tag:blogger.com,1999:blog-37414104180967168272018-10-30T22:24:46.166-07:00Unizor - 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.Zor Shekhtmannoreply@blogger.comBlogger348125tag: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>.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@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 />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@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 />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@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>Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-332377736873635682018-09-20T10:35:00.001-07:002018-09-20T10:35:19.798-07:00Unizor - Physics4Teens - Mechanics - Statics - Point-Objects<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>Equilibrium of Point-Objects</u><br /><br /><br /><br />The state of <i>equilibrium</i> in a case of possibility of a <br />point-object to move in some direction implies that the position of this<br /> object does not change, which, in turn, implies that the vector sum of <br />all forces acting on it is a null-vector:<br /><br /><span style="font-size: large;">Σ</span><b><i><span style="text-decoration: overline;">F<sub>i</sub></span> = <span style="text-decoration: overline;"> 0 </span></i></b>.<br /><br /><br /><br />In three-dimensional world each force <b><i><span style="text-decoration: overline;">F<sub>i</sub></span></i></b><br /> has three components - the projections of this force on X-, Y- and <br />Z-axis. In order for a sum of vectors to be equal to a null-vector, it <br />is necessary and sufficient for a sum of their X-components to be equal <br />to <b><i>0</i></b>, same for Y- and Z-components.<br /><br />Therefore, we can construct three equations, one for each axis, to determine forces participating in the state of equilibrium:<br /><br /><span style="font-size: large;">Σ</span><b><i>F<sub>i</sub><sup>(x)</sup> = 0</i></b><br /><br /><span style="font-size: large;">Σ</span><b><i>F<sub>i</sub><sup>(y)</sup> = 0</i></b><br /><br /><span style="font-size: large;">Σ</span><b><i>F<sub>i</sub><sup>(z)</sup> = 0</i></b><br /><br /><br /><br />The fact that only three equations determine the conditions on forces to<br /> be in equilibrium in a three-dimensional world means that unique <br />determination of these forces is possible only if the number of unknowns<br /> does not exceed three. If four or more unknown variables participate in<br /> the state of equilibrium, we will have infinite number of solution of <br />the system of three linear equations with four or more unknowns.<br /><br />Most likely, for our purposes we will consider situations with only unique solutions.<br /><br /><br /><br />As a simple example of an equilibrium in three-dimensional space, <br />consider a chandelier of some weight hanging from a ceiling on three <br />threads attached to a ceiling at three different points. For simplicity,<br /> assume that a chandelier is a point-object and three points of <br />attachment of the threads to a ceiling are making an equilateral <br />triangle and the length of all threads is the same.<br /><br /><br /><br />The weight of a chandelier <b><i>W</i></b> is balanced by three equal in magnitude <b><i>T</i></b>,<br /> but acting at different angles, tension forces along each thread. <br />Knowing the weight of a chandelier and an angle of each thread with a <br />vertical <b><i>φ</i></b> (the same for each thread in our simple case), we can easily determine the magnitude of the tension force of each thread:<br /><br /><b><i>3·T·cos(φ) = W</i></b><br /><br /><b><i>T = W <span style="font-size: medium;">/</span></i></b> [<b><i>3·cos(φ)</i></b>]<br /><br />Here we only constructed the equation along the vertical Z-axis because <br />in a horizontal plane (along X- and Y-axis) the tension forces balance <br />each other.<br /><br /><br /><br />The example above is a perfect illustration to our next discussion.<br /><br /><br /><br /><i>Stable Equilibrium</i><br /><br /><br /><br />We call an equilibrium <i>stable</i> if, after a small movement of an object from its position of equilibrium, it returns back to this position.<br /><br /><br /><br />A small push of a chandelier will result in its light swinging and eventual return to the original position of equilibrium.<br /><br /><br /><br />Consider a cup in a form of semi-sphere and a small ball at the bottom of this cup. It lies in the state of <i>stable equilibrium</i>. If we push it off this position, it will eventually return back because the sum <b><i>F</i></b> of reaction force <b><i>R</i></b> of the cup and the weight <b><i>W</i></b> of a ball always act in a direction of the lowest point of this semi-spherical cup.<br /><br /><img src="http://www.unizor.com/Pictures/EquilibriumStable.png" style="height: 140px; width: 200px;" /><br /><br />Let's formulate the definition of a <i>stable equilibrium</i> in more precise terms.<br /><br />For this we have to consider a position of an object in equilibrium and <br />its immediate neighborhood. Since we consider the equilibrium to be <br />stable, we have to consider the deviation of an object from a position <br />of equilibrium.<br /><br />Basically, we would like to say that the equilibrium is stable, if an <br />object returns to it from any neighboring point. How big is the <br />neighborhood, from which this return takes place is, obviously, a <br />subject of a definition.<br /><br />The rigorous definition of a <i>stable equilibrium</i> defines the neighborhood as any, however small, one.<br /><br /><br /><br />So, if exists a real positive value <b><i>ε</i></b> such that from any point within <b><i>ε</i></b>-neighborhood<br /> of a position of equilibrium an object, placed at this point, returns <br />to a position of equilibrium, the equilibrium is <i>stable</i>.<br /><br /><br /><br /><i>Unstable Equilibrium</i><br /><br /><br /><br />The <i>unstable equilibrium</i> is the opposite of a stable one. That <br />means that any, however small, deviation of the object's position from <br />the position of equilibrium results in moving of this object away from <br />the equilibrium point.<br /><br />Obvious example is a small ball on the top of a sphere. If its position <br />is exactly at the North Pole, it's in equilibrium. But any, however <br />small, deviation from the NorthPole results in complete departure from <br />the position of equilibrium.<br /><br /><br /><br /><i>Neutral Equilibrium</i><br /><br /><br /><br />Consider a ball on a horizontal plane. There are two forces acting on it<br /> - gravity vertically down and reaction of the plane vertically up and <br />equal in magnitude to the ball's weight. These two forces neutralize <br />each other, no matter where on the plane our ball is positioned. So, any<br /> position is a position of equilibrium. This is exactly what <i>neutral equilibrium</i> is. <br /><br />Any deviation from the position of <i>neutral equilibrium</i> results in the new equilibrium.<br /><br />Strictly speaking, <i>neutral equilibrium</i> implies existence of some <br />neighborhood around the position of equilibrium such as any position <br />within this neighborhood is a position of equilibrium.<br /><br /><br /><br /><i>Solids and Six Degrees of Freedom</i><br /><br /><br /><br />Before we talked about equilibrium of a point-object, that is a state of<br /> rest, when all forces applied to it are in balance and their sum is <br />null-vector.<br /><br />Let's expand our concept of equilibrium to solids. Obviously, they not <br />only can move in three different directions (forward-backward along <br />X-axis, left-right along Y-axis, up-down along Z-axis), but also can <br />rotate around any of the three coordinate axis.<br /><br />That means that a position of a solid is determined by six parameters - so called <i>six degrees of freedom</i>. Equilibrium of this solid object implies its state of rest along each of these six degrees of freedom.<br /><br /><br /><br />We know how to deal with the first <i>three degrees of freedom</i> associated with translational motion.<br /><br />Conditions of equilibrium related to rotation - the other <i>three degrees of freedom</i> - will be analyzed in the next section.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-76883770629782237612018-09-06T11:00:00.001-07:002018-09-06T11:00:23.778-07:00Unizor - Physics4Teens - Mechanics - Rotational Dynamics - Conservation ...<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>Conservation of<br />Angular Momentum</u><br /><br /><br /><br />We ended up the previous lecture deriving the Law of Conservation of <br />Angular Momentum. In this lecture we will continue discussing this <br />particular conservation law.<br /><br />We will consider a thought experiment suggested by Arkady Kokish in our private discussion of this topic.<br /><br />The results of this thought experiment will confirm the Angular Momentum Conservation Law.