the molecule of water.<br /><br /><i><b>States of Matter</b></i><br /><br />The next topic we would like to address is the <i><b>states of matter</b></i>.<br />When the word <i>matter</i> is used in physics, it means any object or substance that occupies certain space and has certain mass, thus consisting of certain molecules interacting among themselves.<br /><br />There are three major <i>states of matter</i>: <b>solid</b>, <b>liquid</b> and <b>gas</b>.<br /><br />When an object is <b>solid</b> or is in <b>solid</b> state, it means that it retains its shape and form regardless of surrounding environment, not intended to change its form. The molecules of this object are strongly connected to each other. Their movement relative to each other is rather restricted. This movement can be oscillating around some point, maintaining an orderly three-dimensional structure, for <i>crystal</i> (or <i>crystalline</i> type of) solids or just slow movement, changing their relative position, but not changing the overall form for <i>amorphous</i> (<i>non-crystalline</i>) type.<br />Examples of solid objects are ice (crystal), steel (crystal), plastic cup (amorphous).<br /><br />When an object or substance is <b>liquid</b> or is in <b>liquid</b> state, it takes the shape of a vessel or reservoir it's in. The connection between the molecules in case of liquid is strong enough to hold the molecules together, but not strong enough to preserve the overall shape.<br />Examples of liquid substances are water, mercury ("quick silver"), oil.<br /><br />When an object or substance is <b>gas</b> or is in <i>gaseous</i> state, it takes as much space as it is available. Connections between the molecules are weak and they fly in all directions in completely chaotic fashion. Examples of gases are air, helium, oxygen.<br /><br />Some examples above represent objects or substances that contain only one type of molecules, like ice or mercury, or helium. Some other examples are objects or substances that contain more than one type of molecules mixed together, like steel, oil or air.<br /><br /><i><b>Nature of Heat</b></i><br /><br />Now we are in position to talk about <b><i>heat</i></b>.<br /><i>Heat</i> is the <i>energy of molecular motion</i> inside any object or substance.<br />As we mentioned, molecules are in constant motion inside any object. The more intense this motion is - the more <i>heat</i> this object possesses. This implies that <i>heat</i> is <i>mechanical energy</i>of molecules inside the object or <i>internal energy</i> of the object or substance.<br /><br />As we know, mechanical energy can be transferred from one object to another, like during the collision of two billiard balls. Similarly, mechanical energy of one molecule can be transferred to its neighbors, from them - to their neighbors etc. This is a process of <i>dissipation of heat</i>. All what's necessary for this is the relative proximity of the molecules. This is exactly the way how heat is transferred from one body to another, from flame to pot, from pot to water, from water to vegetables in it, making soup.<br /><br />Since <i>heat</i> or <i>internal energy</i> of an object is related to motion of its molecules, and increased heat means faster movement of the molecules, and, as we see, different states of matter are related to the strength of connection between the molecules, we can expect that the state of an object (solid, liquid, gas) might change with increasing or decreasing its internal energy by supplying or taking away the source of energy.<br />Indeed, it's true. Heat the ice - it will transform into water. Heat the water - it will transform into vapor. Heat the steel - it will melt. Freeze the helium - it will transform into liquid helium. Freeze the mercury - it will solidify.<br /><br />As we see, the same molecules can form objects in different states. It only depends on the amount of internal energy, that is amount of heat, the object possesses.<br /><br />Even without transformation from one state to another, heat causes certain changes in the object visible without any special instruments. We all know that mercury thermometer is working based on the property of mercury to expand as the temperature is rising.<br />This is a general property of most of the objects - to change physical dimensions with increase or decrease of amount of heat (internal energy) carried by their molecules during their constant motion.<br />This property is the principle, on which measuring of the intensity of molecular movement is based.<br /><br /><i><b>Heat and Temperature</b></i><br /><br />Now let's address the issue of measuring the heat, that is amount of internal energy inside any object.<br /><i>The term <b>temperature</b> is related to average intensity of the molecular movement inside an object or a substance. So, when we say that the <i>temperature</i> of an object has increased or decreased, we mean that average intensity of the molecular movement in it has increased or decreased correspondingly</i>.<br /><br />Our obvious task now is to quantitatively evaluate the temperature, thus measuring the intensity of the molecular movement inside an object.<br /><br />It would be great, if we knew kinetic energy of each molecule at each moment of time and average it up to get the temperature in the units of energy. Alas, it's impossible. We have to find some easier method, not necessarily 100% accurate, but sufficient for day-to-day practical purposes.<br /><br />Convenient instrument for this is a classic thermometer, whose indications are directly related to a change in physical size of objects with change of intensity of the molecular motion inside them.<br /><br />A simple thermometer consists of a small reservoir with mercury and thin tube coming from it, so the mercury level in the tube will go up with increase of intensity of the molecular motion of the mercury or down, when the intensity decreases.<br /><br />If we want to measure the temperature of any object, we bring it in contact with our thermometer and, when the temperatures equalize, which might take some time, the level of mercury in a tube of a thermometer will correspond to intensity of the molecular movement inside the object.<br /><br />All, which is left to establish is the scale and units of measure.<br />There are three major systems of measurement of temperature: Celsius, Fahrenheit and Kelvin. Celsius system is used everywhere, except United States and its territories. Fahrenheit system is used in United States and its territories. Kelvin system is used everywhere in scientific research and equations of Theoretical Physics.<br /><br />The unit of measurement in each system is called a <i>degree</i>and the temperature is written with an indication of the system as follows:<br />0°C, 20°C, -40°C for temperatures in Celsius system;<br />32°F, 68°F, -40°F for temperatures in Fahrenheit system;<br />273.15°K, 293.15°K, 233.15°K for temperatures in Kelvin system.<br />Above are the examples of three different temperatures in three different measurement systems, correspondingly.<br /><br />The conversion formulas are:<br />X°F = 5(X−32)/9°C<br />X°C = (X+273.15)°K<br /><br />In the Celsius system the temperature 0°C corresponds to the temperature of melting ice at the sea level on Earth. Temperature of 100°C is the temperature of boiling water at the sea level on Earth. This range is divided into 100 degrees making up a scale.<br />Degrees in the Fahrenheit system are also connected to some natural processes. 0°F is the temperature of freezing of some chemical solution, while 100°F is approximately the temperature of a human body.<br />Finally, in Kelvin system 0°K is so-called <i>absolute zero</i>temperature - the temperature of outer space far from any source of energy. The unit of one degree in Kelvin system equals to that of one degree in Celsius.</n>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-81429910006220681492019-04-02T19:04:00.001-07:002019-04-11T05:00:52.589-07:00Unizor - Physics4Teens - Energy - Potential Energy - Gravity<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/WwG9DBUplko" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Energy of Gravity</u><br /><br />An object of mass <b><i>m</i></b> is raised on certain height <b><i>h</i></b> above the surface of the Earth.<br />Analyze its potential energy, as a function of height <b><i>h</i></b>.<br />Do not assume that the force of gravity is constant, but rather use the Newton's Law of Universal Gravitation.<br />Assume that the radius of Earth is <b><i>R</i></b> and its mass is <b><i>M</i></b>.<br /><br /><i>Solution</i>:<br /><br />If <b><i>x</i></b> is the distance from the center of the Earth to an object, the force of gravity equals to<br /><b><i>F(x) = G·M·m<span style="font-size: medium;">/</span>x²</i></b><br />(where <b><i>G</i></b> is a <i>gravitational constant</i> approximately equaled to <i>6.674·10<sup>-11</sup>N·m<sup>2</sup>·kg<sup>−2</sup></i>).<br /><br />Potential energy of an object at height <b><i>h</i></b> above the ground is the work performed by the force of gravity as an object falls down to the surface of the Earth from the distance <b><i>R+h</i></b> from the center to the distance <b><i>R</i></b>.<br /><br />Since the force of gravity varies, we should use the calculus to calculate this work.<br />An infinitesimal increment of work <i>d<b>W(x)</b></i>, assuming the force <b><i>F(x)</i></b> acts at a distance <i>d<b>x</b></i>, equals to<br /><i>d<b>W(x) = F(x)·</b>d<b>x</b></i><br /><br />Integrating this by <i><b>x</b></i> from <i><b>R+h</b></i>to <i><b>R</b></i>, we get the full work and the potential energy of an object at height <i><b>h</b></i> above the ground:<br /><b><i>E<sub>pot</sub> = W(h) = <span style="font-size: large;">∫</span><sup><sup>R</sup></sup><sub><sub>R+h</sub></sub>F(x)·</i></b><i>d<b>x =<br />= G·M·m·</b></i>[<b><i>1/R − 1/(R+h)</i></b>]<br /><br />Obviously, when an object is at the ground level (<b><i>h=0</i></b>), its potential energy equals to zero and, as it moves higher (<i><b>h</b></i> is increasing), its potential energy grows to its maximum value of <b><i>G·M·m<span style="font-size: medium;">/</span>R</i></b>, as the object moves farther and farther from the ground.<br /><br />When the height of the object <b><i>h</i></b>is significantly smaller than the radius of the Earth <nobr>(<b><i>h << R</i></b>),</nobr> our formula for potential energy can be approximated with a simpler expression:<br /><b><i>E<sub>pot</sub> = W(h) =<br />= G·M·m·</i></b>[<b><i>1/R − 1/(R+h)</i></b>]<b><i> =<br />= G·M·m·h<span style="font-size: medium;">/</span></i></b>[<b><i>R·(R+h)</i></b>]<b><i> ≅<br />≅ G·M·m·h<span style="font-size: medium;">/</span>R² = m·g·h</i></b>,<br />where <b><i>g=G·M<span style="font-size: medium;">/</span>R²</i></b> is the free fall acceleration at the ground level, approximately equaled to<br /><b><i>g ≅ 9.81 m/sec²</i></b><br />and <b><i>P = m·g</i></b> is the weight of an object at the ground level, assumed constant for small heights <b><i>h</i></b>.<br />In this case the potential energy is simply a product of weight (force of gravity) and distance - a classical expression for work performed by a constant force:<br /><b><i>E<sub>pot</sub> = W(h) ≅ P·h</i></b><br /><br />Let's also analyze the kinetic energy of the object.<br />Firstly, we assume that our object starts its free falling from the height <i><b>H</b></i> above the ground with zero initial velocity.<br /><br />Its potential energy at any height <i><b>h</b></i> we know from the above calculation:<br /><b><i>E<sub>pot</sub> = G·M·m·</i></b>[<b><i>1<span style="font-size: medium;">/</span>R − 1<span style="font-size: medium;">/</span>(R+h)</i></b>]<br />Its kinetic energy depends on its speed <i><b>V</b></i>:<br /><i><b>E<sub>kin</sub> = m·V²<span style="font-size: medium;">/</span>2</b></i><br />The problem is, we don't know how the speed depends on the height <i><b>h</b></i>.<br /><br />What we do know is the dependency of the force of gravity <i><b>F</b></i> on the height <i><b>h</b></i>:<br /><i><b>F = G·M·m<span style="font-size: medium;">/</span>(R+h)²</b></i><br />We also know that speed <i><b>V</b></i> is the first derivative of height <i><b>h</b></i>by time <i><b>t</b></i> and the acceleration is the second derivative:<br /><i><b>V(t) = h'(t)</b></i><br /><i><b>a(t) = h"(t)</b></i><br />Therefore, according to the Newton's Second Law,<br /><i><b>F = m·a = G·M·m<span style="font-size: medium;">/</span>(R+h)²</b></i><br />Now we have an expression for acceleration <i><b>a</b></i>:<br /><i><b>a = G·M<span style="font-size: medium;">/</span>(R+h)²</b></i><br />Let's underscore that acceleration is the second derivative of distance <i><b>h</b></i>, as a function of time <i><b>t</b></i>:<br /><i><b>h"(t) = G·M<span style="font-size: medium;">/</span></b></i>[<b><i>R+h(t)</i></b>]<b><i>²</i></b><br /><br />We will not attempt to solve this differential equation to get a function <i><b>h(t)</b></i>, take its derivative to get the speed and substitute into a formula for kinetic energy.<br />Instead, we will apply a clever trick.<br />Let's multiply the last expression by <i><b>2h'(t)</b></i>:<br /><i><b>2h'(t)h"(t) =<br />= 2G·M·h'(t)<span style="font-size: medium;">/</span></b></i>[<b><i>R+h(t)</i></b>]<b><i>²</i></b><br />The reason for this operation is the following.