<br /><br /><br /><br />Consider a point-object of mass <b><i>M</i></b> with no external forces (like gravity) present and two threads, short of length <b><i>r</i></b> and long of length <b><i>R</i></b> tied to it.<br /><br />A person holds both threads and starts spinning an object with angular velocity <b><i>ω<sub>1</sub></i></b>. It will rotate along a circular trajectory of a radius <b><i>r</i></b> (the length of a shorter thread).<br /><br />Let's analyze what happens if this person lets a shorter thread go, while holding the long one.<br /><br /><img src="http://www.unizor.com/Pictures/AngularMomentum.png" style="height: 200px; width: 200px;" /><br /><br />On the picture above the short thread is let go when an object was at <br />the top. Then, since nothing holds it on a circular trajectory, it will <br />fly along a tangential line maintaining the same linear speed it had <br />when it was rotating, that is <b><i>r·ω<sub>1</sub></i></b>.<br /><br /><br /><br />But our object will not fly too far. The longer thread will stop it, when it will be on a distance <b><i>R</i></b> from a center of rotation.<br /><br />Here is a picture of this particular moment, when the long thread is fully extended.<br /><br /><img src="http://www.unizor.com/Pictures/AngularMomentumVectors.png" style="height: 162px; width: 200px;" /><br /><br />At this moment the linear speed of an object flying (on the picture) left to right equals to <b><i>V=r·ω<sub>1</sub></i></b>.<br /><br />Let's represent this vector as a sum of two vectors: <b><i>V<sub>t</sub></i></b> perpendicular to a line representing the fully extended long thread (tangential to a future circular trajectory of a radius <b><i>R</i></b>) and <b><i>V<sub>r</sub></i></b> going along that thread line (radial).<br /><br /><br /><br />Assuming an angle between a short thread at a moment it was let go and a<br /> long one at the moment it is fully extended, stopping tangential <br />movement of an object, is <b><i>φ</i></b>, it is obvious that <b><i>r=R·cos(φ)</i></b>.<br /><br /><br /><br />At the moment when the long thread is fully extended the tension of the <br />thread instantly (actually, during a very short interval of time Δ<b><i>t</i></b>) nullifies the radial speed <b><i>V<sub>r</sub></i></b>. To be exact, its moment of inertia <b><i>M·V<sub>r</sub></i></b> will be equal to an impulse of the tension force <b><i>T</i></b> during interval of time Δ<b><i>t</i></b>:<br /><br /><b><i>T·</i></b>Δ<b><i>t = M·V<sub>r</sub></i></b><br /><br />This tension force decelerates the radial speed to zero during the interval of time Δ<b><i>t</i></b>. The shorter this interval - the more ideal our experiment is.<br /><br />In an ideal case of non-stretchable thread that can withstand infinite tension the interval Δ<b><i>t</i></b> is infinitely small, and deceleration is infinitely large in magnitude and short in time.<br /><br />In this case the radial speed <b><i>V<sub>r</sub></i></b> will be brought down to zero instantaneously.<br /><br /><br /><br />At the same time the magnitude of tangential vector of speed <b><i>V<sub>t</sub></i></b> is not affected by the tension since this vector is perpendicular to a tension force.<br /><br /><br /><br />We see that the angle between vectors <b><i>V=r·ω<sub>1</sub></i></b> and <b><i>V<sub>t</sub></i></b> is the same as between threads <b><i>φ</i></b>.<br /><br /><br /><br />Therefore, <b><i>V<sub>t</sub> = r·ω<sub>1</sub>·cos(φ)</i></b><br /><br />But <b><i>cos(φ) = r/R</i></b><br /><br />from which follows<br /><br /><b><i>V<sub>t</sub> = r·ω<sub>1</sub>·r/R</i></b><br /><br />Tangential linear speed <b><i>V<sub>t</sub></i></b> is related to angular speed of rotation <b><i>ω<sub>2</sub></i></b> of our object on a new radius <b><i>R</i></b> as <b><i>V<sub>t</sub>=R·ω<sub>2</sub></i></b>.<br /><br />Hence,<br /><br /><b><i>R·ω<sub>2</sub> = r·ω<sub>1</sub>·r/R</i></b><br /><br /><b><i>R²·ω<sub>2</sub> = r²·ω<sub>1</sub></i></b><br /><br /><b><i>M·R²·ω<sub>2</sub> = M·r²·ω<sub>1</sub></i></b><br /><br />Expression <b><i>M·r²</i></b> represents the moment of inertia <b><i>I<sub>1</sub></i></b> of an object of mass <b><i>M</i></b> rotating on a radius <b><i>r</i></b>.<br /><br />Analogously, expression <b><i>M·R²</i></b> represents the moment of inertia <b><i>I<sub>2</sub></i></b> of the same object of mass <b><i>M</i></b> rotating on a radius <b><i>R</i></b>.<br /><br /><br /><br />Therefore, we have derived the Law of Conservation of the Angular Momentum:<br /><br /><b><i>I<sub>1</sub>·ω<sub>1</sub> = I<sub>2</sub>·ω<sub>2</sub></i></b><br /><br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-3512292736197087922018-08-27T14:36:00.001-07:002018-08-27T14:37:48.264-07:00Unizor - Physics4Teens - Mechanics - Rotational Dynamics - Problems<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/dbPkBE0mSbI" 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<br />Rotational Dynamics</u><br /><br /><br /><br /><i>Problem 1</i><br /><br /><br /><br />Consider a device that consists of a wheel of radius <b><i>r</i></b> and mass <b><i>M</i></b>, freely rotating on some fixed axis, a thread rolled around it a few times and a point-object of mass <b><i>m</i></b> hanging off that thread.<br /><img src="http://www.unizor.com/Pictures/TorqueProblem1.png" style="height: 200px; width: 200px;" /><br /><br />Assume that all mass of a wheel is concentrated in its rim, while spokes are weightless.<br /><br />There is no friction in the rotating wheel.<br /><br />A thread is weightless and unstrechable.<br /><br />The force of gravity pulls the object down with free fall acceleration <b><i>g</i></b>.<br /><br />What will be the acceleration of an object?<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.<br /><br />The only force acting tangentially on a wheel is the tension of a tread <b><i>T</i></b> trying to rotate it.<br /><br /><br /><br />Therefore, we can equate its momentum <b><i>τ=T·r</i></b> (<i>torque</i>) to a product of its <i>moment of inertia</i> <b><i>I</i></b> and <i>angular acceleration</i> <b><i>α</i></b>:<br /><br /><b><i>τ = T·r = I·α</i></b><br /><br /><br /><br />The fact that all mass of a wheel is concentrated in its rim allows us <br />to easily calculate its moment of inertia. We can imagine a rim <br />consisting of a large number <b><i>N</i></b> of point-objects of mass <b><i>M/N</i></b> each. The moment of inertia of each is <b><i>(M/N)·r²</i></b> and combined moment of inertia of a wheel is <b><i>N·(M/N)·r²=M·r²</i></b>. If the mass is not concentrated in the rim, analogous logic would lead us to more involved calculations of inertial mass.<br /><br /><br /><br />Considering, <b><i>I=M·r²</i></b> and <b><i>r·α=a</i></b> (where <b><i>a</i></b> is a <i>linear acceleration</i> of a thread) we can write the following equation:<br /><br /><b><i>T·r = M·r²·a/r = M·r·a</i></b><br /><br />Radius cancels out and we get<br /><br /><b><i>T = M·a</i></b><br /><br />Notice that this is exactly the same equation as the Newton's Second Law, as if an object of mass <b><i>M</i></b> is pulled along a straight line by a force <b><i>T</i></b> with linear acceleration <b><i>a</i></b>.<br /><br /><br /><br />Another important detail is that this formula is independent of a radius<br />of a wheel - a direct consequence of the fact that the mass of a wheel <br />is concentrated in its rim.<br /><br /><br /><br />Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration <b><i>a</i></b> down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,<br /><br /><b><i>m·g − T = m·a</i></b><br /><br /><br /><br />Now we have a system of two equations with two unknowns <b><i>T</i></b> and <b><i>a</i></b>, which is easy to solve.<br /><br />Substitute <b><i>T</i></b> from the first equation to the second:<br /><br /><b><i>m·g − M·a = m·a</i></b><br /><br /><b><i>m·g = (M + m) · a</i></b><br /><br /><b><i>a = m·g <span style="font-size: medium;">/</span> (M + m)</i></b><br /><br /><br /><br /><i>Problem 2</i><br /><br /><br /><br />Calculate a moment of inertia of disk of mass <b><i>M</i></b> and radius <b><i>R</i></b> rotating around an axis going through its center perpendicularly to its surface.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Divide a disk into a set of concentric rings of infinitesimal width <i>d<b>r</b></i>. Let the inner radius of a particular ring be <b><i>r</i></b> and it outer radius be <b><i>r+</i></b><i>d<b>r</b></i>.<br /><br />Moment of inertia of each ring is<br /><br /><b><i>I(r) = m·r²</i></b><br /><br />where <b><i>m</i></b> is its mass (see the previous Problem 1 for explanation).<br /><br />Mass of a ring <b><i>m</i></b> is a mass of a disk <b><i>M</i></b> multiplied by a ratio of a ring's area <b><i>2πr·</i></b><i>d<b>r</b></i> to the area of an entire disk <b><i>πR²</i></b>.