<br />The left side of this equation is the derivative of the square of the speed:<br /><i>d/dt</i>{[<i><b>h'(t)</b></i>]²}<i><b> = 2h'(t)·h"(t)</b></i><br />and on the right side of this equation we find another derivative:<i>d/dt</i>{<i><b>2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>]}<i><b> =<br />= 2G·M·h'(t)<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>]<i><b>²</b></i><br />Therefore, we have the equality of the derivatives:<br /><i>d/dt</i>{[<i><b>h'(t)</b></i>]²}<i><b> =</b></i><br />= <i>d/dt</i>{<i><b>2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>]}<br /><br />If derivatives are equal, the functions differ by a constant:<br />[<i><b>h'(t)</b></i>]²<i><b> = 2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>] <i><b>+ C</b></i><br />The constant <i><b>C</b></i> can be determined, using the conditions at the beginning of motion at time <i><b>t=0</b></i>:<br /><i><b>h(0) = H</b></i> and <i><b>h'(0) = 0</b></i><br />Therefore,<br />[<i><b>h'(0)</b></i>]²<i><b> = 2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(0)</b></i>] <i><b>+ C</b></i><br />[<i><b>0</b></i>]²<i><b> = 2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+H</b></i>] <i><b>+ C</b></i><br /><i><b>C = − 2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+H</b></i>]<br />Now we can write<br />[<i><b>h'(t)</b></i>]²<i><b> =<br />= 2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>]<i><b>−2G·M<span style="font-size: medium;">/</span></b></i>[<i><b>R+H</b></i>]<br />From this we can express the kinetic energy<br /><i><b>E<sub>kin</sub> = m·V²(t)<span style="font-size: medium;">/</span>2 =<br />= m·</b></i>[<i><b>h'(t)</b></i>]<i><b>²(t)<span style="font-size: medium;">/</span>2 =<br />= G·M·</b></i>{<i><b>1<span style="font-size: medium;">/</span></b></i>[<i><b>R+h(t)</b></i>]<i><b>−1<span style="font-size: medium;">/</span></b></i>[<i><b>R+H</b></i>]}<br /><br />Since our task was to express the energy in terms of height above the ground, we can simply write it as<br /><i><b>E<sub>kin</sub> = G·M·</b></i>[<i><b>1<span style="font-size: medium;">/</span></b></i>(<i><b>R+h</b></i>)<i><b>−1<span style="font-size: medium;">/</span></b></i>(<i><b>R+H</b></i>)]<br /><br />Finally, let's find the full mechanical energy of an object falling from the sky.<br /><b><i>E<sub>pot</sub> = G·M·m·</i></b>[<b><i>1<span style="font-size: medium;">/</span>R−1<span style="font-size: medium;">/</span>(R+h)</i></b>]<br /><i><b>E<sub>kin</sub> = G·M·</b></i>[<i><b>1<span style="font-size: medium;">/</span></b></i>(<i><b>R+h</b></i>)<i><b>−1<span style="font-size: medium;">/</span>(R+H)</b></i>]<br /><b><i>E<sub>full</sub> = E<sub>pot</sub> + E<sub>kin</sub> =<br />= G·M·</i></b>[<i><b>1<span style="font-size: medium;">/</span>R−1<span style="font-size: medium;">/</span></b></i>(<i><b>R+H</b></i>)]<br />As we see, the full mechanical energy is constant, it does not depend on the height. As an object falls from the sky, its potential energy is decreasing, but kinetic energy increasing, and the sum of both types of energy remains constant.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-64530033689238085032019-02-25T20:57:00.001-08:002019-02-25T20:57:38.229-08:00Unizor - Physics4Teens - Energy - Potential Energy of a Spring<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/RTL8ESfVkas" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Energy of Spring</u><br /><br />The energy discussed in the previous introductory lecture was an energy related to constant force acting on an object. In this lecture we will address a more difficult case of a variable force of a spring acting horizontally on an object according to the Hooke's Law.<br />Let's remind the Hooke's Law. It states that the force of a spring, fixed on one end, onto a point-object attached to its free end is proportional to the length the spring is extended or compressed from its neutral position.<br />The corresponding formula for the force <b><i>F</i></b> of a spring onto an object is<br /><b><i>F = −k·L</i></b><br />where <b><i>L</i></b> is a displacement of the spring's free end and <b><i>k</i></b> is a <i>coefficient of elasticity</i> - a specific characteristic of "stiffness" of each spring.<br />The minus sign signifies that the direction of the force is opposite to a displacement. The frame of reference is associated with the neutral position of a spring and the X-axis is directed towards spring's stretching. Then, for positive displacement <b><i>L</i></b>(stretching) the force is directed towards neutral point (that is, negative), while for negative displacement (compression) the force also directed towards a neutral position, (that is, positive).<br /><br />At this point we recommend to refresh the material in the "Work and Elasticity" lecture of the "Mechanical Work" chapter of the "Mechanics" part of this course.<br />As stated in that lecture, the total amount of work <b><i>W</i></b> to compress (or stretch) a spring with elasticity <b><i>k</i></b> by length <b><i>L</i></b>equals<br /><b><i>W = k·L²/2</i></b><br />regardless of how exactly we compress or stretch a spring.<br /><br />Assume now that we have compressed a spring by the length <b><i>L</i></b>, performing work equaled to <b><i>W=k·L²/2</i></b>. At a spring's end we have attached an object of mass <b><i>m</i></b>. Then we let the spring go by itself with the object attached to its end.<br /><br />Intuitively, we think that, at first, the spring should return back to its neutral position. During this time the object's speed will increase because the force of elasticity of a spring will push it.<br /><br />But then, when a spring reaches its neutral position, the object, having certain speed, will continue its move by inertia, stretching the spring. During this time the force of elasticity of a spring will slow down the movement of the object until it stops at the end of the stretching process.<br /><br />At that time the force of elasticity will pull the object back to a neutral position, increasing its speed.<br /><br />At the neutral position object continues moving by inertia, compressing a string, until the force of elasticity stops its movement when the spring will be in the compressed position, as in the beginning of our process.<br /><br />At this point the process repeats itself and can continue like this indefinitely.<br /><br />Well, our intuition is correct, a spring with an object attached to it will oscillate indefinitely. Let's analyze quantitatively these oscillations.<br /><br />Our first goal is to find the dependency of the object's position on time. This can be done using the Newton's Second Law, taking into account the initial position of the object at the end of a compressed spring with no initial speed.<br />Let <b><i>S(t)</i></b> be a function describing the position of the object at the end of a spring relatively to its neutral position with stretching being a positive direction. Then <b><i>S(0)=−L</i></b> and <b><i>S'(0)=0</i></b>. The acceleration is the second derivative of <b><i>S(t)</i></b>, that is <nobr><b><i>a=S"(t)</i></b>.</nobr> The force of elasticity is proportional to <b><i>S(t)</i></b> as <b><i>F=−k·S(t)</i></b>. Now the Newton's Second Law <b><i>F=m·a</i></b> looks like this differential equation:<br /><b><i>−k·S(t) = m·S"(d)</i></b><br />with two initial conditions:<br /><b><i>S(0) = −L</i></b><br /><b><i>S'(0) = 0</i></b><br />This differential equation is fully solvable and its general solution is<br /><b><i>x(t) = C<sub>1</sub>·cos(t·√<span style="text-decoration-line: overline;">k/m</span>) +<br />+ C<sub>2</sub>·sin(t·√<span style="text-decoration-line: overline;">k/m</span>)</i></b><br /><br />Constants <b><i>C<sub>1</sub></i></b> and <b><i>C<sub>2</sub></i></b> can be determined using the initial conditions:<br /><b><i>C<sub>1</sub> = −L</i></b><br /><b><i>C<sub>2</sub> = 0</i></b><br />Therefore,<br /><b><i>S(t) = −L·cos(t·√<span style="text-decoration-line: overline;">k/m</span>)</i></b><br /><br />This function describes simple sinusoidal oscillations with amplitude <b><i>L</i></b> and period<br /><b><i>T=2π√<span style="text-decoration-line: overline;">m/k</span></i></b>.<br /><br />We can also calculate the speed of an object as the first derivative of distance:<br /><b><i>V(t) = S'(t) = </i></b><i>d<b>S(t)/</b>d<b>t =<br />= L·√<span style="text-decoration-line: overline;">k/m</span>·sin(t·√<span style="text-decoration-line: overline;">k/m</span>)</b></i><br /><br />During the first quarter of this time an object accelerates under the influence of the elasticity of a spring, moving from the extremely compressed position on a spring to its neutral position. During the next quarter of the period the spring's elasticity slows it down, while it moves to extremely stretched position of a spring. During the next quarter of the period elasticity of a spring accelerates it again to a string's neutral position. Finally, the elasticity slows the object down, as it moves to the initial most compressed position.<br /><br />Let's calculate the work that spring performs during the first quarter of its period from complete compression to a neutral point by accelerating the object from initial speed <b><i>V(0)=0</i></b>to its maximum at a spring's neutral position <b><i>V(T/4)</i></b>.<br />It can be done by two methods.<br /><br />1. By integrating by displacement of a spring's end from value <b><i>−L</i></b> (full compression) to value <b><i>0</i></b> (neutral position)<br /><b><i>W<sub>[−L,0]</sub> =<br />= <span style="font-size: large;">∫</span><sup><sup>0</sup></sup><sub><sub>−L</sub></sub>F(S)·</i></b><i>d<b>S =<br />= <span style="font-size: large;">∫</span><sup><sup>0</sup></sup><sub><sub>−L</sub></sub>(−k·S)·</b>d<b>S =<br />= −k·S²/2<span style="font-size: large;">|</span><sup><sup>0</sup></sup><sub><sub>−L</sub></sub> = k·L²/2</b></i><br /><br />2. By integrating by time from <b><i>t=0</i></b> to <b><i>t=T/4=(π/2)·√<span style="text-decoration-line: overline;">m/k</span></i></b>(calculations are more complex but lead to the same result)<br /><b><i>W<sub>[0,T/4]</sub> =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>F(S(t))·</i></b><i>d<b>S(t) =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>F(S(t))·(</b>d<b>S(t)/</b>d<b>t)·</b>d<b>t =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>F(S(t))·V(t)·</b>d<b>t =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>k·L·cos(t·√<span style="text-decoration-line: overline;">k/m</span>)·<br />·L·√<span style="text-decoration-line: overline;">k/m</span>·sin(t·√<span style="text-decoration-line: overline;">k/m</span>)·</b>d<b>t =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>k·L²·cos(t·√<span style="text-decoration-line: overline;">k/m</span>)·<br />·sin(t·√<span style="text-decoration-line: overline;">k/m</span>)·</b>d<b>(t·√<span style="text-decoration-line: overline;">k/m</span>) =<br />= <span style="font-size: large;">∫</span><sub><sub>0</sub></sub><sup><sup>T/4</sup></sup>k·L²·sin(t·√<span style="text-decoration-line: overline;">k/m</span>)·<br />·</b>d<b>(sin(t·√<span style="text-decoration-line: overline;">k/m</span>)) =<br />= k·L²sin²((1/2)π√<span style="text-decoration-line: overline;">m/k</span>·√<span style="text-decoration-line: overline;">k/m</span>)/2</b></i><br />After cancellation of factors and taking into account that <b><i>sin(π/2)=1</i></b> we obtain the same expression for work:<br /><b><i>W<sub>[0,T/4]</sub> = k·L²/2</i></b><br /><br />What we have proven now is that the same amount of work done to compress a spring (<b><i>W=k·L²/2</i></b>) is performed by a spring when it returns to its neutral position.<br /><br />That's why we say that the spring accumulates <i>potential energy</i>, when it's compressed, and releases it, when it returns back to neutral position.<br /><br />Now, let's see where this energy, released by a spring during its return to a neutral position, goes. Our assumption is that it is transformed into kinetic energy of the object attached to its free end. Let's check it out.<br /><b><i>E<sub>kin</sub> = m·V²(T/4)/2 =<br />= m·L²·(k/m)·sin²(π/2)/2 =<br />= k·L²/2</i></b><br />So, kinetic energy of the object, when a spring is in the neutral position, exactly equals to a spring's potential energy when it's fully compressed.<br />Compressing a spring results in transferring some external energy to its potential energy. Then, as the spring returns to its neutral position, it releases this potential energy into a kinetic energy of the object attached to its end. At the neutral position the potential energy of a spring is zero, but kinetic energy of the object equals to potential energy of a spring in a compressed position.<br /><br />Our last step is to calculate <i>full energy</i> of the system (<i>potential energy</i> of a spring and <i>kinetic energy</i> of the object attached to it) at any moment of time <b><i>t</i></b>.<br /><b><i>E<sub>pot</sub>(t) = k·S²(t)/2 =<br />= k·L²·cos²(t·√<span style="text-decoration-line: overline;">k/m</span>)/2</i></b><br /><b><i>E<sub>kin</sub> = m·V²(t)/2 =<br />= m·L²·k/m·sin²(t·√<span style="text-decoration-line: overline;">k/m</span>)/2</i></b><br /><b><i>E<sub>full</sub> = E<sub>pot</sub> + E<sub>kin</sub> = k·L²/2</i></b><br />(since <i>sin²(φ)+cos²(φ)=1</i>)<br /><br />As we see, the <i>full energy</i>remains constant, it does not depend on the attached object's mass, only on the properties of the spring - coefficient of <i>elasticity</i> <b><i>k</i></b> and its initial displacement from the neutral position.