<br /><br />So, the moment of inertia of our ring of radius <b><i>r</i></b> and infinitesimal width <i>d<b>r</b></i> is<br /><br /><b><i>I(r) = </i></b>[<b><i>M·2πr·</i></b><i>d<b>r/(πR²)</b></i>]<b><i>·r²</i></b><br /><br />Integrating this from <b><i>r=0</i></b> to <b><i>r=R</i></b>, we get<br /><br /><b><i>I<sub>disk</sub>(M, R) =<br /><br />= <span style="font-size: large;">∫</span></i></b>[<b><i>M·2πr·</i></b><i>d<b>r/(πR<sup>2</sup>)</b></i>]<b><i>·r<sup>2</sup> =<br /><br />= (2M/R<sup>2</sup>)<span style="font-size: large;">∫</span>r<sup>3</sup></i></b><i>d<b>r =<br /><br />= (2M/R<sup>2</sup>)·(r<sup>4</sup>/4) =<br /><br />= M·R<sup>2</sup>/2</b></i><br /><br /><br /><br /><i>Problem 3</i><br /><br /><br /><br />Consider Problem 1, but change a wheel with all mass concentrated in the<br />rim with a wheel with mass evenly distributed inside the circumference,<br />making it a disk.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />We follow the same logic as in Problem 1.<br /><br />The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.<br /><br />The only force acting tangentially on a wheel is the tension of a tread <b><i>T</i></b> trying to rotate it.<br /><br /><br /><br />Therefore, we can equate its momentum <b><i>τ=T·r</i></b> (<i>torque</i>) to a product of its <i>moment of inertia</i> <b><i>I</i></b> and <i>angular acceleration</i> <b><i>α</i></b>:<br /><br /><b><i>τ = T·r = I·α</i></b><br /><br /><br /><br />The fact that all mass of a wheel is evenly distributed within its <br />circumference allows us to easily calculate its moment of inertia using <br />the Problem 2 above:<br /><b><i>I=M·r²/2</i></b><br />Linear acceleration <b><i> a</i></b> and angular acceleration <b><i>α</i></b> are related:<br /><br /><b><i>r·α=a</i></b><br /><br />Therefore,<br /><br /><b><i>T·r = </i></b>[<b><i>M·r²/2</i></b>]<b><i>·(a/r) = M·r·a/2</i></b><br /><br />Radius cancels out and we get<br /><br /><b><i>T = M·a/2</i></b><br /><br /><br /><br />An important detail is that this formula is independent of a radius of a wheel.<br /><br /><br /><br />Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration <b><i>a</i></b> down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,<br /><br /><b><i>m·g − T = m·a</i></b><br /><br /><br /><br />Now we have a system of two equations with two unknowns <b><i>T</i></b> and <b><i>a</i></b>, which is easy to solve.<br /><br />Substitute <b><i>T</i></b> from the first equation to the second:<br /><br /><b><i>m·g − M·a/2 = m·a</i></b><br /><br /><b><i>m·g = (m + M/2) · a</i></b><br /><br /><b><i>a = m·g <span style="font-size: medium;">/</span> (m + M/2)</i></b><br /><br /><br /><br />If we compare this formula with the one in Problem 1, we see that final acceleration is greater because denominator is smaller.<br /><br />So, the disk in this problem will rotate faster than a wheel with empty middle part from Problem 1.<br /><br /><br /><br /><i>Problem 4</i><br /><br /><br /><br />Let's rotate a small ball of mass <b><i>M</i></b> within a horizontal plane on a thread of length <b><i>D</i></b>. The thread will make certain angle <b><i>φ</i></b> with the horizon. Experiments show that the angle will be smaller if a ball rotates faster.<br /><br />Determine the relationship between angular speed of rotation <b><i>ω</i></b>, mass of a ball <b><i>M</i></b>, length of a thread <b><i>D</i></b> and angle <b><i>φ</i></b>, ignoring friction and air resistance.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />The tension of a thread <b><i>T</i></b> keeps a ball on its orbit. Radius of an orbit is <b><i>R=D·cos(φ)</i></b>.<br /><br />The tension of a thread <b><i>T</i></b> serves dual purpose - its vertical component <b><i>T·sin(φ)</i></b> acts against gravity <b><i>M·g</i></b>, its horizontal component <b><i>T·cos(φ)</i></b> acts as a centripetal force and should be equal to <b><i>M·V<sup>2</sup>/R</i></b>, where <b><i>V</i></b> is a linear speed of a ball, which is equal, in turn, to <b><i>R·ω</i></b>, where <b><i>ω</i></b> is angular speed of rotation of a ball.<br /><br /><br />So, we have the following equations:<br /><b><i>T·sin(φ) = M·g</i></b><br /><br /><b><i>T·cos(φ) = M·(V<sup>2</sup>)/R = M·R·ω<sup>2</sup> = M·D·cos(φ)·ω<sup>2</sup></i></b><br /><br />The second equation is simplified to<br /><b><i>T = M·D·ω<sup>2</sup> </i></b><br /><br />Substituting it to the first equation,<br /><b><i>D·ω<sup>2</sup>·sin(φ) = g</i></b><br /><br />Therefore,<br /><b><i>sin(φ) = g/[D·ω<sup>2</sup>]</i></b><br /><br />For a given angular speed we can find an angle of a thread to horizon <b><i>φ</i></b>. It in inversely proportional to a length of a thread <b><i>D</i></b> and to a square of angular speed <b><i>ω</i></b>, which seems to be reasonable.<br /><br />Interestingly, this angle is independent of the mass <b><i>M</i></b>.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20071601754902461762018-08-21T12:00:00.001-07:002018-08-21T12:00:58.866-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Angular Momentum<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/B3cY6I4sqho" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Angular Momentum</u><br /><br /><br /><br />Recall the rotational equivalent of the Newton's Second Law:<br /><br /><b><i>τ = I·α</i></b><br /><br />where<br /><br /><b><i>τ = F·r</i></b> - a <i>torque</i>, a product of tangentially applied force <b><i>F</i></b> and the distance from the point of application of force to the axis of rotation <b><i>r</i></b>;<br /><br /><b><i>I = m·R²</i></b> - a <i>moment of inertia</i>, a product of inertial mass of a point-object by a square of its distance from the axis of rotation <b><i>R</i></b> (might be different from the distance <b><i>r</i></b> above);<br /><br /><b><i>α</i></b> - <i>angular acceleration</i> of rotation.<br /><br /><br /><br />It is very important to correspond straight line translational elements of motion with their rotational counterparts:<br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Force</b><br /><i>F</i></td><td><b>Torque</b><br /><i>τ = F·r</i></td></tr><tr><td><b>Linear Acceleration</b><br /><i>a</i></td><td><b>Angular Acceleration</b><br /><i>α = a/r</i></td></tr><tr><td><b>Inertial<br />Mass</b><br /><i>m</i></td><td><b>Moment<br />of Inertia</b><br /><i>I = m·r²</i></td></tr><tr><td><b>Newton's<br />Second Law</b><br /><i>F=m·a</i></td><td><b>Rotational<br />Equivalent</b><br /><i>τ = I·α</i></td></tr></tbody></table><br /><br />Now we will consider a rotational equivalent of a familiar equation <br />between impulse and momentum in translational motion along a straight <br />line.<br /><br /><br /><br />As is known from the previous material, there is a correspondence between these quantities: impulse of the force <b><i>F</i></b> exhorted during an infinitesimal time period <i>d<b>t</b></i> equals to an increment of the momentum <i>d<b>(m·v)</b></i>, where <b><i>m</i></b> is mass and <b><i>v</i></b> - velocity of an object:<br /><br /><b><i>F·</i></b><i>d<b>t = </b>d<b>(m·v)</b></i><br /><br /><br /><br />Let's see if the corresponding rotational characteristics of motion have similar dependency.<br /><br /><br /><br />We know the rotational equivalent of the Newton's Second Law <b><i>τ=I·α</i></b>.<br /><br />Multiplying it by <i>d<b>t</b></i>, we obtain<br /><br /><b><i>τ·</i></b><i>d<b>t = I·α·</b>d<b>t</b></i><br /><br /><br /><br />Since a product of angular acceleration <b><i>α</i></b> and infinitesimal time interval <i>d<b>t</b></i> is an infinitesimal increment of angular speed <i>d<b>ω</b></i>, there is an equality<br /><br /><b><i>τ·</i></b><i>d<b>t = I·</b>d<b>ω = </b>d<b>(I·ω)</b></i><br /><br /><br /><br />The latter fully corresponds to an equation between impulse and momentum<br /> of translational movement. It represents the relationship between <i>rotational impulse</i> <i><b>τ·</b>d<b>t</b></i> of <i>torque</i> <b><i>τ</i></b> during time interval <i>d<b>t</b></i> and increment of <i>rotational (angular) momentum</i> <i>d<b>(I·ω)</b></i> of an object with a <i>moment of inertia</i> <b><i>I</i></b> and <i>angular speed</i> <b><i>ω</i></b>.<br /><br /><br /><br />We can continue our table of correspondence between translational motion<br /> along a straight line and rotation along a circular trajectory.<br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Impulse</b><br /><i>F·d<b>t</b></i></td><td><b>Rotational Impulse</b><br /><i>τ·d<b>t</b></i></td></tr><tr><td><b>Momentum</b><br /><i>d<b>(m·v)=F·</b>d<b>t</b></i></td><td><b>Rotational Momentum</b><br /><i>d<b>(I·ω)=τ·</b>d<b>t</b></i></td></tr></tbody></table><br /><br />Trivial consequence of an equation<br /><br /><b><i>τ·</i></b><i>d<b>t = </b>d<b>(I·ω)</b></i><br /><br />is that<br /><br /><b><i>τ = </i></b><i>d<b>(I·ω)/</b>d<b>t = (I·ω)'</b></i><br /><br />From this follows that <b>if the balance of all torques acting on an object is zero, the rotational momentum remains constant</b>.<br /><br />This is the <b>Law of Conservation of the rotational (angular) momentum</b>.