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-92198475701089309432019-02-25T13:09:00.001-08:002019-02-25T13:09:02.740-08:00Unizor - Physics4Teens - Energy - Potential Energy - Introduction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/R0kvn8nnHKA" width="480"></iframe><br /><br /><i><br /></i><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Potential Energy - Introduction</u><br /><br /><i>Potential energy</i> is a quantitative characteristic of an object's position relative to other objects or environment that are a source of some forces. These forces would result in motion of the object under consideration (if no counteracting forces present) and, therefore, would result in performing certain work related to this motion.<br /><br />Typical example of <i>potential energy</i> is related to a position of some object in the gravitational field.<br />Since a planet Earth attracts all objects, an object raised to a certain height above the ground will fall, if no support is provided. The motion of falling object is the result of work of the forces of gravity. Therefore, by definition, an object raised above the ground has certain amount of <i>potential energy</i>equal to work performed by the forces of gravity to move it down to the ground.<br /><br />In a simplified case of a small object of mass <b><i>m</i></b>, raised on small (relatively to the size of the Earth) height <b><i>h</i></b>, the force of gravity, acting on this object, is constant and is equal to the object's weight <nobr><b><i>P = m·g</i></b></nobr>, where <nobr><b><i>g ≅ 9.8m/sec²</i></b></nobr> is the acceleration of the free fall near the Earth's surface.<br />If no support is provided, the object will fall straight down. Thus, the constant force of gravity <b><i>P</i></b> would perform work, pushing the object down at a distance <b><i>h</i></b>, thus performing work<br /><b><i>W = P·h = m·g·h</i></b><br /><br />Let's examine the relationship between an object's <i>potential energy</i> and its <i>kinetic energy</i> as it falls down from certain initial height.<br />Assume an inertial frame of reference associated with Earth and X-coordinate directed straight up perpendicularly to the ground with ground level be at coordinate <b><i>X=0</i></b>.<br />Initial position of the object is, therefore, at X-coordinate equal to <b><i>X=h</i></b>. The falling object moves with constant acceleration <b><i>−g</i></b> (negative, since it moves from positive <b><i>X=h</i></b> to <b><i>X=0</i></b>). Initial speed of the object is zero.<br />Therefore, the time-dependent motion <b><i>X(t)</i></b> can be described by the kinematic equation<br /><b><i>X(t) = h − g·t²/2</i></b><br />The object's velocity is directed vertically down and, as a function of time, is the first derivative of the <b><i>X(t)</i></b>:<br /><b><i>V(t) = X'(t) = −g·t</i></b><br />(which is obvious from the fact that the motion is with constant acceleration <b><i>−g</i></b> and zero initial speed)<br /><br />As object falls down, its height diminishes, which means that its <i>potential energy</i> also diminishes. At any given moment of time <b><i>t</i></b> this energy equals to<br /><b><i>E<sub>pot</sub>(t) = m·g·X(t) = m·g·(h−g·t²/2)</i></b><br /><br />At the same time the <i>kinetic energy</i> of the object is increasing, since its speed <b><i>V(t)</i></b>is increasing. At any given moment of time <b><i>t</i></b> this energy equals to<br /><b><i>E<sub>kin</sub>(t) = m·V²(t)/2 = m·g²·t²/2</i></b><br /><br />Remarkably, the <i>full mechanical energy</i> of this object, which is a sum of its <i>potential energy</i> and <i>kinetic energy</i> is constant:<br /><b><i>E<sub>full</sub> = E<sub>pot</sub> + E<sub>kin</sub> = m·g·h</i></b>,<br />which is its <i>potential energy</i> in the beginning of motion (when its initial speed and, therefore, <i>kinetic energy</i> are zero).<br /><br />The total time in motion from the initial position above the ground at height <b><i>h</i></b> to the ground can be obtained by resolving the equation <b><i>X(t)=0</i></b> for variable <b><i>t</i></b>:<br /><b><i>X(t) = h − g·t²/2 = 0</i></b><br /><b><i>t<sub>fall</sub> = √<span style="text-decoration-line: overline;">2h/g</span></i></b><br />At that time of touching the ground the <i>potential energy</i> of an object equals to zero and its <i>kinetic energy</i> equals to<br /><b><i>E<sub>kin</sub>(t<sub>fall</sub>) = m·V²(t<sub>fall</sub>)/2 = m·g²t²<sub>fall</sub>/2 = m·g·h</i></b><br /><br />In the beginning the potential energy is at its maximum and kinetic energy at zero. At the end the potential energy is at zero, while kinetic energy at its maximum, but their sum is always the same. So, it appears that, as an object falls, its potential energy is transformed into kinetic energy, preserving their sum - the <i>full mechanical energy</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-10341770902094528212019-02-22T12:39:00.001-08:002019-02-22T12:39:55.771-08:00Unizor - Physics4Teens - Energy - Problems on Kinetic Energy<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/Hp8BHnRkO1A" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems on Kinetic Energy</u><br /><br /><i>Problem A</i><br /><br />A car engine accelerates a car of mass <b><i>m</i></b> from the state at rest to some maximum speed during the time <b><i>T</i></b> with constant acceleration <b><i>a</i></b> along a straight line.<br />Ignore loss of mass due to burning fuel.<br />What is the kinetic energy <b><i>E<sub>kin</sub></i></b>of a car at the end of this period of acceleration and what is the work <b><i>W</i></b> performed by the car engine during this time?<br /><br /><i>Solution</i>:<br /><br /><b><i>V</i></b><i><sub>end</sub><b> = a·T</b></i><br /><b><i>E</i></b><i><sub>kin</sub><b> = m·V²</b><sub>end</sub><b> /2 = m·a²·T²/2</b></i><br /><br /><b><i>W = F·S = (m·a) · (a·T²/2) =<br />= m·a²·T²/2</i></b><br /><b><i>W = E</i></b><i><sub>kin</sub><b> = m·a²·T²/2</b></i><br /><br /><i>Problem B</i><br /><br />A driver of a car, going with speed <b><i>V</i></b>, sees an obstacle and strongly presses on breaks, so its wheels stop spinning. The car slows down to complete stop by the force of friction only at a distance <b><i>S</i></b> from the spot when it started to break.<br />What is the coefficient of kinetic friction <b><i>μ</i></b> of car's wheels against the ground?<br />Assume, the free fall acceleration is <b><i>g</i></b>.<br /><br /><i>Solution</i><br /><br />Let <b><i>m</i></b> be an unknown mass of the car.<br />Let <b><i>W</i></b> be a work of the force of friction to stop the car.<br />Let <b><i>E</i></b><i><sub>kin</sub><b></b></i> be a kinetic energy of the car in the beginning of breaking. The ending kinetic energy is zero, since the speed at the end is zero. So, the initial kinetic energy should be equal to the work required to stop the car.<br />Let <b><i>F</i></b> be a constant force of friction equal to the car's weight multiplied by a coefficient of kinetic friction.<br /><b><i>E</i></b><i><sub>kin</sub><b> = m·V²/2 = W</b></i><br /><b><i>W = F</i></b><i><sub friction="" sub=""><b>·S</b></sub></i><br /><b><i>F = W/S = m·g·μ</i></b><br /><b><i>μ = W/(m·g·S) = m·V²/(2·m·g·S)</i></b><br /><b><i>μ = V²/(2·g·S)</i></b><br />Notice independence of the car's mass.<br /><br /><br /><i>Problem C</i><br /><img src="http://www.unizor.com/Pictures/RotationalEnergy.png" style="height: 270px; width: 200px;" /><br />A point-object <b><i>A</i></b> of mass <b><i>m</i></b> is freely rotating on a thread of a length <b><i>L</i></b> that forms an angle <b><i>φ</i></b>with vertical.<br />What is its kinetic energy?<br /><br /><i>Solution</i><br /><br /><b><i>T</i></b> is a force of tension of the thread. We can represent it as a sum of two components - <b><i>T<sub>v</sub></i></b>that balances the weight <b><i>P</i></b> and <b><i>T<sub>h</sub></i></b> that keeps an object on a circular trajectory (centripetal force).<br /><b><i>T<sub>v</sub> = m·g</i></b><br /><b><i>T<sub>h</sub> = T<sub>v</sub>·tan(φ) = m·g·tan(φ)</i></b><br />This centripetal force causes centripetal acceleration<br /><b><i>a = V²/R</i></b><br />(see a chapter <i>Mechanics - Superposition of Forces - Resultant Forces - Example 2</i> in this course)<br />where <b><i>V</i></b> is a linear speed of an object and <b><i>R</i></b> is a radius of a circular trajectory, which is equal to <b><i>L·sin(φ)</i></b>.<br />Therefore,<br /><b><i>T<sub>h</sub> = m·a = m·V²/R</i></b><br />Hence,<br /><b><i>m·g·tan(φ) = m·V²/R</i></b><br />Kinetic energy is<br /><b><i>E = m·V²/2 =<br />= m·g·tan(φ)·R/2 =<br />= m·g·L·tan(φ)·sin(φ)/2</i></b><br /><br /><br /><i>Problem D</i><br /><br />An ideal pendulum of mass <b><i>m</i></b>with a thread length <b><i>L</i></b> performs small <i>harmonic</i> oscillations with its angle of deviation from the vertical <b><i>φ</i></b> conforming to the equation<br /><b><i>φ(t) = φ<sub>0</sub>·cos(√<span style="text-decoration-line: overline;">g/L </span>·t)</i></b><br />where <b><i>φ<sub>0</sub></i></b> is the initial deviation from the vertical, <b><i>g</i></b> - free fall acceleration and <b><i>t</i></b> - time.<br />What is its kinetic energy at the lowest point of its trajectory?<br /><br /><i>Solution</i>:<br /><br />We recommend to refresh the properties of a pendulum in the "Mechanics" part of this course in the chapter "Pendulum, Spring".<br />Let's find the period of oscillations.<br />Since function <b><i>y=cos(x)</i></b> has a period <b><i>2π</i></b>, function <b><i>y=cos(k·x)</i></b>has period <b><i>2π/k</i></b>.<br />In our case the period of oscillation <b><i>T</i></b> is equal to<br /><b><i>T = 2π·√<span style="text-decoration-line: overline;">L/g </span></i></b><br />The angular velocity is a derivative of the angle of deviation <b><i>φ(t)</i></b>:<br /><b><i>ω(t) = </i></b><i>d<b>φ(t)/</b>d<b>t =<br />= −φ<sub>0</sub>·√<span style="text-decoration-line: overline;">g/L </span>·sin(√<span style="text-decoration-line: overline;">g/L </span>·t)</b></i><br />At the lowest point of a trajectory the time equals to 1/4 of a period.<br />Therefore, we can calculate the angular velocity at this point by substituting<br /><b><i>t = T/4 = (1/4)·2π·√<span style="text-decoration-line: overline;">L/g </span>=<br />= (π/2)·√<span style="text-decoration-line: overline;">L/g </span></i></b><br />which gives the angular speed at this point as<br /><b><i>ω(T/4) =<br />= −φ<sub>0</sub>·√<span style="text-decoration-line: overline;">g/L </span>·sin(√<span style="text-decoration-line: overline;">g/L </span>·T/4) =<br />= −φ<sub>0</sub>·√<span style="text-decoration-line: overline;">g/L </span>·<br />·sin(√<span style="text-decoration-line: overline;">g/L </span>·(π/2)·√<span style="text-decoration-line: overline;">L/g </span>) =<br />= −φ<sub>0</sub>·√<span style="text-decoration-line: overline;">g/L </span></i></b><br />The linear speed of a rotating object equals to its angular speed, multiplied by a radius.<br />Therefore, linear speed <b><i>V</i></b> at the lowest point equals to<br /><b><i>V = L·ω(T/4) =<br />= −φ<sub>0</sub>·√<span style="text-decoration-line: overline;">g·L </span></i></b><br />The kinetic energy at this point is<br /><b><i>E = m·V²/2 = m·g·L·φ<sub>0</sub>²/2</i></b><br /><b>IMPORTANT NOTE</b>:<br />The following equation for small <i>harmonic</i> oscillation used in this problem<br /><b><i>φ(t) = φ<sub>0</sub>·cos(√<span style="text-decoration-line: overline;">g/L </span>·t)</i></b><br />was obtained in "Mechanics" part of this course in chapter "Pendulum, Spring" by simplification of the differential equation that described the motion of the pendulum. In particular, it was assumed that for small oscillations the angle of deviation from the vertical <b><i>φ</i></b>and its sine <b><i>sin(φ)</i></b> are approximately equal. The exact value of the kinetic energy of a pendulum at its lowest point can be derived using its <i>potential energy</i> at the top position, which we will address in the next chapter, and which is equal to <b><i>2m·g·L·sin²(φ<sub>0</sub>/2)</i></b>. If in this expression we replace <b><i>sin(φ<sub>0</sub>)</i></b>with <b><i>φ<sub>0</sub></i></b> (which is acceptable for small angles), we will get the formula for kinetic energy we obtained above.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-62303294637641957662019-01-22T12:57:00.001-08:002019-01-22T12:57:08.197-08:00Unizor - Physics4Teens - Mechanics - Work and Gravity<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Work and Gravity</u><br /><br /><br /><br />Let's consider a force to lift an object above the planet against the <br />force of gravity. In simple cases, when the height is relatively small <br />compared with the size of a planet, the force of gravity is considered <br />to be constant. In this case we will assume that we lift to substantial <br />height and have to take into consideration the Law of Gravity that tells<br /> that the force of gravity is proportional to masses of objects involved<br /> (a planet of mass <b><i>M</i></b> and an object of mass <b><i>m</i></b> that we lift) and inversely proportional to a distance <b><i>r</i></b> between the objects:<br /><br /><b><i>F = G·M·m <span style="font-size: medium;">/</span>r²</i></b><br /><br />where <b><i>G</i></b> - a <i>gravitational constant</i>.<br /><br /><br /><br />Our task is to find the work <b><i>W</i></b> needed to lift an object of mass <b><i>m</i></b> from a surface of a planet of mass <b><i>M</i></b> and radius <b><i>R</i></b> to height <b><i>H</i></b> above its surface.