<br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-80733364675585339312018-08-20T13:21:00.001-07:002018-08-20T13:21:58.934-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Moment of Inertia<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/sVrlohiKqog" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Moment of Inertia</u><br /><br /><br /><br />Recall the rotational equivalent of the Newton's Second Law:<br /><br /><b><i>τ = I·α</i></b><br /><br />where<br /><br /><b><i>τ = F·r</i></b> - a <i>torque</i>, a product of tangentially applied force <b><i>F</i></b> and the distance from the point of application of force to the axis of rotation <b><i>r</i></b>;<br /><br /><b><i>I = m·R²</i></b> - a <i>moment of inertia</i>, a product of inertial mass of a point-object by a square of its distance from the axis of rotation <b><i>R</i></b> (might be different from the distance <b><i>r</i></b> above);<br /><br /><b><i>α</i></b> - <i>angular acceleration</i> of rotation.<br /><br /><br /><br />It is very important to correspond straight line translational elements of motion with their rotational counterparts:<br /><br /><br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Force</b><br /><i>F</i></td><td><b>Torque</b><br /><i>τ = F·r</i></td></tr><tr><td><b>Linear Acceleration</b><br /><i>a</i></td><td><b>Angular Acceleration</b><br /><i>α = a/r</i></td></tr><tr><td><b>Inertial<br />Mass</b><br /><i>m</i></td><td><b>Moment<br />of Inertia</b><br /><i>I = m·r²</i></td></tr><tr><td><b>Newton's<br />Second Law</b><br /><i>F=m·a</i></td><td><b>Rotational<br />Equivalent</b><br /><i>τ = I·α</i></td></tr></tbody></table><br /><br />In this lecture we will study the <i>moment of inertia</i> and its properties. We will also calculate the moment of inertia in a few simple cases.<br /><br /><br /><br />The first very important property of the <i>moment of inertia</i> is that it is <b>additive</b>.<br /> It means that the moment of inertia of two rigidly connected <br />point-objects rotating around the same axis of rotation equals to sum of<br /> their individual <i>moments of inertia</i>.<br /><br />Let's explain why.<br /><br /><br /><br />Consider a rigid weightless disk rotating around an axis going through <br />its center perpendicularly to it and two point-objects of mass <i><b>m<sub>1</sub></b></i> and <i><b>m<sub>2</sub></b></i> fixed on its surface at distances, correspondingly, <i><b>r<sub>1<sub></sub></sub></b></i> and <i><b>r<sub>2</sub></b></i>.<br /><br />Assume that this disk with both fixed on it objects is rotating with angular acceleration <b><i>α</i></b> caused by some force <b><i>F</i></b> acting on a distance <b><i>R</i></b> from its center tangentially to a circle of a radius <b><i>R</i></b> within a plane of this disk.<br /><br />The picture below illustrates this.<br /><br /><img src="http://www.unizor.com/Pictures/MomentInertia.png" style="height: 150px; width: 200px;" /><br /><br /><br /><br />Our purpose is to explain that the combined <i>moment of inertia</i> of both objects equals to a sum of their corresponding individual <i>moments of inertia</i>.<br /><br /><br /><br />The fact that object of mass <i><b>m<sub>1</sub></b></i> on a distance <i><b>r<sub>1</sub></b></i> from the axis of rotation has angular acceleration <i><b>α</b></i> implies that there is a force <i><b>f<sub>1</sub></b></i> acting on it tangentially to its trajectory such that (from rotational equivalent of the Newton's Second Law):<br /><br /><i><b>f<sub>1</sub>·r<sub>1</sub> = I<sub>1</sub>·α</b></i><br /><br />where <i><b>I<sub>1</sub>=m<sub>1</sub>·r<sub>1</sub>²</b></i> - <i>moment of inertia</i> of this object.<br /><br /><br /><br />Similarly, the fact that object of mass <i><b>m<sub>2</sub></b></i> on a distance <i><b>r<sub>2</sub></b></i> from the axis of rotation has angular acceleration <i><b>α</b></i> implies that there is a force <i><b>f<sub>2</sub></b></i> acting on it tangentially to its trajectory such that (from rotational equivalent of the Newton's Second Law):<br /><br /><i><b>f<sub>2</sub>·r<sub>2</sub> = I<sub>2</sub>·α</b></i><br /><br />where <i><b>I<sub>2</sub>=m<sub>2</sub>·r<sub>2</sub>²</b></i> - <i>moment of inertia</i> of this object.<br /><br /><br /><br />Behavior of this system depends only on parameters specified above - <br />masses, distances from the center of a disk and forces. So, if we change<br /> the initial position of masses without changing their distances from <br />the center, the system will work exactly the same.<br /><br /><br /><br />Our first modification then will be to shift the position of mass <i><b>m<sub>2</sub></b></i> along its circular trajectory of radius <i><b>r<sub>2</sub></b></i> to a position, where it is on the same radius as <i><b>m<sub>1</sub></b></i>. So, both masses and a center of a disk are on one line.<br /><br /><br /><br />Similarly, let's shift the point of application of force <b><i>F</i></b> to a point on the same line where our two objects are now located without changing the distance <b><i>R</i></b><br /> from the point of application of the force to a center. Now both masses<br /> and a point of application of main force that rotates the disk are on <br />the same line with a center of the disk, while their distances from a <br />center are still the same as before.<br /><br /><br /><br />As we know, the angular acceleration of a rotating object depends on its<br /> mass, its distance from an axis of rotation and a torque of a force <br />rotating this object.<br /><br />If force <i><b>f<sub>1</sub></b></i> applied directly to an object of mass <i><b>m<sub>1</sub></b></i> located on a distance <i><b>r<sub>1</sub></b></i> from the axis of rotation caused the angular acceleration <i><b>α</b></i>, any other force having the same torque <i><b>τ<sub>1</sub>=f<sub>1</sub>·r<sub>1</sub></b></i> would result in the same angular acceleration.<br /><br />Let's replace then the force <i><b>f<sub>1</sub></b></i> acting on a distance <i><b>r<sub>1</sub></b></i> from the axis with force <i><b>g<sub>1</sub>=f<sub>1</sub>·r<sub>1</sub><span style="font-size: medium;">/</span>R</b></i>, acting on a distance <b><i>R</i></b> at the point where our main force <b><i>F</i></b> was applied. There will be no change in angular acceleration since the torque acting on our object is the same:<br /><br /><i><b>τ'<sub>1</sub> = (f<sub>1</sub>·r<sub>1</sub><span style="font-size: medium;">/</span>R)·R = f<sub>1</sub>·r<sub>1</sub> = τ<sub>1</sub></b></i><br /><br /><br /><br />Same trick with the second object results in using force <i><b>g<sub>2</sub>=f<sub>2</sub>·r<sub>2</sub><span style="font-size: medium;">/</span>R</b></i>, acting on a distance <b><i>R</i></b> at the point where our main force <b><i>F</i></b> was applied, with no change in the behavior of the system.<br /><br /><i><b>τ'<sub>2</sub> = (f<sub>2</sub>·r<sub>2</sub><span style="font-size: medium;">/</span>R)·R = f<sub>2</sub>·r<sub>2</sub> = τ<sub>2</sub></b></i><br /><br /><br /><br />We have brought all forces (<i><b>g<sub>1</sub></b></i>, <i><b>g<sub>2</sub></b></i> and <i><b>F</b></i>)<br /> to the same point of application. They act in the same direction - <br />tangentially to a circular trajectory. We know that the effect of main <br />force <b><i>F</i></b> applied to this point is equivalent to a combined effect of forces <i><b>g<sub>1</sub>=f<sub>1</sub>·r<sub>1</sub><span style="font-size: medium;">/</span>R</b></i> and <i><b>g<sub>2</sub>=f<sub>2</sub>·r<sub>2</sub><span style="font-size: medium;">/</span>R</b></i> applied to the same point in the same direction. Therefore, sum of forces <i><b>g<sub>1</sub></b></i> and <i><b>g<sub>2</sub></b></i> must be equal to force <i><b>F</b></i>:<br /><br /><b><i>g<sub>1</sub> + g2 = F</i></b><br /><br /><br /><br />Now what remains is simple algebra.<br /><br /><i><b>f<sub>1</sub>·r<sub>1</sub><span style="font-size: medium;">/</span>R + f<sub>2</sub>·r<sub>2</sub><span style="font-size: medium;">/</span>R = F</b></i><br /><br /><i><b>f<sub>1</sub>·r<sub>1</sub> + f<sub>2</sub>·r<sub>2</sub> = F·R</b></i><br /><br /><b><i>τ<sub>1</sub> + τ<sub>2</sub> = F·R</i></b><br /><br /><b><i>I<sub>1</sub>·α + I<sub>2</sub>·α = F·R</i></b><br /><br /><b><i>(I<sub>1</sub> + I<sub>2</sub>)·α = F·R</i></b><br /><br /><br /><br />The expression on the right is the torque of the main force <b><i>F</i></b> acting tangentially to a circle of rotation on a distance <b><i>R</i></b> from the axis:<br /><br /><b><i>τ = F·R</i></b><br /><br />Therefore, the expression on the left, according to rotational equivalent of the Newton's Second Law, must be a <i>moment of inertia</i> of the entire system <b><i>I</i></b> multiplied by the system's angular acceleration <b><i>α</i></b>:<br /><br /><b><i>I·α = τ = F·R</i></b><br /><br />from which follows the equality<br /><br /><b><i>I·α = (I<sub>1</sub> + I<sub>2</sub>)·α</i></b><br /><br />and, finally,<br /><br /><b><i>I = I<sub>1</sub> + I<sub>2</sub></i></b><br /><br /><br /><br />That means that the <b>combined <i>moment of inertia</i> of two objects equals to a sum of their individual <i>moments of inertia</i></b>.<br /><br />In short, <i>moment of inertia</i> is <b>additive</b>.<br /><br />The most important consequence of this is that, to calculate a moment of<br /> inertia of a rotating solid object of some complicated geometric shape,<br /> we can divide it in small (strictly speaking, infinitesimally small) <br />pieces, calculate a moment of inertia of each piece and add all these <br />moments of inertia together (strictly speaking, integrating them) to get<br /> a moment of inertia of the whole rotating solid object.<br /><br /><br /><br />Let's calculate a moment of inertia of some solids, using this approach.<br /><br /><br /><br />1. <i>Rod rotating around its edge</i><br /><br /><br /><br />Consider a solid thin rod of a length <b><i>L</i></b> and mass <b><i>m</i></b> rotating around an axis that is perpendicular to it and going through its one end. <br /><br /><img src="http://www.unizor.com/Pictures/MomentRod.png" style="height: 150px; width: 200px;" /><br /><br />To calculate its moment of inertia, we divide it in small pieces of the length <i>d<b>r</b></i>, located on a distance <b><i>r</i></b> from the axis and having mass <i>d<b>m=(m/L)·</b>d<b>r</b></i>.<br /><br /><br /><br />Each such piece has a moment of inertia<br /><br /><i>d<b>I = (</b>d<b>m)·r² = (m/L)·r²·</b>d<b>r</b></i><br /><br /><br /><br />Since moment of inertia is additive, all we have to do now is to integrate this by <b><i>r</i></b> from <b><i>0</i></b> to <b><i>L</i></b>.<br /><br />The indefinite integral of function <b><i>r²</i></b> is <b><i>r³/3</i></b>. So, by Newton-Leibniz formula the result of integration on an interval [<b><i>0,L</i></b>] is<br /><br /><b><i>L³/3 − 0³/3 = L³/3</i></b>.<br /><br />So, the total moment of inertia of a rod is<br /><br /><i><b>I</b><sub>rod</sub><b> = (m/L)·L³/3 = m·L²/3</b></i><br /><br /><br /><br />2. <i>Rod rotating around its center</i><br /><br /><br /><br />Let's calculate the moment of inertia, if the rod rotates around its center point.<br /><br />We can consider now this rod as a system of two small half-rods, each being a half of the original one, with length <b><i>L/2</i></b> and mass <b><i>m/2</i></b>.<br /><br />Each half-rod rotates around its end, so the formula above is applicable to each half-rod.<br /><br />Its moment of inertia is<br /><br /><i><b>I</b><sub>half</sub><b> = (m/2)·(L/2)²/3 = m·L²/24</b></i><br /><br />The whole rod, being a system of two half-rods, has a momentum of <br />inertia equal to a sum of momentums of its two halves, so its value is <br />double of the above moment of inertia for each half-rod:<br /><br /><i><b>I</b><sub>whole</sub><b> = m·L²/12</b></i>Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-78669495315164055262018-08-15T13:21:00.001-07:002018-08-15T13:21:59.706-07:00Unizor - Physics4Teens - Mechanics - Rotational Dynamics<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/2TrypKxqJFc" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Torque</u><br /><br /><br /><br />So far we were mostly considering <i>translational</i> motion of point-objects - a motion along a straight line with or without external <i>forces</i><br /> acting upon this object. We have specified three Newton's Laws of this <br />motion and derived a lot of interesting facts based on these laws.<br /><br /><br /><br />Let's recall these laws.<br /><br /><br /><br />The Newton's First Law is the familiar Law of Inertia that states that <br />an object at rest stays at rest and an object in uniform motion stays in<br /> this uniform motion, unless acted upon by unbalanced forces.<br /><br /><br /><br />The Newton's Second Law brings quantitative relationship to vector of force (<b><i><span style="text-decoration: overline;">F</span></i></b>), mass (<b><i>m</i></b>) and vector of acceleration (<b><i><span style="text-decoration: overline;">a</span></i></b>):<br /><br /><b><i><span style="text-decoration: overline;">F</span> = m·<span style="text-decoration: overline;">a</span></i></b><br /><br /><br /><br />The Newton's Third Law states that for every action there is an equal in magnitude and oppositely directed reaction.<br /><br /><br /><br /><i>Rotational</i> motion obeys the rules in many respects analogous to the laws of <i>translational</i> motion, except we have to change linear movement to rotation, which, in essence, is an <i>angular</i> movement with constant radius.<br /><br /><br /><br />Consider a point-object of some mass <b><i>m</i></b> connected by a weightless rigid rod of the length <b><i>r</i></b> to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.<br /><br />The picture below illustrates such a movement and also indicates the position of <i>angular velocity <b>ω</b></i><br /> of a rotation, which was explained earlier, when we studied the <br />Kinematics of rotation (we assume your familiarity with this topic and <br />strongly recommend to refresh it before proceeding any further).<br /><br /><img src="http://www.unizor.com/Pictures/AngularVelocity.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />Let's discuss the similarities and differences between <i>translational</i> movement along a straight line and <i>rotation</i> around an axis within a <i>plane of rotation</i> perpendicular to this axis.<br /><br /><br /><br />Obviously, the <i>time</i> concept remains the same in both types of motion.<br /><br /><br /><br />There is a clear rotational analogy to the <b>Newton's First Law</b>.<br /><br /><br /><b>An object at rest stays at rest and an object in uniform rotation<br /> stays in this uniform rotation, unless acted upon by unbalanced forces.</b><br /><br /><br /><br />The <b>Newton's Third Law</b> is not really specific for a form of motion, so there is no need to address it separately at this moment.<br /><br /><br /><br />The <b>Newton's Second Law</b> requires certain modification to be applied to rotation.<br /><br /><br /><br />Let's address the main concept of Dynamics - the <i>force</i> - in connection to rotation.<br /><br /><br /><br />The very important characteristic of <i>force</i> in Dynamics of <i>translational</i> movement along a straight line is that it can be measured by its effect on objects of certain <i>inertial mass</i>. Thus, a measure of force that gives <i>linear acceleration</i> of <nobr><i>1 m/sec²</i></nobr> to an object of <i>inertial mass</i> of <nobr><i>1 kg</i></nobr> is <nobr><i>1 newton</i></nobr>. The same force applied to an object of <nobr><i>0.5 kg</i></nobr> of <i>inertial mass</i> will cause <nobr><i>2 m/sec²</i></nobr> <i>linear acceleration</i>.<br /><br />The same force applied to different objects of the same <i>inertial mass</i> will cause the same <i>linear acceleration</i> etc.<br /><br /><br /><br />The situation with <i>rotational</i> movement is not the same.<br /><br /><br /><br />Consider a simple experiment of opening a door. If you apply a force to <br />open a door at its edge opposite to hinges, where the handle is usually <br />located, it opens relatively faster than if you apply exactly the same <br />force in the middle of a door. The closer a point of application of the <br />same force to hinges - the slower a door opens, if the force applied to <br />it is the same. In an extreme case, when we apply the pressure where the<br /> hinges are, a door will not open at all.<br /><br />As a continuation of this experiment, if we want to achieve the same <br />speed of the opening of a door by applying the force at different <br />distances from the hinges, we need more efforts for a point of force <br />application located closer to the hinges.<br /><br /><br /><br />We can measure the force, the distance from the hinges of a point of application of this force and an <i>angular acceleration</i> of the door and experimentally come up with the fact that for the fixed force the <i>angular acceleration</i> of the door is proportional to the distance of a point of application of the force from the hinges.<br /><br />Moreover, leaving the point of application of the force the same and changing the force, we can determine that <i>angular acceleration</i> is proportional to the force.<br /><br /><br /><br />What it means is that in a case of rotation a force by itself does not <br />determine the final motion of a rotating object. It's a product of a <br />force <i><b>F</b></i> and radius to a point this force is applied <i><b>r</b></i>, that determines the final effect. This product <nobr><b><i>τ = F·r</i></b></nobr> is called the <i>torque</i> and it is <i>rotational</i> equivalent of a <i>force</i> in <i>translational</i> movement.