<br /><br /><br /><br />The easiest approach is to represent all parameters as functions of the distance <b><i>r</i></b> between a center of a planet and an object.<br /><br />Then the force of gravity as a function of <b><i>r</i></b> is<br /><br /><b><i>F(r) = G·M·m <span style="font-size: medium;">/</span>r²</i></b><br /><br />An infinitesimal increment (<i>differential</i>) of work equals<br /><br /><i>d<b>W(r) = F(r)·</b>d<b>r =<br /><br />= G·(M·m <span style="font-size: medium;">/</span>r²)·</b>d<b>r</b></i><br /><br /><br /><br />Let's integrate the differential of work on an interval of <b><i>r</i></b> [<b><i>R,R+H</i></b>]:<br /><br /><b><i><span style="font-size: large;">∫</span></i></b><sub><sub>[R,R+H]</sub></sub><i>d<b>W(r) =<br /><br />= G·M·m·<span style="font-size: large;">∫</span></b></i><sub><sub>[R,R+H]</sub></sub><b><i>(1/r²)</i></b><i>d<b>r =<br /><br />= G·M·m·</b></i>[<b><i>1/R − 1/(R+H)</i></b>]<b><i> =<br /><br />= G·M·m·H <span style="font-size: medium;">/</span> </i></b>[<b><i>R·(R+H)</i></b>]<br /><br /><br /><br />For small (relatively to radius of a planet <b><i>R</i></b>) height <b><i>H</i></b> the expression in curly brackets above is approximately equal to<br /><br /><b><i>G·M·m <span style="font-size: medium;">/</span></i></b><b><i>R² = m·g</i></b><br /><br />where <b><i>g = G·M <span style="font-size: medium;">/</span></i></b><b><i>R²</i></b> is an acceleration of the free falling on a planet's surface.<br /><br />By definition, <b><i>m·g</i></b> is the weight <b><i>P</i></b> of an object on a surface of a planet, which we consider a constant force in this approximation.<br /><br />So, our formula for work for small height above the planet is reduced to simple expression<br /><br /><b><i>W = P·H</i></b><br /><br />which is a base "force times distance" expression that defines the work <br />in simple case of constant force acting along a trajectory.<br /><br /><br /><br />For large height <b><i>H</i></b> we cannot ignore the change of gravity <br />as an object moves far from the planet, and the exact formula must be <br />used to calculate the work.<br /><br /><br /><br />As in other cases, the work depends only on characteristics of <br />interacting objects (their masses in this case) and the result of work <br />(lifting on certain height), not the way how we achieve this result.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-8647100944692850352019-01-22T12:34:00.001-08:002019-01-22T12:34:27.458-08:00Unizor - Physics4Teens - Mechanics - Work and Elasticity<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Work and Elasticity</u><br /><br /><br /><br />Let's consider a slightly more complicated case of a variable force.<br /><br />A good example of this is the work performed by a force to stretch or <br />compress a spring, in which case a force linearly depends on a length a <br />spring is stretched or compressed.<br /><br /><br /><br />Consider a spring with elasticity coefficient <b><i>k</i></b> that force <b><i>F</i></b> compresses by length <b><i>S</i></b>.<br /><br />Consider further that a spring is compressed by the same length in equal<br /> time interval, that is its end, where the force is applied, is <br />uniformly moves under the force of compression with constant speed <b><i>V</i></b>.<br /><br />Our task is to find the amount of work needed to compress a spring by the length <b><i>L</i></b>.<br /><br /><br /><br />In this case we can calculate the length of compression as a function of time<br /><br /><b><i>S(t) = V·t</i></b><br /><br />According to the Hook's Law, the force of compression is proportional to the length of compression:<br /><br /><b><i>F = k·S</i></b><br /><br />Since the force and the length of compression are time-dependent, this can be written as<br /><br /><b><i>F(t) = k·S(t) = k·V·t</i></b><br /><br /><br /><br />An infinitesimal increment (<i>differential</i>) of work equals<br /><br /><i>d<b>W(t) = F(t)·</b>d<b>S(t) = F(t)·V·</b>d<b>t =<br /><br />= k·V·t·V·</b>d<b>t = k·V²·t·</b>d<b>t</b></i><br /><br /><br /><br />The time interval <b><i>T</i></b> needed to compress a spring by length <b><i>L</i></b> with speed of compression <b><i>V</i></b> equals to <b><i>T = L<span style="font-size: medium;">/</span>V</i></b><br /><br />Let's integrate the differential of work on a time interval [<b><i>0,T</i></b>]:<br /><br /><b><i><span style="font-size: large;">∫</span></i></b><sub><sub>[0,T]</sub></sub><i>d<b>W(t) = k·V²·<span style="font-size: large;">∫</span></b></i><sub><sub>[0,T]</sub></sub><b><i>t·</i></b><i>d<b>t =<br /><br />= k·V²·T²<span style="font-size: medium;">/</span>2 = k·L²<span style="font-size: medium;">/</span>2</b></i><br /><br /><br /><br />Remarkably, the work does not depend on speed <b><i>V</i></b> of compression, only on the length <b><i>L</i></b> the spring is compressed and its elasticity <b><i>k</i></b>.<br /><br /><br /><br />We can easily generalize this by getting rid of dependency on the speed of compression in our calculations.<br /><br />Instead of all parameters being functions of time <b><i>t</i></b>, let's use the length of compression <b><i>S</i></b> as a base variable.<br /><br />Since <b><i>F(S) = k·S</i></b>,<br /><br /><i>d<b>W(S) = F(S)·</b>d<b>S = k·S·</b>d<b>S</b></i><br /><br />We can integrate it on an interval <b><i>S</i></b>∈[<b><i>0,L</i></b>] getting<br /><br /><b><i><span style="font-size: large;">∫</span></i></b><sub><sub>[0,L]</sub></sub><i>d<b>W(S) = k·<span style="font-size: large;">∫</span></b></i><sub><sub>[0,L]</sub></sub><b><i>S·</i></b><i>d<b>S =<br /><br />= k·L²<span style="font-size: medium;">/</span>2</b></i><br /><br /><br /><br />As you see, this is a more general (and simpler!) derivation of the same<br /> formula that does not depend at all on how we compress the spring, only<br /> on its elasticity and a length of compression.<br /><br /><br /><br />So, total amount of work <b><i>W</i></b> to compress a spring with elasticity <b><i>k</i></b> by length <b><i>L</i></b> equals<br /><br /><b><i>W = k·L²<span style="font-size: medium;">/</span>2</i></b><br /><br />regardless of how exactly we compress a spring.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-1247834147562597282019-01-14T12:07:00.001-08:002019-01-14T12:07:24.467-08:00Unizor - Physics4Teens - Energy - Kinetic Energy - Introduction<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Kinetic Energy - Introduction</u><br /><br /><br /><br />Kinetic energy is a quantitative characteristic of an object's motion, <br />that signifies that this motion can result in some work, when our object<br /> interacts with surrounding environment.<br /><br /><br /><br />As an example, consider an object of mass <b><i>m</i></b> uniformly moving within some inertial frame along a straight line trajectory with speed <b><i>v</i></b>.<br /><br /><br /><br />Assume that at some moment of time there appears a constant force <b><i>F</i></b> that acts against its motion. For example, a constant air resistance.<br /><br /><br /><br />As a result of this interaction with air molecules, our object will slow<br /> down because of air resistance acting against its motion until it <br />stops. So, the motion of our object caused some work - moving molecules <br />of air away from the trajectory, after which the motion of our object no<br /> longer exists, it stops completely.<br /><br />Obviously, instead of air resistance, we can consider friction or <br />gravity, or any other force, assumed constant for this experiment.<br /><br /><br /><br />Let's calculate the work done by this force, as it acts against the motion of our object and causes its deceleration.<br /><br /><br /><br />First of all, we calculate the deceleration <b><i>a</i></b>:<br /><br /><b><i>a = F/m</i></b><br /><br />Then, knowing deceleration, we get the time <b><i>t</i></b> our force acts until an object stops:<br /><br /><b><i>t = v/a = m·v/F</i></b><br /><br />Now the distance <b><i>S</i></b> our object travels until a full stop:<br /><br /><b><i>S = a·t²/2 = F·t²/(2m) =<br /><br />= F·m²·v²/(2m·F²) =<br /><br />= m·v²/(2F)</i></b><br /><br />Finally, the work <b><i>W</i></b> performed by our force <b><i>F</i></b> during the distance <b><i>S</i></b>:<br /><br /><b><i>W = F·S = m·v²/2</i></b><br /><br /><br /><br />What's most remarkable about this formula is that the work performed by force <b><i>F</i></b> does not depend on this force, but only on the characteristic of an object (mass <b><i>m</i></b>) and its motion (speed <b><i>v</i></b>).<br /><br /><br /><br />So, it appears that there is something specific for an object (its <i>mass</i>) and its motion (<i>speed</i>) - the amount of work to bring this object to a state of rest, and this quantity of work equals to <b><i>m·v²/2</i></b>.<br /><br />This quantitative characteristic of an object in motion is called its <b>kinetic energy</b>.<br /><br /><br /><br />Let's slightly complicate the problem. The constant force <b><i>F</i></b> slows down our object not to a full stop, but to final speed <b><i>v<sub>end</sub></i></b>. What will be the work force <b><i>F</i></b> would perform?<br /><br /><br /><br />As before,<br /><br /><b><i>a = F/m</i></b><br /><br /><b><i>t = (v−v<sub>end</sub>)/a = m·(v−v<sub>end</sub>)/F</i></b><br /><br /><b><i>S = v·t − a·t²/2 =<br /><br />= m·v·(v−v<sub>end</sub>)/F −<br /><br />− F·m²·(v−v<sub>end</sub>)²/(2m·F²) =<br /><br />= m·v·(v−v<sub>end</sub>)/F −<br /><br />− m·(v−v<sub>end</sub>)²/(2F) =<br /><br />= (m/2F)·(v²−v²<sub>end</sub>)</i></b><br /><br />Finally, the work <b><i>W</i></b> performed by our force <b><i>F</i></b> during the distance <b><i>S</i></b>:<br /><br /><b><i>W = F·S = m·(v²−v²<sub>end</sub>)/2 =<br /><br />= m·v²/2 − m·v²<sub>end</sub>)/2</i></b><br /><br /><br /><br />This formula indicates that the work needed to change the state of a moving object from a state with one value of its <b>kinetic energy</b> to another equals to a difference between the values of its <b>kinetic energy</b> at these two states.<br /><br /><br /><br />Absolutely analogous calculations can prove that the work of a constant force that accelerates an object of mass <b><i>m</i></b> from a state of rest to speed <b><i>v</i></b> equals to<br /><br /><b><i>W = m·v²/2</i></b> for any force,<br /><br />and the work needed to accelerated an object from speed <b><i>v<sub>beg</sub></i></b> to <b><i>v</i></b> equals to<br /><br /><b><i>W = m·v²/2 − m·v²<sub>beg</sub>/2</i></b><br /><br /><br /><br />In other words, <b>work</b>, that results from the interaction of moving object with surrounding environment, and its <b>kinetic energy</b> are intimately related. One can be converted into another and vice versa.<br /><br /><br /><br />Work performed by different forces is, by definition, <i>additive</i>.<br /><br />It means that, if force <b><i>F<sub>1</sub></i></b> acted on a distance <b><i>S<sub>1</sub></i></b> performing work <b><i>W<sub>1</sub>=F<sub>1</sub>·S<sub>1</sub></i></b> and force <b><i>F<sub>2</sub></i></b> acted on a distance <b><i>S<sub>2</sub></i></b> performing work <b><i>W<sub>2</sub>=F<sub>2</sub>·S<sub>2</sub></i></b>, then the total amount of work performed by both forces is a sum of their individual work.<br /><br /><br /><br />Immediate consequence of this is that <b>kinetic energy</b> of a system of objects, each having its own mass and moving with its own speed, equals to a sum of their individual <b>kinetic energies</b>.<br /><br />That is, <b>kinetic energy</b> is <i>additive</i>.<br /><br /><br /><br />If <b><i>N</i></b> objects of masses <b><i>m<sub>1</sub></i></b>, <b><i>m<sub>2</sub></i></b>, ... <b><i>m<sub>N</sub></i></b> are moving with speeds <b><i>v<sub>1</sub></i></b>, <b><i>v<sub>2</sub></i></b>, ... <b><i>v<sub>N</sub></i></b> then their <b>total kinetic energy</b> equals to<br /><br /><b><i>W = </i><span style="font-size: medium;">Σ</span></b><sub>i∈[1,N]</sub><b>(<i>m</i></b><i><sub>i</sub><b>·v²</b><sub>i</sub></i><b>)/<i>2</i></b><br /><br /><br /><br />As a final example, consider a case when an object of mass <b><i>m</i></b> initially moves along the X-axis. The X-component of its velocity vector is <b><i><span style="text-decoration: overline;">V<sub>x</sub></span></i></b> and its Y-component of a velocity vector <b><i><span style="text-decoration: overline;">V<sub>y</sub></span></i></b> is zero. Then its speed <b><i>V</i></b> (a scalar) equals to its X-component <b><i>V<sub>x</sub></i></b> (a scalar) and its <b>kinetic energy</b> is<br /><br /><b><i>E<sub>k</sub> = m·V²<span style="font-size: medium;">/</span>2</i></b><br /><br />where <b><i>V = V<sub>x</sub></i></b><br /><br /><br /><br />Assume that the force <b><i>F</i></b> acts at an angle <b><i>φ</i></b> to the X-axis.<br /><br />In this case we can represent the vector of force <b><i><span style="text-decoration: overline;">F</span></i></b> as a sum of two perpendicular vectors of force: <b><i><span style="text-decoration: overline;">F<sub>x</sub></span></i></b> acting along the X-axis and <b><i><span style="text-decoration: overline;">F<sub>y</sub></span></i></b> acting along the Y-axis.<br /><br />Obviously,<br /><br /><b><i><span style="text-decoration: overline;">F </span></i></b> = <b><i><span style="text-decoration: overline;">F<sub>x</sub></span></i></b> + <b><i><span style="text-decoration: overline;">F<sub>y</sub></span></i></b><br /><br /><br /><br />These two perpendicular to each other forces, applied to our object, <br />give it a vector of acceleration that also can be represented as a sum <br />of two perpendicular vectors<br /><br /><b><i><span style="text-decoration: overline;">a<sub>x</sub></span> = <span style="text-decoration: overline;">F<sub>x</sub></span> <span style="font-size: medium;">/</span>m</i></b><br /><br /><b><i><span style="text-decoration: overline;">a<sub>y</sub></span> = <span style="text-decoration: overline;">F<sub>y</sub></span> <span style="font-size: medium;">/</span>m</i></b><br /><br /><br /><br />Let's assume that the force <b><i><span style="text-decoration: overline;">F </span></i></b> acts for a duration of time <b><i>t</i></b>.<br /><br />During this time the force <b><i><span style="text-decoration: overline;">F </span></i></b><br /> performs certain work and the velocity of an object will change. We <br />will compare the amount of work performed by the force with a change in <br />the kinetic energy of an object.<br /><br /><br /><br />Considering the initial speed of an object along the X-axis was <b><i>V<sub>x</sub></i></b> and acceleration <b><i>a<sub>x</sub></i></b>, the distance covered by our object along the X-axis equals to<br /><br /><b><i>S<sub>x</sub> = V<sub>x</sub>·t + a<sub>x</sub>·t² <span style="font-size: medium;">/</span>2</i></b><br /><br />At the same time our object moved along the Y-axis with initial speed <b><i>0</i></b> and acceleration <b><i>a<sub>y</sub></i></b>, covering the distance<br /><br /><b><i>S<sub>y</sub> = a<sub>y</sub>·t² <span style="font-size: medium;">/</span>2</i></b><br /><br /><br /><br />In terms of components of the force <b><i><span style="text-decoration: overline;">F </span></i></b> the distances along the coordinates are<br /><br /><b><i>S<sub>x</sub> = V<sub>x</sub>·t + F<sub>x</sub>·t² <span style="font-size: medium;">/</span>(2m)</i></b><br /><br /><b><i>S<sub>y</sub> = F<sub>y</sub>·t² <span style="font-size: medium;">/</span>(2m)</i></b><br /><br /><br /><br />The total work performed by force <b><i><span style="text-decoration: overline;">F </span></i></b>, considering work is additive and can be summarized in each direction, is<br /><br /><b><i>W = F<sub>x</sub>·S<sub>x</sub> + F<sub>y</sub>·S<sub>y</sub> =<br /><br />= F<sub>x</sub>·V<sub>x</sub>·t + F<sub>x</sub>²·t²<span style="font-size: medium;">/</span>(2m) +<br /><br />+ F<sub>y</sub>²·t²<span style="font-size: medium;">/</span>(2m)</i></b><br /><br /><br /><br />Now let's calculate the kinetic energy of an object at the end of time period <b><i>t</i></b>.<br /><br />The X-component of the velocity will be equal to<br /><br /><b><i>V<sub>x</sub><sup>t</sup> = V<sub>x</sub> + a<sub>x</sub>·t = V<sub>x</sub> + F<sub>x</sub>·t <span style="font-size: medium;">/m</span></i></b><br /><br />The Y-component of the velocity will be equal to<br /><br /><b><i>V<sub>y</sub><sup>t</sup> = a<sub>y</sub>·t = F<sub>y</sub>·t <span style="font-size: medium;">/m</span></i></b><br /><br /><br /><br />The object's kinetic energy at the end of the time period <b><i>t</i></b>, equals to<br /><br /><b><i>E<sub>k</sub><sup>t</sup> = m·(V<sup>t</sup>)² <span style="font-size: medium;">/2</span></i></b><br /><br />where <b><i>V<sup>t</sup></i></b> is the speed at time <b><i>t</i></b>.<br />Using the Pythagorean Theorem, we can represent <b><i>(V<sup>t</sup>)²</i></b> as<br /><br /><b><i>(V<sup>t</sup>)² = (V<sub>x</sub><sup>t</sup>)² + (V<sub>y</sub><sup>t</sup>)²</i></b>.<br />Therefore, the kinetic energy can be summarized from X- and Y-components and can be calculated as a sum of:<br /><br /><b><i>E<sub>x</sub><sup>t</sup> = m·(V<sub>x</sub><sup>t</sup>)² <span style="font-size: medium;">/2</span></i></b><br /><br /><b><i>E<sub>y</sub><sup>t</sup> = m·(V<sub>y</sub><sup>t</sup>)² <span style="font-size: medium;">/2</span></i></b><br /><br /><br /><br />The increment of the kinetic energy is<br /><br />Δ<b><i>E = E<sub>x</sub><sup>t</sup> + E<sub>y</sub><sup>t</sup> − E<sub>k</sub></i></b><br /><br />All we have to do is to compare amount of work <b><i>W</i></b> performed by force <b><i><span style="text-decoration: overline;">F </span></i></b> and increment of kinetic energy Δ<b><i>E</i></b> and make sure that they are equal.<br /><br /><br /><br />Indeed,<br /><br />Δ<b><i>E = m·(V<sub>x</sub>+F<sub>x</sub>·t <span style="font-size: medium;">/</span>m)²<span style="font-size: medium;">/</span>2 +<br /><br />+ m·(F<sub>y</sub>·t <span style="font-size: medium;">/</span>m)²<span style="font-size: medium;">/</span>2 − m·V<sub>x</sub>²<span style="font-size: medium;">/</span>2 =<br /><br />= F<sub>x</sub>·V<sub>x</sub>·t + F<sub>x</sub>²·t²/(2m) +<br /><br />+ F<sub>y</sub>²·t²/(2m) = W</i></b><br /><br /><br /><br /><b>Therefore, the work performed by a force acting on an object during <br />certain period of time equals to an increment of a kinetic energy of <br />this object</b>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-11104726460567048342018-12-27T12:44:00.001-08:002018-12-27T12:44:23.577-08:00Unizor - Physics4Teens - Mechanics - Problems on Power<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on Mechanical Power</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />A car engine accelerates a car of mass <b><i>m</i></b> from the state at rest to speed <b><i>V</i></b> during the time <b><i>T</i></b> with constant acceleration along a straight line.<br /><br />Ignore loss of mass due to burning fuel.<br /><br />(a) What is the power of the car engine as a function of time?<br /><br />(b) If engine's power is proportional to amount of fuel supplied to it <br />in a unit of time (fuel burring speed), how the fuel burning speed will <br />change over time to assure the kind of motion described in this problem?<br /><br /><br /><i>Solution</i>:<br /><br /><br /><br />(a) <b><i>a = V/T</i></b><br /><br /><b><i>F = m·a = m·V/T</i></b><br /><br /><b><i>S(t) = a·t²/2 = (V/T)·t²/2</i></b><br /><br /><b><i>W(t) = F·S(t) = (m·V/T)·(V/T)·t²/2 = m·V²·t²/(2T²)</i></b><br /><br /><b><i>P(t) = </i></b><i>d<b>W(t)/</b>d<b>t = (m·V²/T²)·t</b></i><br /><br /><br /><br />(b) The fuel burning speed is linearly increasing with time. The faster a<br /> car goes, as it accelerates, - the more power should be supplied to <br />assure the constant acceleration, and the faster fuel is burning.<br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />A car engine supplies constant power <b><i>P</i></b> to a car of mass <b><i>m</i></b>.<br /><br />Find its speed and acceleration as a function of time.<br /><br />Is speed linearly increasing with time?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />We will use the general formula for power as a function of mass, speed and acceleration:<br /><br /><b><i>P(t) = m·V(t)·a(t)</i></b><br /><br />or, since <b><i>a(t)=</i></b><i>d<b>V(t)/</b>d<b>t</b></i>,<br /><br /><b><i>P(t) = m·V(t)·</i></b><i>d<b>V(t)/</b>d<b>t</b></i><br /><br />Therefore,<br /><br /><b><i>P(t)·</i></b><i>d<b>t = m·V(t)·</b>d<b>V(t)</b></i><br /><br />In our case <b><i>P(t)</i></b> is constant <b><i>P</i></b>. Therefore,<br /><br /><b><i>P·</i></b><i>d<b>t = m·V(t)·</b>d<b>V(t)</b></i><br /><br />Integrating this by time from <b><i>t=0</i></b> to <b><i>t</i></b>,<br /><br /><b><i>P·t = m·V²(t)/2</i></b><br /><br />From this we can find speed <b><i>V(t)</i></b>:<br /><br /><b><i>V(t) = √<span style="text-decoration: overline;">2P·t/m</span></i></b><br /><br />Acceleration is<br /><br /><b><i>a(t) = </i></b><i>d<b>V(t)/</b>d<b>t = √<span style="text-decoration: overline;">P·/(2m·t)</span></b></i><br /><br />Speed is not linearly increasing with time, it is proportional to a <br />square root of time, which means that acceleration is monotonically <br />diminishes, while speed increases.<br /><br /><br /><br /><br /><br /><i>Problem C</i><br /><br /><br /><br />A car engine supplies constant power <b><i>P</i></b> to a car of mass <b><i>m</i></b>.<br /><br />How long would it take for a car to reach speed <b><i>V<sub>max</sub></i></b>?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />As we know,<br /><br /><b><i>P(t) = m·V(t)·a(t)</i></b><br /><br />Since power <b><i>P(t)=P</i></b> - is constant, and acceleration is the first derivative of speed by time, this can be written as<br /><br /><b><i>P = m·V(t)·</i></b><i>d<b>V(t)/</b>d<b>t</b></i><br /><br />Therefore,<br /><br /><b><i>P·</i></b><i>d<b>t = m·V(t)·</b>d<b>V(t)</b></i><br /><br />Integrating this by time from <b><i>t=0</i></b> to <b><i>t</i></b>,<br /><br /><b><i>P·t = m·V²(t)/2</i></b><br /><br />From this we can find time <b><i>t</i></b> as a function of speed:<br /><br /><b><i>t = m·V²(t)/(2P)</i></b><br /><br />Therefore, the time at the moment the speed is equal to <b><i>V<sub>max</sub></i></b> equals to<br /><br /><b><i>t<sub>max</sub> = m·V²<sub>max</sub> /(2P)</i></b><br /><br />Obviously, the more powerful an engine is - the shorter will be the time interval it takes to achieve needed speed.<br /><br /><br /><br />As an example, consider a car 2012 Tesla Model S 85 kWh .<br /><br />It has a mass of 2108 kg and its engine develops the power of 310 Kw.<br /><br />According to the formula above, the time it takes for this car to reach the speed of 60 miles/hour (26.8 m/sec) equal to<br /><br /><b><i>2108·26.8²/(2·310000) = 2.44(sec)</i></b><br /><br />So, in theory, this car is capable to reach the speed of 60 miles/hour in just under 2.5 sec.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-85544329080122466282018-12-26T15:56:00.001-08:002018-12-26T15:56:51.646-08:00Unizor - Physics4Teens - Mechanics - Power<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Definition of Mechanical Power</u><br /><br /><br /><br />To analyze the motion, we often use a concept of <i>speed</i>.<br /><br />Let's assume that an object moves in some inertial reference frame, and <br />the distance covered by it from its initial position along its <br />trajectory is a function of time <b><i>S(t)</i></b>.<br /><br />Recall the definition of <i>speed</i> of an object as the rate, at which this object covers the <i>distance</i> along its motion along a trajectory.<br /><br />In case of a <i>uniform</i> motion we can simply divide the distance <b><i>S</i></b>, covered during time <b><i>t</i></b>, by the time <b><i>t</i></b> to get the speed:<br /><br /><b><i>V = S/t</i></b><br /><br />In case of <i>non-uniform</i> motion the speed changes and at any particular moment of time <b><i>t</i></b> an instantaneous speed can be calculated using differentials:<br /><br /><b><i>V(t) = </i></b><i>d<b>S(t)/</b>d<b>t</b></i><br /><br /><br /><br />To analyze the mechanical work performed to achieve certain results, we often use a concept of <i>power</i>.<br /><br />Let's assume that something or someone performs certain work and, as the time goes by, the work performed is a function of time <b><i>W(t)</i></b>.<br /><br />The <i><b>power</b></i> is the rate, at which the <i>work</i> is performed.<br /><br />If during the time <b><i>t</i></b> the <i>work</i> performed is <b><i>W</i></b>, we define the <i>average power</i> of the whoever or whatever performs the work as<br /><br /><b><i>P = W/t</i></b><br /><br />Most likely, at equal in length but different time intervals the amount <br />of work performed will be different. For example, when a car starts, its<br /> engine should give a car an acceleration, which requires more work per <br />unit of time than to maintain a constant speed on a smooth straight <br />road.<br /><br />In cases like this we can talk about an instantaneous <i><b>power</b></i> as a function of time that can be calculated using differentials:<br /><br /><b><i>P(t) = </i></b><i>d<b>W(t)/</b>d<b>t</b></i><br /><br /><br /><br />Consider an example of an object in uniform motion against the force of friction with a constant speed <b><i>V</i></b>.<br /><br />The force <b><i>F</i></b> that moves it forward must be equal in <br />magnitude and opposite in direction to the force of friction to maintain<br /> the constant speed. Since the friction is constant, the force <b><i>F</i></b> must be constant as well.<br /><br />The distance <b><i>S</i></b> covered as a function of time <b><i>t</i></b> is<br /><br /><b><i>S(t) = V·t</i></b><br /><br />Therefore, the work performed by the force <b><i>F</i></b> during the time <b><i>t</i></b> is<br /><br /><b><i>W(t) = F·S(t) = F·V·t</i></b><br /><br />From the definition of <i>power</i> follows that the power this force <b><i>F</i></b> exhorts is<br /><br /><b><i>P(t) = </i></b><i>d<b>W(t)/</b>d<b>t = F·V</b></i><br /><br />As an example, the car engine exhorts the same power and consumes the <br />same amount of gas per unit of time, if the car uniformly moves along a <br />straight road. This power is used to generate a force sufficient to <br />overcome the friction of wheels and air resistance.<br /><br /><br /><br />Consider a more general case, when the motion is not uniform.<br /><br />Assume, an object of mass <b><i>m</i></b> moves as a result of action of force <b><i>F(t)</i></b>, where <b><i>t</i></b> is time. The distance it covers is <b><i>S(t)</i></b>.<br /><br />Then during an infinitesimal time interval from <b><i>t</i></b> to <b><i>t+</i></b><i>d<b>t</b></i> the work performed by this force will be<br /><br /><i>d<b>W(t) = F(t)·</b>d<b>S(t)</b></i><br /><br />Considering the Newton's second law,<br /><br /><b><i>F(t) = m·a(t)</i></b>,<br /><br />where <b><i>a(t)</i></b> is acceleration.