<br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Force</b><br /><i>F</i></td><td><b>Torque</b><br /><i>τ = F·r</i></td></tr></tbody></table><br /><br />Returning to a picture above, we can apply some force <b><i>F</i></b> to<br /> any point on the rigid rod, connecting our point-object to an axis of <br />rotation, directing this force perpendicularly to the rod, and observe <br />that the resulting angular acceleration of the object is proportional to<br /> both the force <b><i>F</i></b> and the distance <b><i>r</i></b> from the axis to a point of application of this force, thus proportional to torque <nobr><b><i>τ = F·r</i></b></nobr>.<br /><br /><br /><br />Now we have concluded that an <i>angular acceleration</i> <b><i>α</i></b> of rotational motion is proportional to a <i>torque</i> <b><i>τ</i></b>. This is analogous to <i>linear acceleration</i> <b><i>a</i></b> of translational motion being proportional to a <i>force</i> <b><i>F</i></b>. The coefficient of proportionality for translational motion is <i>inertial mass</i> <b><i>m</i></b> of an object (this is the Newton's Second Law <b><i>F=m·a</i></b>).<br /><br /><br /><br />The obvious question is, what is the coefficient of proportionality between <i>angular acceleration</i> and <i>torque</i>?<br /><br />Answer to this question will result in rotational equivalent of the Newton's Second Law.<br /><br /><br /><br />We have experimentally established that equal torques produce equal angular accelerations.<br /><br />Consider a rotation illustrated on the picture above. Assume that the point of application of force <b><i>F</i></b> is exactly at the point-object of mass <b><i>m</i></b> rotating around an axis at a distance <b><i>r</i></b><br /> from it on a rigid weightless rod. Assume further that our force acts <br />within a plane of rotation and directed perpendicularly to the rod.<br /><br /><br /><br />During an infinitesimal time interval <i>d<b>t</b></i> the motion of an object can be considered as linear and, therefore, the Newton's Second Law can be applied, giving <b><i>F=m·a</i></b>.<br /><br />Now we can express it in terms of <i><b>angular acceleration</b></i> and <i><b>torque</b></i> as follows:<br /><br /><b><i>a = r·α</i></b><br /><br /><b><i>τ = F·r</i></b><br /><br />Hence,<br /><br /><b><i>F·r = (m·a)·r = m·r²·α</i></b><br /><br />Finally,<br /><br /><b><i>τ = (m·r²)·α</i></b><br /><br /><br /><br />One more logical step is needed. We started from a force applied on an object itself at a distance <b><i>r</i></b><br /> from an axis. But we have experimentally established that equal torques<br /> produce equal actions. It means that some other force applied to some <br />other point will produce the same effect, causing the same angular <br />acceleration, as long as the torque is the same.<br /><br />So, the equality <b><i>τ=(m·r²)·α</i></b> is universal, regardless of point of application of force since it depends not on force, but on torque.<br /><br /><br /><br />The above equality represents the rotational analogue of the Newton's Second Law.<br /><br /><br /><br />Some generalization can be applied to the above.<br /><br />What if the force, acting within a plane of rotation, is not perpendicular to a rod?<br /><br />Obvious solution is to replace vector <b><i><span style="text-decoration: overline;">F</span></i></b><br /> with its projection onto a line on the rotation plane that is <br />perpendicular to a radius and to multiply the product of two scalars <b><i>F·r</i></b> by a sine of an angle between corresponding vectors, effectively using a <b>vector product</b> <b><i><span style="text-decoration: overline;">F</span></i></b>⨯<b><i><span style="text-decoration: overline;">r</span></i></b>.<br /><br />So, more general definition of a <i>torque</i> is a vector (or, more precisely, pseudo-vector)<br /><br /><b><i><span style="text-decoration: overline;">τ</span> = <span style="text-decoration: overline;">F</span></i></b>⨯<b><i><span style="text-decoration: overline;">r</span></i></b><br /><br />whose direction, like a direction of an <i>angular acceleration</i>, is along an axis of rotation.<br /><br />So, for rotational movement the vectors of <i>torque</i> <b><i><span style="text-decoration: overline;">τ</span></i></b> and <i>angular acceleration</i> <b><i><span style="text-decoration: overline;">α</span></i></b> are collinear, similarly to collinearity of vectors of <i>force</i> <b><i><span style="text-decoration: overline;">F</span></i></b> and <i>linear acceleration</i> <b><i><span style="text-decoration: overline;">a</span></i></b> for translational movement.<br /><br /><br /><br />We'd like to note that for purposes of simplicity in this course we will<br /> rarely deal with forces not perpendicular to a radius of rotation.<br /><br /><br /><br />IMPORTANT TERMINOLOGY POINTS<br /><br /><br /><br />1. The torque <b><i>τ</i></b> is often called the <i>moment of force</i>.<br /><br /><br /><br />2. Recall the expression tying together a torque <b><i>τ</i></b> and an angular acceleration <b><i>α</i></b><br /><br /><b><i>τ = (m·r²)·α</i></b><br /><br />The expression <b><i>m·r²</i></b> is called <i>moment of inertia</i>, is symbolized by letter <b><i>I</i></b>, which allows to specify the formula above as<br /><br /><b><i>τ = I·α</i></b><br /><br />where <b><i>I=m·r²</i></b> is a <i>moment of inertia</i>, playing the same role in this equation as <i>inertial mass</i> <b><i>m</i></b> in the Newton's Second Law and representing resistance to a rotational force.<br /><br /><br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Force</b><br /><i>F</i></td><td><b>Torque</b><br /><i>τ = F·r</i></td></tr><tr><td><b>Acceleration</b><br /><i>a</i></td><td><b>Angular Acc.</b><br /><i>α = a/r</i></td></tr><tr><td><b>Inertial<br />Mass</b><br /><i>m</i></td><td><b>Moment<br />of Inertia</b><br /><i>I = m·r²</i></td></tr><tr><td><b>Newton's<br />Second Law</b><br /><i>F=m·</i></td><td><b>Rotational<br />Equivalent</b><br /><i>τ = I·α</i></td></tr></tbody></table>Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-27375773765727996672018-08-13T12:46:00.001-07:002018-08-13T12:46:27.467-07:00Unizor - Physics4Teens - Mechanics - Rotational Kinematics<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/9c0VeYZmTuw" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Rotational Kinematics</u><br /><br /><br /><br />So far we were mostly considering <i>translational</i> motion of point-objects - a motion along a straight line with or without external <i>forces</i> acting upon this object.<br /><br /><br /><br /><i>Rotational</i> motion obeys the rules in many respects analogous to the laws of <i>translational</i> motion, except we have to change linear movement to rotation.<br /><br /><br /><br />Consider a point-object <b><i>m</i></b> connected by a rigid rod of the length <b><i>r</i></b> to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.<br /><br />The picture below illustrates such a movement and also indicates the position of <i>angular velocity <b>ω</b></i> of a rotation, which we will explain later.<br /><br /><img src="http://www.unizor.com/Pictures/AngularVelocity.png" style="height: 120px; width: 200px;" /><br /><br /><br /><br />Let's discuss the similarities and differences between <i>translational</i> movement along a straight line and <i>rotation</i> around an axis within a plane perpendicular to this axis (a <i>plane of rotation</i>).<br /><br /><br /><br />The first main concept of translational motion is <i>position</i> or <i>distance from the beginning of motion</i> (for a straight line movement) as a function of time. In <i>rotational</i> motion its equivalent is <i>angle of rotation</i> from some original position as a function of time.<br /><br /><br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Distance</b><br /><i>s(t)</i></td><td><b>Angle</b><br /><i>φ(t)</i></td></tr></tbody></table><br /><br />The next concept is <i>speed</i> or (better) <i>velocity</i> of translational motion. This is a first derivative of <i>position</i> (or <i>distance</i>) by time:<br /><br /><i>v(t) = s'(t)</i>.<br /><br />Its equivalent for rotational motion is <i>angular speed</i>, which is a first derivative of <i>angle of rotation</i> by time:<br /><br /><i>ω(t) = φ'(t)</i><br /><br /><br /><br />While vector character of <i>speed</i> of translational motion is obvious and is reflected in the term <i>velocity</i>, vector character of <i>angular speed</i> is less obvious.<br /><br />The <i>angle of rotation</i> from the first glance is a scalar function of time. But only from the first glance.<br /><br /><br /><br />In theory, rotational motion always assumes existence of an axis of <br />rotation and a plane of rotation. To reflect these characteristics and a<br /> magnitude of angular speed, an angular speed is represented by a vector<br /> from a center of rotation along an axis of rotation perpendicularly to a<br /> plane of rotation with a magnitude equal to a value of angular speed.<br /><br /><br /><br />This allows to represent the rotation in its full spectrum of <br />characteristics - magnitude, axis, plane of rotation. The picture above <br />represents angular speed as a vector <i><b>ω(t)</b></i>, which we may call <i>angular velocity</i> vector.<br /><br /><br /><br />There is one more characteristic of rotational motion not yet discussed - its direction. It is also reflected in <i>angular velocity</i><br /> as a vector by its direction. In theory, we can choose two different <br />directions along the axis of rotation. The direction chosen is such <br />that, if we look from its end onto a plane of rotation, the rotation is <br />counterclockwise. Another interpretation of this is the "rule of the <br />right hand" because if you put you right hand on a plane of rotation <br />such that your finger go around the axis of rotation pointing to a <br />direction of rotation, your thumb points to a direction of the <i>angular velocity</i> vector.<br /><br />So, <i>angular velocity</i> vector represents axis, plane, direction of rotation as well as magnitude of <i>angular speed</i>.<br /><br /><br /><br />To be more precise, since this vector representation of <i>angular velocity</i> is a little unusual, it is customary to call it "pseudo-vector" instead of "vector".<br /><br /><br /><br />During infinitesimal time interval <i>d<b>t</b></i> an object rotating around an axis on a radius <i><b>r</b></i> turns by an angle <i>d<b>φ(t)</b></i>, covering the distance <i>d<b>s(t)=r</b>·d<b>φ(t)</b></i> (this is the length of an arc of radius <i>r</i> and angle <i>dφ</i>, according to a known formula of geometry).<br /><br />From this follows:<br /><br /><i>d<b>s(t)/</b>d<b>t = r</b>·d<b>φ(t)/</b>d<b>t</b></i> or<br /><br /><i><b>v(t) = r·ω(t)</b></i><br /><br /><br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Speed</b><br /><i>v(t)=s'(t)<br />v(t)=r·ω(t)</i></td><td><b>Angular Speed</b><br /><i>ω(t)=φ'(t)<br />ω(t)=v(t)/r</i></td></tr></tbody></table><br /><br />The next concept is <i>acceleration</i> that needs its rotational analogue. Obviously, it's the first derivative of <i>angular velocity</i> or the second derivative of an <i>angle of rotation</i> by time.<br /><br />Using the vector interpretation of <i>angular velocity</i>, we can consider <i>angular acceleration</i> as a vector as well. It is also directed along the axis of rotation.<br /><br />During infinitesimal time interval <i>d<b>t</b></i> an <i>angular velocity</i> <i><b>ω(t)</b></i> changes by <i>d<b>ω(t)</b></i>.<br /><br />From this follows relationship between <i>linear acceleration</i> <i><b>a</b></i> and <i>angular acceleration</i> <i><b>α</b></i>:<br /><br /><i><b>a(t) = </b>d<b>v(t)/</b>d<b>t = r</b>·d<b>ω(t)/</b>d<b>t</b></i> or<br /><br /><i><b>a(t) = r·α(t)</b></i><br /><br /><br /><br /><table border="1"><tbody><tr><td><i>Translation</i></td><td><i>Rotation</i></td></tr><tr><td><b>Acceleration</b><br /><i>a(t)=v'(t)<br />a(t)=r·α(t)</i></td><td><b>Angular Acc.</b><br /><i>α(t)=ω'(t)<br />α(t)=a(t)/r</i></td></tr></tbody></table><br /><br />Obviously, integrating the definitions of <i>angular velocity</i> <b><i>ω</i></b> and <i>angular acceleration</i> <b><i>α</i></b> for <u>constant</u> <i>angular acceleration</i>, we come up with formulas similar to those familiar from <i>translational</i> movement:<br /><br /><b><i>ω(t) = ω(0) + α·t</i></b><br /><br /><b><i>φ(t) = φ(0) + ω(0)·t + α·t²/2</i></b><br /><br /><br /><br /><i>Angular acceleration</i> as a vector (as a pseudo-vector, to be exact), is colinear to the axis of rotation, because <i>angular velocity</i> is.<br /><br />If rotation goes as on the above picture and the speed of rotation increases, the <i>angular acceleration</i> would be directed upwards, the same way as the <i>angular velocity</i>.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-60227063500202532542018-08-06T11:58:00.001-07:002018-08-06T11:58:11.369-07:00Unizor - Physics4Teens - Mechanics - Statics - Equilibrium<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/bubkT2py5yE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Equilibrium</u><br /><br /><br /><br /><i>Statics</i> is a part of Mechanics that studies forces not as <br />quantitative measure of a motion they cause (that is a subject of <br />Dynamics), but from the more fundamental viewpoint of whether these <br />forces are or are not balanced, that might or might not cause the motion<br /> of objects these forces act upon.<br /><br /><br /><br />Historically, studies of static aspects of forces precede quantitative <br />studies of motion in Dynamics. People, first of all, were concerned with<br /> how to build bridges and buildings, so that they stay in place and not <br />destroyed by gravity, wind or load. And a concept of <i>equilibrium</i> is a central point of Statics, it's goal and purpose.<br /><br /><br /><br /><b><i>Equilibrium</i>, as it is understood in Statics, is a state of forces, that result in an object acted upon to be in the state of rest.</b><br /><br /><br /><br />A person standing on a floor is at rest because two main forces that act<br /> upon him, the gravity and the reaction of the floor are equal in <br />magnitude and opposite in direction, they balance each other, which <br />results in a state of <i>equilibrium</i>.<br /><br /><br /><br />A person standing on the weights to check his weight is in a state of <i>equilibrium</i><br /> because its weight is balanced by elastic force inside the weights that<br /> is equal in magnitude and opposite in direction to the force of <br />gravity. The equality in magnitude allows to measure the weight by <br />measuring the elasticity inside the weights.<br /><br /><br /><br />A building is in a state of <i>equilibrium</i> because its each part's weight is balanced by equal in magnitude and opposite in direction force of reaction.<br /><br /><br /><br />An airplane, flying horizontally on its route, is in a state of <i>vertical equilibrium</i><br /> because its weight is balanced by the lifting power of the air under <br />its wings, where the air pressure is larger because of the wings' shape.<br /><br /><br /><br />Let's study forces applied to the same object and balance each other causing this object to be in <i>equilibrium</i>.<br /><br />As we know, force is a vector. <b>All forces acting on the same point-object can be added as vectors</b>, using the rules of the Vector Algebra.<br /><br /><b>If the result is a null-vector, we have an <i>equilibrium</i>.</b><br /><br /><br /><br />So, if there are <i>N</i> forces <b><i><span style="text-decoration: overline;">F<sub>i</sub></span></i></b> (where <nobr><i>1 ≤ <b>i</b> ≤ N</i>)</nobr> acting simultaneously on the same point-object, the <b>condition of equilibrium</b> is<br /><br /><span style="font-size: large;">Σ</span><b><i><span style="text-decoration: overline;">F<sub>i</sub></span> = <span style="text-decoration: overline;"> 0 </span></i></b>.<br /><br /><br /><br />Let's consider a few examples.<br /><br /><br /><br />1. An object of weight <b><i>W</i></b> is hanging on three threads as pictured below.<br /><br /><img src="http://www.unizor.com/Pictures/Equilibrium_1.png" style="height: 180px; width: 200px;" /><br /><br />Thread <b><i>a</i></b> is horizontal, thread <b><i>b</i></b> is at angle <b><i>φ</i></b> to horizon.<br /><br />The system is in equilibrium. <br /><br />What is the magnitude of tension forces <b><i>T<sub>a</sub></i></b> and <b><i>T<sub>b</sub></i></b> on threads <b><i>a</i></b> and <b><i>b</i></b>?<br /><br /><br /><br /><i>Solution</i><br /><br />There are three forces acting at the point where all threads are connected:<br /><br />vertical down - the weight <b><i>W</i></b> of the object;<br /><br />tension <b><i>T<sub>a</sub></i></b> horizontally to the left along the <b><i>a</i></b> thread;<br /><br />tension <b><i>T<sub>b</sub></i></b> at angle <b><i>φ</i></b> to horizon along the <b><i>b</i></b> thread.<br /><br />Since the system is in equilibrium, this point where threads come <br />together does not move in any direction. In particular, it does not move<br /> in a vertical, nor horizontal direction.<br /><br />This is sufficient to determine magnitude of tension vectors <b><i>T<sub>a</sub></i></b> and <b><i>T<sub>b</sub></i></b>.<br /><br />In the horizontal direction the force of gravity is irrelevant, so the <br />only two forces acting on out object in horizontal direction are tension<br /> <b><i>T<sub>a</sub></i></b> acting to the left and a projection of tension <b><i>T<sub>b</sub></i></b> on the horizontal line acting to the right.<br /><br />That gives the first equation about magnitudes of tension forces:<br /><br /><b><i>T<sub>a</sub> = T<sub>b</sub>·cos(φ)</i></b><br /><br />Now consider the vertical direction. The thread <b><i>a</i></b> is irrelevant. So, the only two forces acting vertically, are weight <b><i>W</i></b> pulling down and a vertical component of <b><i>T<sub>b</sub></i></b> that is equal to <b><i>T<sub>b</sub>·sin(φ)</i></b> pulling upwards.