<br /><br />Increment of distance is<br /><br /><i>d<b>S(t) = V(t)·</b>d<b>t</b></i>,<br /><br />where <b><i>V(t)</i></b> is an object's speed.<br /><br />Also, by definition of acceleration,<br /><br /><b><i>a(t) = </i></b><i>d<b>V(t)/</b>d<b>t</b></i><br /><br />Therefore,<br /><br /><i>d<b>W(t) = (m·</b>d<b>V(t)/</b>d<b>t)·V(t)·</b>d<b>t =<br /><br />= m·V·</b>d<b>V(t)</b></i><br /><br />Power exhausted by this force is, therefore<br /><br /><b><i>P(t) = </i></b><i>d<b>W(t)/</b>d<b>t =<br /><br />= m·V·</b>d<b>V(t)/</b>d<b>t = <br /><br />= m·V(t)·a(t)</b></i><br /><br /><br /><br />From the definition of <i>power</i> as amount of work per unit of time or, more precisely, the first derivative of work by time<br /><br /><b><i>P(t) = </i></b><i>d<b>W(t)/</b>d<b>t</b></i><br /><br />follows that the unit of measurement of power is <i>joule/sec</i> called <i>watt</i>.<br /><br />Expanding the definition of <i>joule</i> as <i>newton·meter</i>,<br /><br /><i><b>1 watt = 1J/sec = 1N·m/sec</b></i><br /><br />Obvious extensions of unit of power <i>watt</i> are<br /><br /><i>kilowatt = 1,000 watt</i> and<br /><br /><i>megawatt = 1,000,000 watt</i>.<br /><br /><br /><br />There is an old unit of power called <i>horsepower</i>.<br /><br /><i>Metric horsepower</i>, derived from lifting up against a force of gravity on Earth a weight of mass 75 <i>kg</i> with a constant speed of 1 <i>m/sec</i>, is related to <i>watt</i> unit as<br /><br /><b><i>1(metric HP) =<br /><br />=75(kg)·9.8(m/sec²)·1(m/sec)≅<br /><br />≅ 735.5(W)</i></b><br /><br />For historical reasons there is also a <i>mechanical horsepower</i>, defined as 33,000 <i>pound-feet per minute</i>, related to <i>watt</i> unit as<br /><br /><b><i>1(mechanical HP) ≅ 745.7(W)</i></b><br /><br /><br /><br />So, a car engine of 200 <i>mechanic horsepower</i> has the power of about 149,140 <i>watt</i>.<br /><br /><br /><br /><i>Watt</i>, as a unit of measurement, was called in honor of James <br />Watt, an 18th century Scottish scientist who was one of the first to <br />research a concept of power, developed steam engines and measured the <br />power of a horse.<br /><br /><br /><br />Now let's address the concept of <i>power</i> in a case of rotation with constant angular speed. An example is lifting a bucket of water from a well.<br /><br />Assume, a bucket of water has a mass <b><i>m</i></b> and we lift it with constant linear speed <b><i>V</i></b> with an angular speed of the well's wheel <b><i>ω=V/R</i></b>, where <b><i>R</i></b> - radius of a well's wheel.<br /><br /><br /><br />Since the speed is constant, the force <b><i>F</i></b> that acts on a bucket equals to <b><i>m·g</i></b>, where <b><i>g</i></b> - acceleration of the free fall.<br /><br />At the same time, if the wheel is turned by some motor and <b><i>R</i></b> is the radius of its shaft, the motor manufacturer provides technical characteristic not only of the <i>power</i>, but also of a <i>torque</i> of a motor.<br /><br />Remember that the <i>torque</i> equals to<br /><br /><b><i>τ = F·R</i></b><br /><br /><br /><br />So, on one hand, we have expressed the power of a motor <b><i>P</i></b> in terms of unknown force <b><i>F</i></b> and linear speed of a bucket:<br /><br /><b><i>P = F·V = F·R·ω</i></b><br /><br />On another hand, we expressed the torque of this motor in terms of the same unknown force <b><i>F</i></b> and a radius of its shaft:<br /><br /><b><i>τ = F·R</i></b><br /><br />Substituting torque <b><i>τ</i></b> for <b><i>F·R</i></b> in the formula for power, we can find the relationship between the power of a motor and is torque:<br /><br /><b><i>P = τ·ω</i></b><br /><br /><br /><br />Notice the similarity between the formula for power in case of uniform motion along a straight line <b><i>P=F·V</i></b> and formula for power in case of rotation with constant angular speed <b><i>P=τ·ω</i></b>. Instead of force <b><i>F</i></b> in case of straight line motion, we use torque <b><i>τ</i></b> for rotation and, instead of linear speed <b><i>V</i></b> for straight line motion, we use angular speed <b><i>ω</i></b> for rotation.<br /><br /><br /><br />Let's check this with real data about a particular engine.<br /><br />Below is a graph representing the power and torque of Ford Motor Company 6.7L Power Stroke diesel V-8.<br /><br /><img src="http://www.unizor.com/Pictures/PowerTorque.jpg" style="height: 130px; width: 200px;" /><br /><br />As you see, the power and torque grow relatively monotonically until <br />some engine limitations start playing significant role. While in the <br />area of monotonic growth, we can take a particular angular speed, say, <br />1400 RPM (revolutions per minute) and see that the power of an engine <br />equals approximately 174 HP (mechanical horsepower) and the torque is <br />about 650 Lb-Ft (pound-feet).<br /><br />Let's check if the relationship between power and torque derived above is held in this case.<br /><br />First of all, we transform all units into standard physical measures defined in SI:<br /><br /><i>1 RPM (revolutions per minutes) =<br /><br />= 2π radian per minute =<br /><br />= 2π/60 radians/sec =<br /><br />= 0.1048 rad/sec</i><br /><br /><i>1 HP = 745.7 W = 745.7 J/sec</i><br /><br /><i>1 Lb-Ft = 1.35582 J</i><br /><br />Substituting all the above, we see that<br /><br />angular speed is equal to<br /><br /><b><i>1400·0.1048 = 146.6 rad/sec</i></b><br /><br />power is<br /><br /><b><i>174·745.7 = 129,751.8 J/sec</i></b><br /><br />torque equals to<br /><br /><b><i>650·1.35582 = 881.3 J</i></b><br /><br /><br /><br />Now we can check the relationship between the power, torque and angular speed<br /><br /><b><i>P = τ·ω</i></b><br /><br />Indeed,<br /><br /><b><i>881.3·146.6 = 129,198.6</i></b><br /><br />As we see, the difference between power and a product of torque by <br />angular speed is minimal, attributable to imprecise measurement of the <br />parameters and graph reading.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-40926153168876130552018-12-11T13:58:00.001-08:002018-12-11T13:58:00.703-08:00Unizor - Physics4Teens - Mechanics - Work - Problems<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on Mechanical Work</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />An object of mass <b><i>m</i></b> is pushed up a plane inclined to horizon at angle <b><i>φ</i></b> with constant acceleration <b><i>a</i></b> to a height <b><i>h</i></b> against a friction with coefficient of kinetic friction <b><i>μ</i></b>.<br /><br />(a) What is the amount of work necessary to achieve this goal?<br /><br /><img src="http://www.unizor.com/Pictures/WorkProblem1.png" style="height: 130px; width: 200px;" /><br /><br />(b) Consider the following real conditions of this experiment:<br /><br /><i><b>m = 10kg</b>(kilograms)</i><br /><br /><i><b>g = 9.8m/sec²</b>(meters per sec²)</i><br /><br /><i><b>h = 0.5m</b>(meters)</i><br /><br /><i><b>φ = 30°</b>(degrees)</i><br /><br /><i><b>μ = 0.2</b></i><br /><br /><i><b>a = 0.1m/sec²</b>(meters per sec²)</i><br /><br />The work <b><i>W=F·S</i></b> is measured in <i>N·m=kg·m²/sec²</i> units, called <i>joule</i> and abbreviated as <b><i>J</i></b>.<br /><br />Calculate the work <b><i>W</i></b> performed in this experiments in <i>joules</i>.<br /><br /><br /><br /><i>Solution</i><br /><br />Let <b><i>F</i></b> be the force that pushes the object up the slope, <b><i>P=m·g</i></b> is the object's weight, <b><i>R</i></b> is the force of resistance from the friction, <b><i>S</i></b> is the distance this object moves to reach the height <b><i>h</i></b>.<br /><br />(a) <b><i>F−P·sin(φ)−R = m·a</i></b><br /><br /><b><i>R = P·cos(φ)·μ</i></b><br /><br /><b><i>W = F·S</i></b><br /><br /><b><i>S = h/sin(φ)</i></b><br /><br /><b><i>F = P·sin(φ) + R + m·a =<br /><br />= m·[g·sin(φ) + g·cos(φ)·μ + a]</i></b><br /><br /><b><i>W = F·h/sin(φ) =<br /><br />=m·h[g+g·cot(φ)·μ+a/sin(φ)]</i></b><br /><br />(b)<br /><i><b>W = 67J</b>(joules)</i><br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />An object of mass <b><i>m</i></b> is pushed by a constant force <b><i>F</i></b> down a slope of a plane inclined to horizon at angle <b><i>φ</i></b>. Initially, it's at rest, the final speed is <b><i>V</i></b>. There is a friction with coefficient of kinetic friction <b><i>μ</i></b>.<br /><br />(a) What is the time <b><i>t</i></b> from the beginning to the end of the object's motion?<br /><br />(b) What is the distance <b><i>S</i></b> this object moved until reaching the final speed <b><i>V</i></b>?<br /><br />(c) What is the amount of work <b><i>W</i></b> performed by this force?<br /><br />(d) Consider the following real conditions of this experiment:<br /><br /><i><b>F = 20N</b>(newtons)</i><br /><br /><i><b>m = 10kg</b>(kilograms)</i><br /><br /><i><b>g = 9.8m/sec²</b>(meters per sec²)</i><br /><br /><i><b>V = 0.5m/sec</b>(meters per sec)</i><br /><br /><i><b>φ = 5°</b>(degrees)</i><br /><br /><i><b>μ = 0.7</b></i><br /><br />Calculate the work <b><i>W</i></b> performed in this experiments in <i>joules</i>, distance <b><i>S</i></b> and time <b><i>t</i></b> of motion.<br /><br /><br /><br /><i>Solution</i><br /><br /><b><i>P=m·g</i></b> is the object's weight,<br /><br /><b><i>R = P·cos(φ)·μ</i></b> is the force of resistance from the friction,<br /><br /><b><i>S</i></b> is the distance this object moves to reach the speed <b><i>V</i></b>,<br /><br /><b><i>a</i></b> is the acceleration of this object.<br /><br />Then from the Newton's Second Law<br /><br /><b><i>F+P·sin(φ)−R = m·a</i></b><br /><br />(we added <b><i>P·sin(φ)</i></b> because an object moves down a slope)<br /><br />Now we can find an acceleration of our object:<br /><br /><b><i>F+m·g·sin(φ)−m·g·cos(φ)·μ =<br /><br />= m·a</i></b><br /><br /><b><i>a = F/m + g·sin(φ) − g·cos(φ)·μ</i></b><br /><br />Knowing acceleration <b><i>a</i></b> and the correspondence between acceleration, final speed and time <b><i>V=a·t</i></b>, we can determine time:<br /><br />(a) <b><i>t = V/a</i></b><br /><br />Now we can find the distance<br /><br />(b) <b><i>S = a·t²/2 = V·t/2</i></b><br /><br />The work performed by the force <b><i>F</i></b> equals to<br /><br />(c) <b><i>W = F·S</i></b><br /><br />(d) <b><i>a = 0.066(m/sec²)</i></b><br /><br /><b><i>t = 7.576(sec)</i></b><br /><br /><b><i>S = 1.894(m)</i></b><br /><br /><b><i>W = 37.879(joules))</i></b><br /><br /><br /><br /><br /><br /><i>Problem C</i><br /><br /><br /><br />An object of mass <b><i>m</i></b> is dropped down from a certain height above a surface of some planet. At the moment it hits the ground its speed is <b><i>V</i></b>.<br /><br />What is the work <b><i>W</i></b> performed by the force of gravity?<br /><br /><br /><br /><i>Solution</i><br /><br />Let the acceleration of free falling on this planet is <b><i>a</i></b>, the time of falling <b><i>t</i></b> and the height <b><i>S</i></b>.<br /><br />Then<br /><br /><b><i>F = m·a</i></b><br /><br /><b><i>V = a·t</i></b><br /><br /><b><i>t = V/a</i></b><br /><br /><b><i>S = a·t²/2 = V²/(2·a)</i></b><br /><br /><b><i>W = F·S = m·V²/2</i></b><br /><br />Notice, the work depends only on mass and final speed - the results of <br />action, not on the height, nor acceleration of free falling, nor on <br />time.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-79701968385328358152018-10-12T11:06:00.001-07:002018-10-12T11:06:40.981-07:00Unizor - Physics4Teens - Mechanics - Work - Golden Rule of Mechanics<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Golden Rule of Mechanics</u><br /><br /><br /><br />Let's summarize what we have learned about mechanical <b>work</b> in the previous lecture.<br /><br /><br /><br />1. In a simple case of motion along a straight line with a constant force <b><i>F</i></b> acting along a trajectory, the most important parameter that quantifies the result of the action is <b>work</b> defined as <b><i>W=F·S</i></b>, that fully characterizes and is fully characterized by speed <b><i>V</i></b> of an object:<br /><br /><b><i>W = F·S = m·V²/2</i></b><br /><br />In particular, it means that increasing the force by a factor of <b><i>N</i></b> and decreasing the distance it acts by the same factor of <b><i>N</i></b> would result in the same final speed of an object.<br /><br />So, we can "win" in distance, but we will "lose" in force and vice versa.<br /><br />This is the first example of the <i>Golden Rule of Mechanics</i> - there<br /> are many ways to achieve the result, you can reduce your distance, but <br />you will have to increase the force or you can reduce the force, but you<br /> will have to increase the distance.<br /><br />In short, as we mentioned above, <i>whatever you win in distance you lose in force and vice versa</i>.<br /><br /><br /><br />2. In case of a constant force acting at an angle to a straight line trajectory, the difference is only a factor <b><i>cos(φ)</i></b>, where <b><i>φ</i></b> is an angle between a force and a direction of a trajectory. In vector form it represents the scalar product <b><i>(F·S)</i></b>.<br /><br />So, the <i>Golden Rule</i> works exactly as above.<br /><br /><br /><br />3. Recall the formula for work of a force <b><i>F</i></b> pushing an object of weight <b><i>P</i></b> up along an inclined plane of the length <b><i>S</i></b> making angle φ with horizon to the height <b><i>H</i></b>:<br /><br /><b><i>W = F·S = P·H</i></b><br /><br />We've proven this in the previous lecture and, as you see, it's <br />independent of the slope of an inclined plane. However, the minimum <br />effort we have to apply as a force to move an object up the slope is <b><i>F=P·sin(φ)</i></b>, while the distance equals to <b><i>S=H/sin(φ)</i></b>.<br /><br />We can reduce the effort (the force <b><i>F</i></b>) by using an incline of a smaller slope, but that would lengthen the distance <b><i>S</i></b> we have to push an object.<br /><br />So, again, we see the <i>Golden Rule</i> in action.<br /><br /><br /><br />4. Consider lifting some heavy object of the weight <b><i>P</i></b> using a lever, applying the force <b><i>F</i></b> to the opposite to an object end of the lever.<br /><br /><img src="http://www.unizor.com/Pictures/WorkLever.jpg" style="height: 106px; width: 200px;" /><br /><br />This is a case of equilibrium in rotational motion and the balance can <br />be achieved by equating the moments of two forces acting against each <br />other:<br /><br /><b><i>F·L<sub>f</sub> = P·L<sub>p</sub></i></b><br /><br />If <b><i>S<sub>f</sub></i></b> is the distance the force <b><i>F</i></b> acts down and <b><i>S<sub>p</sub></i></b> is the distance our object moves up,<br /><br /><b><i>S<sub>f</sub>/S<sub>p</sub> = L<sub>f</sub>/L<sub>p</sub></i></b> and<br /><br /><b><i>F·S<sub>f</sub> = P·S<sub>p</sub></i></b><br /><br />By using a lever with longer arm <b><i>L<sub>f</sub></i></b> for the application of force, we can proportionally reduce the force <b><i>F</i></b><br /> achieving the same result - lifting an object to certain height. An <br />inverse is true as well - we can shorten the arm and proportionally <br />increase the force.<br /><br />In any case, the <i>Golden Rule of Mechanics</i> is observed: "winning" <br />in force - proportionally "losing" in distance or "winning" in distance"<br /> - proportionally "losing" in force.<br /><br /><br /><br />All the above examples emphasize the importance of the concept of <b>mechanical work</b> as the quantitative measure and characteristic of the purpose and the result of applying a force. The so-called <i>Golden Rule of Mechanics</i> is just a catchy term that underscores the importance of the concept of <b>work</b>.<br /><br /><br /><br /><b>Acting with the force to achieve certain goal necessitates performing<br /> certain amount of work that depends on the goal, not on how we achieve <br />it</b>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-17339257276692859422018-10-10T13:10:00.001-07:002018-10-10T13:10:37.815-07:00Unizor - Physics4Teens - Mechanics - Work - Definition<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Definition of Mechanical Work</u><br /><br /><br /><br />Why do we apply a force to an object?<br /><br />Usually, to achieve certain result, like to accelerate it to a certain speed or to move it from one point to another.<br /><br /><br /><br /><i>Straight line motion with a force acting along a trajectory</i><br /><br /><br /><br />Consider an acceleration of a car by pressing the gas pedal. As a <br />result, a car will reach certain speed and then, in absence of the <br />traffic lights, disregarding friction and air resistance, it goes by <br />inertia, maintaining this constant speed.<br /><br /><br /><br />A car with a more powerful engine will need a shorter distance to reach <br />certain speed. A car with a weaker engine accelerates on a longer <br />distance to reach the same speed.<br /><br />It seems, there is some relationship between the force, the distance it <br />is applied to an object and the final speed reached by this object as a <br />result of the acting force.<br /><br /><br /><br />Let's consider the final speed of a constant force <b><i>F</i></b> acting on an object of mass <b><i>m</i></b>, initially at rest, acting on certain distance <b><i>S</i></b>.<br /><br /><br /><br />The acceleration of this object, according to the Second Newton's Law is<br /><br /><b><i>a = F/m</i></b><br /><br />The time <b><i>t</i></b> to cover distance <b><i>S</i></b> with acceleration <b><i>a</i></b> is based on the formula of Kinematics<br /><br /><b><i>S = a·t²/2</i></b> and is equal to<br /><br /><b><i>t = √<span style="text-decoration: overline;">2S/a </span> = √<span style="text-decoration: overline;">2S·m/F </span></i></b><br /><br />From this we derive the final speed at the end of acceleration<br /><br /><b><i>V = a·t = (F/m)·√<span style="text-decoration: overline;">2S·m/F </span> =<br /><br />= √<span style="text-decoration: overline;">2F·S/m </span></i></b><br /><br /><br /><br />As we see, the final speed of an object of mass <b><i>m</i></b> depends on the product of the force and the distance this force is acting.<br /><br />This product that characterizes the result of applying a force on a certain distance is called the <b>work</b> performed by a given force acting on a given distance:<br /><br /><b><i>W = F·S</i></b><br /><br />We can reduce by half the force and double the distance - the resulting speed will be the same.<br /><br />Using this definition of <b>work</b>, we can represent the final speed as<br /><br /><b><i>V = √<span style="text-decoration: overline;">2W/m </span></i></b><br /><br />Resolving for <b>work</b> as a function of final speed, we obtain<br /><br /><b><i>W = m·V²/2</i></b><br /><br /><br /><br />So, given a final speed, we can determine how much work it takes to <br />achieve it and, given amount of work, we can calculate the speed this <br />work will cause.<br /><br /><br /><br />This concept of <b>work</b> is closely tied with such less precisely <br />defined concepts of result, purpose, effect etc. While we not always can<br /> compare the results or effect of two physical experiments, we can <br />always compare the <b>work</b> done by the forces participating in these experiments.<br /><br />Thus, the <b>work</b> can be used to measure the results or effect of physical experiment.<br /><br /><br /><br /><i>Straight line motion with a force acting at an angle to a trajectory</i><br /><br /><br /><br />Consider a slightly more complicated example with force acting at an angle to a trajectory.<br /><br />A toy train of mass <b><i>m</i></b> is pulled by a child, who stands on a side of a track, with a constant force <b><i>F</i></b>, so the force acts at an angle to a straight line trajectory, making an angle <b><i>φ</i></b> with it.<br /><br /><img src="http://www.unizor.com/Pictures/WorkAngle.png" style="height: 140px; width: 200px;" /><br /><br />Representing this force <b><i>F</i></b> (in blue on a picture above) as a<br /> sum of two forces (in red), one acting along a track, forcing the train<br /> to speed up, and another acting perpendicularly to a track (force <b><i>N</i></b>), which is balanced by the reaction of the track <b><i>R</i></b> (in brown), we see that the force that pushes a train forward equals to <b><i>F·cos(φ)</i></b>, while the force acting perpendicularly to a track can be simply ignored as being balanced by an opposite reaction force.<br /><br /><br /><br />Exactly the same considerations as above leads us to a more universal formula for relationship between the final speed and the <b>work</b>:<br /><br /><b><i>V = √<span style="text-decoration: overline;">2W/m </span></i></b>,<br /><br />where <b><i>W = F·S·cos(φ)</i></b><br /><br /><br /><br />Obviously, if the force acts along a straight line of a trajectory, angle <b><i>φ</i></b> is zero, its cosine is 1 and the formula corresponds to the one derived earlier.<br /><br /><br /><br />Resolving the formula above for <b>work</b>, we obtain<br /><br /><b><i>W = m·V²/2</i></b><br /><br />which indicates that the <b>work</b> depends on the result of an action <br />only (final speed), not the way how we achieve this result, using a <br />stronger force on a shorter distance or a weaker force on a longer <br />distance.<br /><br /><br /><br /><i>Motion against gravity along an inclined plane</i><br /><br /><br /><br />Let's consider a completely different goal of a physical experiment and show the importance of a concept of <b>work</b>.<br /><br />This time our purpose is to lift an object of mass <b><i>m</i></b> to a certain height <b><i>H</i></b> above the ground along a frictionless inclined plane making angle <b><i>φ</i></b> with horizon.<br /><br /><img src="http://www.unizor.com/Pictures/WorkInclined.png" style="height: 165px; width: 200px;" /><br /><br />The force <b><i>F</i></b> needed for this must be equal to a component <b><i>R</i></b> of the weight <b><i>P</i></b> of an object along the inclined because the other component of its weight <b><i>N</i></b>, perpendicular to an inclined, is balanced by a reaction of the plane.<br /><br /><b><i>F = P·sin(φ)</i></b><br /><br />The distance <b><i>S</i></b> this force is acting on equals to<br /><br /><b><i>S = H/sin(φ)</i></b><br /><br />From this we can derive the <b>work</b> performed by force <b><i>F</i></b> along the distance <b><i>S</i></b>:<br /><br /><b><i>W = F·S = P·sin(φ)·H/sin(φ) =<br /><br />= P·H</i></b><br /><br /><br /><br />This is quite a remarkable result. The <b>work</b> does not depend on the angle of an incline, only on the height we lift the object and its weight.<br /><br /><br /><br />As in the previous cases, the <b>work</b> seems to be a characteristic <br />of the result and does not depend on how we have achieved it, using an <br />inclined with bigger or smaller slope.<br /><br /><br /><br /><i>Rotational motion with a force acting tangentially to a trajectory</i><br /><br /><br /><br />Our final example is about rotation.<br /><br />Consider a person starts rotating a carousel of a radius <b><i>r</i></b>, having a moment of inertia <b><i>I</i></b>, from the state of rest to some angular speed <b><i>ω</i></b> with constant angular acceleration <b><i>α</i></b>, applying constant force <b><i>F</i></b> tangentially to the carousel's rim.<br /><br />From the rotational dynamics we can determine angular acceleration, knowing torque <b><i>τ</i></b> of the force and a moment of inertia of a carousel <b><i>I</i></b>:<br /><br /><b><i>τ = F·r = I·α</i></b><br /><br /><b><i>α = F·r/I</i></b><br /><br />Linear acceleration of the point of application of the force equals to <b><i>r·α</i></b>.<br /><br /><br /><br />To achieve the final angular speed <b><i>ω</i></b> with angular acceleration <b><i>α</i></b> we need time<br /><br /><b><i>t = ω/α = ω·I/(F·r)</i></b><br /><br />During this time the point of application of force travels around a circle for a distance of<br /><br /><b><i>S = r·α·t²/2 =<br /><br />= r·(F·r/I)·[ω·I/(F·r)]²/2 =<br /><br />= I·ω²/(2F)</i></b><br /><br /><br /><br />Multiplying by <b><i>F</i></b> both sides, we obtain the relationship between <b>work</b> <b><i>W=F·S</i></b> and final angular speed achieved as a result of application of the force:<br /><br /><b><i>W = F·S = I·ω²/2</i></b><br /><br />This is a rotational equivalent of the analogous formula for straight line movement derived in the beginning of this lecture.<br /><br /><br /><br />So, given a final angular speed, we can determine how much work it takes<br /> to achieve it and, given amount of work, we can calculate the angular <br />speed this work will cause.<br /><br /><br /><br /><i>Definition of work in simple cases presented above</i><br /><br /><br /><br />All the above examples provide a proper basis for the following definition of the <b>work</b>:<br /><br /><b><i> Work of the force F acting at an angle φ to trajectory on the distance S is <nobr>W = F·S·cos(φ)</nobr></i></b>.<br /><br /><br /><br /><i>General case of a motion along a curved trajectory</i><br /><br /><br /><br />To be more precise and to cover a case of a curved trajectory, we have <br />to require that this definition should be applied to infinitesimal <br />amount of <b>work</b> <i>d<b>W</b></i> performed on infinitesimal interval <i>d<b>S</b></i> of trajectory:<br /><br /><i>d<b>W = F·</b>d<b>S·cos(φ)</b></i><br /><br /><br /><br />Using a concept of <i>scalar product</i> of vectors and considering force and interval as vectors, the same definition can be written as<br /><br /><i>d<b>W = (<span style="text-decoration: overline;">F </span>·</b>d<b><span style="text-decoration: overline;">S </span>)</b></i><br /><br /><br /><br />The latter represents the most rigorous definition of <b>work</b>.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-48669735375370020802018-10-03T14:38:00.001-07:002018-10-03T14:38:56.936-07:00Unizor - Physics4Teens - Mechanics - Statics - Problems<br /><br /><br /><br /><iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/7JADXRMPR5U" width="480"></iframe> <br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Problems on Equilibrium</u><br /><br /><br /><br /><i>Problem 1</i><br /><br /><br /><br />Two point-objects of mass <b><i>M</i></b> and <b><i>m</i></b> are hanging at two opposite ends of a weightless rod of length <b><i>L</i></b>.<br /><br />At what point on a rod should we fix a thread, so the system of a rod with two weights is hanging on this thread in equilibrium?<br /><br />Does this state of equilibrium imply that a rod is horizontal?<br /><br /><br /><br /><i>Answer</i><br /><br />At distance <b><i>m·L <span style="font-size: medium;">/</span>(M+m)</i></b> from the end with object of mass <b><i>M</i></b>.<br /><br /><br /><br /><i>Problem 2</i><br /><br /><br /><br />An object of weight <b><i>W</i></b> is positioned on a flat surface. The coefficient of <i>static</i> friction is <b><i>μ</i></b>.<br /><br />What is the minimum pulling force <b><i>P</i></b> to be applied to a <br />rope, attached to this object, to start pulling it forward, if the angle<br /> between a rope and a horizontal flat surface is <b><i>φ</i></b>?<br /><br /><br /><br /><i>Answer</i><br /><br /><br /><br /><b><i>P = μ·W <span style="font-size: medium;">/</span> </i></b>[<b><i>cos(φ)+μ·sin(φ)</i></b>]<br /><br /><br /><br /><i>Problem 3</i><br /><br /><br /><br />Consider the following illustration to this problem.<br /><img src="http://www.unizor.com/Pictures/EquilibriumProblem3.png" style="height: 400px; width: 200px;" /><br /><br />A person of weight <b><i>W</i></b> stands on a platform of weight <b><i>P</i></b>. Pulleys and ropes are arranged as on this illustration.<br /><br />What is the force <b><i>F</i></b> a person should apply to a rope to keep the whole system in equilibrium?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Let <b><i>N</i></b> be a reaction of a platform onto a person. It pushes the person upward.<br /><br />If <b><i>F</i></b> is a force of a person pulling a rope down, and the <br />system is in equilibrium, that is all components are at rest and all <br />forces are balanced, the rope pulls a person up with the same force <b><i>F</i></b>, according to the Third Newton's Law.<br /><br />Since a person is in equilibrium, forces directed upward (<b><i>N+F</i></b>) should be equal in magnitude to forces directed downwards (weight of the person <b><i>W</i></b>):<br /><br /><b><i>N+F = W</i></b><br /><br />The platform is also in equilibrium. The forces that push it down are its weight <b><i>P</i></b> and reaction <b><i>N</i></b> of a person standing on it. The forces pulling the platform up are tensions of two ropes <b><i>T</i></b> on the left and <b><i>F</i></b> on the right:<br /><br /><b><i>P+N = T+F</i></b><br /><br />Finally, a small pulley is in equilibrium. It's pulled down by two tensions <b><i>F</i></b>, from the left and from the right. The tension <b><i>T</i></b> pulls it up:<br /><br /><b><i>T = 2F</i></b><br /><br />We have three equations with three unknowns: <b><i>T</i></b>, <b><i>N</i></b> and <b><i>F</i></b>, which we have to solve for an unknown force <b><i>F</i></b>, with which a person pulls a rope to balance the system:<br /><br />Substituting <b><i>T</i></b> from the last equation into the second one, getting a system of two equations:<br /><br /><b><i>N+F = W</i></b><br /><br /><b><i>P+N = 3F</i></b><br /><br />Solving the first equation for <b><i>N</i></b> and substituting into the second:<br /><br /><b><i>P+W−F = 3F</i></b><br /><br />Resolving for <b><i>F</i></b>:<br /><br /><b><i>F = (P+W)/4</i></b><br /><br /><br /><br /><i>Answer</i><br /><br /><b><i>F = (P+W)/4</i></b><br /><br /><br /><br /><i>Problem 4</i><br /><br /><br /><br />A ladder is standing at some angle to the floor. It's in equilibrium, <br />that is it's not slipping down to the floor. What holds it is the <i>static</i> friction between it and the floor.<br /><br />If the ladder stands almost vertically, the <i>static</i> friction is <br />greater and it easily holds the ladder in the position. As we move the <br />bottom of a ladder away from the vertical, there will be a point when <i>static</i> friction will no longer could hold the ladder in place and the ladder would slip down to the floor.<br /><br />What is the smallest angle the ladder would hold if the coefficient of <i>static</i> friction is <b><i>μ</i></b>?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Consider the following illustration to this problem.<br /><img src="http://www.unizor.com/Pictures/EquilibriumLadder.png" style="height: 200px; width: 200px;" /><br /><br />While the reaction of the wall <b><i>N<sub>1</sub></i></b> is smaller than the <i>static</i> friction <b><i>F</i></b>, the ladder would hold its position.<br /><br />The smallest angle between the ladder and the floor will be when these two forces are equal.<br /><br />To be in equilibrium, vector sum of all forces must be equal to <br />null-vector and a sum of all angular momentums also must be equal to <br />null-vector.<br /><br />Along the horizontal X-axis the sum of all forces equals to zero if the horizontal reaction of the wall <b><i>N<sub>1</sub></i></b> equals to the force of <i>static</i> friction <b><i>F</i></b>, which, in turn, equals to a product of the vertical reaction of the floor <b><i>N<sub>2</sub></i></b> by a coefficient of <i>static</i> friction <b><i>μ</i></b>:<br /><br />(a) <b><i>N<sub>1</sub> = μ·N<sub>2</sub></i></b><br /><br />Along the vertical Y-axis the sum of forces equals to zero if the reaction of the floor <b><i>N<sub>2</sub></i></b> equals to weight of a ladder <b><i>W</i></b>:<br /><br />(b) <b><i>N<sub>2</sub> = W</i></b><br /><br />To analyze angular momentums, let's choose an axis of rotation. It's <br />easier to choose an axis perpendicular to both X- and Y-axis that goes <br />through a point where a ladder touches the floor. In this case angular <br />momentums of forces <b><i>N<sub>2</sub></i></b> and <b><i>F</i></b> are zero because their corresponding radiuses are zero.<br /><br />Let's assume that the length of the ladder is <b><i>D</i></b> and the angle between the ladder and the floor is <b><i>φ</i></b>. The magnitude of the angular momentum of the force of weight <b><i>W</i></b> will then be <b><i>W·D·cos(φ)/2</i></b>. The magnitude of the angular momentum of the reaction of the wall <b><i>N<sub>1</sub></i></b> will be <b><i>N<sub>1</sub>·D·sin(φ)</i></b>. The condition of the sum of angular momentums to be equal to zero results in the equation:<br /><br /><b><i>N<sub>1</sub>·D·sin(φ) = W·D·cos(φ)/2</i></b><br /><br />The length of the ladder can be canceled:<br /><br />(c) <b><i>N<sub>1</sub>·sin(φ) = W·cos(φ)/2</i></b><br /><br />Equations (a), (b) and (c) form a system of three equations with three unknowns:<br /><br />(a) <b><i>N<sub>1</sub> = μ·N<sub>2</sub></i></b><br /><br />(b) <b><i>N<sub>2</sub> = W</i></b><br /><br />(c) <b><i>N<sub>1</sub>·sin(φ) = W·cos(φ)/2</i></b><br /><br />From this<br /><br /><b><i>tan(φ) = 1/(2μ)</i></b><br /><br /><br /><br /><i>Answer</i><br /><br /><br /><br /><b><i>φ = arctan</i></b>[<b><i>1/(2μ)</i></b>]<br /><br />Interestingly, the answer does not depend on the weight or the length of the ladder.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-87507561583900535442018-09-26T11:40:00.001-07:002018-09-26T11:40:52.547-07:00Unizor - Physics4Teens - Mechanics - Statics - Equilibrium of Solids<br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Equilibrium of Solids</u><br /><br /><br /><br />As we know, if a vector sum of forces, acting on a point-object, equals <br />to null-vector, this point-object is at rest or in a state of uniform <br />motion along a straight line in any inertial frame of reference. <br />Alternatively, we can say that a frame of reference associated with this<br /> point-object is <i>inertial</i>.<br /><br /><br /><br />The complexity of a concept of equilibrium of solids lies in the fact <br />that, even if vector sum of forces, acting on a solid, equals to <br />null-vector, this solid might not be in the state of rest in any <br />inertial system. In a simple case of two forces, equal in magnitude and <br />opposite in direction, applied to two different points of a solid, will <br />rotate it, making a frame of reference associated with it <i>non-inertial</i>.<br /><br /><img src="http://www.unizor.com/Pictures/UnbalancedSolid.png" style="height: 200px; width: 200px;" /><br /><br /><br /><br />Any rotation of a solid, however complex, can be represented as a <br />combination of three rotations around three coordinate axes. Therefore, a<br /> solid in a state of equilibrium should be in equilibrium relative to <br />each coordinate axis, and, to study equilibrium of a solid in <br />three-dimensional space, it is sufficient to study it relative to one <br />axis.<br /><br /><br /><br />As we know, in case of rotation around an axis, the main factor replacing a concept of force for translational movement, is <i>torque</i>. Since <i>torque</i> is a cause of rotation, the necessary condition to prevent a rotation is <i>balancing of torques</i>.<br /><br /><br /><br />Torque is a vector equal to a vector (cross) product of a radius-vector <b><i><span style="text-decoration: overline;">r</span></i></b> from the axis of rotation to a point of application of force (perpendicularly to the axis) by a vector of force:<br /><br /><b><i><span style="font-size: small;"><span style="text-decoration: overline;">τ</span></span> = <span style="text-decoration: overline;">r</span><span style="font-size: medium;">×</span><span style="text-decoration: overline;">F </span></i></b><br /><br /><br /><br />Using this concept of torque, we can formulate the following necessary condition for an equilibrium of a solid.<br /><br /><b>Sum of torques of all forces acting on a solid relative to each coordinate axis must be equal to null-vector</b>.<br /><br /><br /><br />In the previous lecture we introduced a concept of a degree of freedom <br />and, in particular, mentioned six degrees of freedom of a solid: three <br />degrees of freedom along each coordinate axis and three degrees of <br />freedom of rotation around each coordinate axis.<br /><br /><br /><br />If a vector sum of all forces acting on a solid is a null-vector, there <br />will be a point that is not moving anywhere or, more precisely, a frame <br />reference associated with this point is inertial. This condition assures<br /> an equilibrium by three degrees of freedom associated with <br />translational motion (a motion <i>along</i> coordinate axes).<br /><br /><br /><br />If a vector sum of all torques relatively to each of the three <br />coordinate axes is a null-vector, there will be no rotation and the <br />system will be in equilibrium by three other degrees of freedom <br />associated with rotational motion (a motion <i>around</i> coordinate axes).<br /><br /><br /><br /><b>So, for a solid to be in equilibrium, the necessary conditions are <br />all of the above - sum of all the forces and sum of all the torques must<br /> be null-vectors</b>.<br /><br /><br /><br />Consider the following example.<br /><br />A weightless rod of a length <b><i>2r</i></b> is hanging horizontally on a thread fixed at its midpoint.<br /><br />On both ends of a rod there are equal weights <b><i>W</i></b> hanging down.<br /><br /><img src="http://www.unizor.com/Pictures/EquilibriumTorque.png" style="height: 200px; width: 200px;" /><br /><br />To balance the translational movement of a system vertically down the force of tension <b><i>T</i></b> on a thread that holds the rod must balance a sum of forces of weight acting in opposite direction:<br /><br /><b><i><span style="text-decoration: overline;">T </span> + <span style="text-decoration: overline;">W </span> + <span style="text-decoration: overline;">W </span> = <span style="text-decoration: overline;">0 </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">T </span> = −2<span style="text-decoration: overline;">W </span></i></b><br /><br /><br /><br />To balance the rotational movement of a system around a midpoint of a <br />rod the torque of one weight should be equal in magnitude to the torque <br />of another weight since the directions of these two torques are <br />opposite:<br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>blue</sub> = <span style="text-decoration: overline;">r </span>×<span style="text-decoration: overline;">W </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>red</sub> = −<span style="text-decoration: overline;">r </span>×<span style="text-decoration: overline;">W </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>red</sub> + <span style="text-decoration: overline;">τ </span><sub>blue</sub> = <span style="text-decoration: overline;">0 </span></i></b><br /><br /><br /><br />Consider a less symmetrical case below.<br /><br />Here a rod of the length <b><i>3r</i></b> is hanging by a thread fixed not in the middle, but at a distance <b><i>r</i></b> from the left edge and, therefore, at a distance <b><i>2r</i></b> from the right edge.<br /><br />At the same time, the weight on the left is twice as heavy as the one on the right.<br /><br /><img src="http://www.unizor.com/Pictures/EquilibriumTorque2.png" style="height: 200px; width: 200px;" /><br /><br />Tension on the thread holding a rod must be greater to hold more weight:<br /><br /><b><i><span style="text-decoration: overline;">T </span> + <span style="text-decoration: overline;">W </span> + 2<span style="text-decoration: overline;">W </span> = <span style="text-decoration: overline;">0 </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">T </span> = −3<span style="text-decoration: overline;">W </span></i></b><br /><br /><br /><br />From the rotational viewpoint, this system is in equilibrium because the<br /> torques on the left of a rod and on the right are equal in magnitude, <br />while opposite in direction:<br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>blue</sub> = 2<span style="text-decoration: overline;">r </span>×<span style="text-decoration: overline;">W </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>red</sub> = −<span style="text-decoration: overline;">r </span>×2<span style="text-decoration: overline;">W </span></i></b><br /><br /><b><i><span style="text-decoration: overline;">τ </span><sub>red</sub> + <span style="text-decoration: overline;">τ </span><sub>blue</sub> = <span style="text-decoration: overline;">0 </span></i></b>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0