<br /><br />This gives the second equation<br /><br /><b><i>T<sub>b</sub>·sin(φ) = W</i></b><br /><br />From these equations we derive:<br /><b><i>T<sub>b</sub> = W/sin(φ) </i></b><br /><br />After which we can find <b><i>T<sub>a</sub></i></b>:<br /><br /><b><i>T<sub>a</sub> = W·cos(φ)/sin(φ)</i></b><br /><br /><br /><br />2. Two objects of mass <b><i>M</i></b> (larger, on an inclined plane) and <b><i>m</i></b> (smaller, hanging freely over the edge of an inclined plane) are connected with a thread that goes over a pulley.<br /><br /><img src="http://www.unizor.com/Pictures/Equilibrium_2.png" style="height: 100px; width: 200px;" /><br /><br />What is the angle of an inclined plane <b><i>φ</i></b> for this system to be in equilibrium?<br /><br />Ignore the friction.<br /><br /><br /><br /><i>Solution</i><br /><br />Let's represent the weight of an object on an inclined plane as a sum of two forces:<br /><br />one is perpendicular to the plane and balanced by a plane's reaction;<br /><br />another is parallel to the plane and balanced by a tension of a thread.<br /><br />To be in equilibrium, the force of weight of an object hanging freely <br />over the edge of an inclined plane must be equal in magnitude to a <br />tension of a thread (same thread, same tension as before).<br /><br />Therefore, a component of the weight of an object on a plane that goes <br />parallel to a plane must be equal in magnitude to a weight of another <br />object.<br /><br /><b><i>M·g·sin(φ) = m·g</i></b><br /><br />From this we derive<br /><br /><b><i>sin(φ) = m/M</i></b><br /><br /><b><i>φ = arcsin(m/M)</i></b><br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-5822486693412560442018-07-30T13:52:00.001-07:002018-07-30T13:52:53.262-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Rockets and Gravitation<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/7feRsZh5-Qg" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Rocket in Gravitational Field</u><br /><br /><br /><br />In the previous lectures we examined the motion of a rocket with no <br />external forces (like gravity or drag) acting on it and came with the <br />Rocket Equation, stating that<br /><br />IF<br /><br /><b><i>m(t)</i></b> is the mass of a rocket (including propellant) at time <b><i>t</i></b> and<br /><br /><b><i>V(t)</i></b> is its speed in some inertial reference frame <br />(related to stars, for example, and positioned in such a way that a <br />rocket moves along one axis in a positive direction) and<br /><br /><b><i>m(t+</i></b>Δ<b><i>t)</i></b> is the mass after time interval Δ<b><i>t</i></b>, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed <b><i>v<sub>e</sub></i></b> and <br /><br /><b><i>V(t+</i></b>Δ<b><i>t)</i></b> is its speed after time interval Δ<b><i>t</i></b> in the same inertial reference frame<br /><br />THEN<br /><br />the maximum increment of the rocket's speed<br /><br />Δ<b><i>V=V(t+</i></b>Δ<b><i>t)−V(t)</i></b> during this interval of time Δ<b><i>t</i></b> is<br /><br />Δ<b><i>V = −v<sub>e</sub></i></b>·<i>ln</i>[<b><i>m(t)<span style="font-size: medium;">/</span>m(t+</i></b>Δ<b><i>t)</i></b>]<br /><br /><br /><br />The equation above should be interpreted as the vector equation.<br /><br />If inertial frame of reference is directed in such a way that the rocket<br /> moves along one axis in positive direction and the exhaust is directed <br />backwards relative to a rocket's movement, the <b><i>v<sub>e</sub></i></b> is negative. The mass during this process decreases, so <b><i>m(t)</i></b> is greater than <b><i>m(t+</i></b>Δ<b><i>t)</i></b> and the logarithm is positive. This results in the positive Δ<b><i>V</i></b>, that is a rocket accelerates.<br /><br />If the exhaust is directed forward relative to a rocket's movement, the <b><i>v<sub>e</sub></i></b> is positive, Δ<b><i>V</i></b> is negative and a rocket decelerates.<br /><br /><br /><br />Now let's add gravity as an external force that acts on a rocket.<br /><br />There are two cases:<br /><br />(a) when a rocket is launched from a planet to an orbit, gravity acts <br />against its movement, thus requiring extra effort by an engine to <br />overcome the gravity;<br /><br />(b) when a rocket returns back to a planet for soft landing, gravity <br />acts in the direction of its movement, but we have to decelerate a <br />rocket using propellant exhausted also in the same direction, so it also<br /> requires extra effort by an engine.<br /><br />So, no matter how rocket moves in the gravitational field, an engine <br />should work harder to either launch it to an orbit or to slow it down <br />for soft landing.<br /><br /><br /><br />We will consider the launching from the Earth case only.<br /><br /><br /><br />Let's follow the same logic as in case of a rocket moving in empty space<br /> with no forces involved and add the effect of gravity in the equation <br />of conservation of momentum.<br /><br /><br /><br />1. At moment <b><i>t</i></b> the momentum of an entire system of a rocket with its propellant was equal to <b><i>m(t)·V(t)</i></b>.<br /><br /><br /><br />2. During the next time interval <b><i></i></b><i>d<b>t</b></i> the rocket has exhausted <b><i>m(t)−m(t+</i></b><i>d<b>t)=−</b>d<b>m(t)</b></i> of propellant with constant relatively to a rocket speed <b><i>v<sub>e</sub></i></b>. Since a rocket moves in some inertial system with speed <b><i>V(t)</i></b> and propellant moves relatively to a rocket with constant speed <b><i>v<sub>e</sub></i></b>, the speed of propellant in the inertial system equals to <b><i>v<sub>e</sub>+V(t)</i></b>. This resulted in the momentum of exhausted propellant at moment <b><i>t+</i></b><i>d<b>t</b></i> to be <b><i>−</i></b><i>d<b>m(t)</b></i>·[<b><i>v<sub>e</sub>+V(t)</i></b>].<br /><br />This equation should be interpreted in the vector form. When a rocket <br />accelerates, velocity vector of its movement and velocity vector of <br />exhausted propellant are opposite in their directions.<br /><br /><br /><br />3. A rocket with remaining propellant at moment <b><i>t+</i></b><i>d<b>t</b></i> has mass <b><i>m(t+</i></b><i>d<b>t)=m(t)+</b>d<b>m(t)</b></i>, velocity <i><b>V(t+</b>d<b>t)=V(t)+</b>d<b>V(t)</b></i> and momentum<br /><br />[<b><i>m(t)+</i></b><i>d<b>m(t)</b></i>]·[<i><b>V(t)+</b>d<b>V(t)</b></i>]<br /><br /><br /><br />4. When rocket leaves the planet, the force of gravity <b><i>F=m(t)·g</i></b> acts against its movement. During time <i>d<b>t</b></i> it reduces the impulse of a rocket by<br /><br /><i><b>F·</b>d<b>t = m(t)·g·</b>d<b>t</b></i>.<br /><br /><br /><br />Now we are ready to apply the Law of Conservation of Momentum.<br /><br />Item 1 above describes the momentum of the system at time <b><i>t</i></b>.<br /><br />At the moment <i><b>t+</b>d<b>t</b></i> the momentum of the system is a combination of the momentum of the exhausted propellant during time <i>d<b>t</b></i><br /> (see item 2 above) plus the momentum of the remaining rocket mass (see <br />item 3 above) plus impulse of the gravitational force (see item 4 <br />above).<br /><br /><br /><br />Equalizing these two momentums at time <i><b>t</b></i> and <i><b>t</b>+d<b>t</b></i>, according to the Law of Conservation of Momentum, we get the following equation:<br /><br /><b><i>m(t)·V(t) = −</i></b><i>d<b>m(t)</b></i>·[<b><i>v<sub>e</sub>+V(t)</i></b>] <b>+</b> [<b><i>m(t)+</i></b><i>d<b>m(t)</b></i>]·[<i><b>V(t)+</b>d<b>V(t)</b></i>]<i><b> + m(t)·g·</b>d<b>t</b></i>.<br /><br /><br /><br />We would like to express the dependency between rocket's speed and the <br />way it exhausts its propellant without mentioning the time parameter. <br />This can be done by using the following:<br /><br /><i><b>m'(t) = </b>d<b>m(t)/</b>d<b>t</b></i> (by definition of the derivative and differential)<br /><br />From this:<br /><br /><i>d<b>t = </b>d<b>m(t)/m'(t)</b></i><br /><br /><br /><br />The rocket equation above can be simplified. After this the equation looks like<br /><br /><b><i>0 = −v<sub>e</sub>·</i></b><i>d<b>m(t) + m(t)·</b>d<b>V(t) + </b>d<b>V(t)·</b>d<b>m(t) + m(t)·g·</b>d<b>t</b></i><br /><br />Ignoring an infinitesimal of a higher order <i>d<b>V(t)·</b>d<b>m(t)</b></i>, the resulting equation looks like<br /><br /><i><b>m(t)·</b>d<b>V(t)+m(t)·g·</b>d<b>t=v<sub>e</sub>·</b>d<b>m(t)</b></i><br /><br /><br /><br />Divide both parts by <i><b>m(t)</b></i> and take into consideration that <i>d<b>m(t)<span style="font-size: medium;">/</span>m(t) = </b>d</i>[<i>ln(<b>m(t)</b>)</i>]. Then the differential equation of a rocket in the gravitational field looks like<br /><br /><i>d<b>V(t) + g·</b>d<b>t = v<sub>e</sub>·</b>d</i>[<i>ln(<b>m(t)</b>)</i>]<br /><br />Replacing <i>d<b>t</b></i> with its equivalent <i>d<b>m(t)/m'(t)</b></i>, we obtain an equivalent equation<br /><br /><i>d<b>V(t)+g·</b>d<b>m(t)/m'(t) = v<sub>e</sub>·</b>d<b>m(t)</b></i><br /><br /><br /><br />Expression on the right is positive because<br /><br />(a) <i><b>m(t)</b></i> is a decreasing function,<br /><br />(b) <i>ln<b>(m(t))</b></i>, therefore, is also a decreasing function,<br /><br />(c) differential of a decreasing function <i>d</i>[<i>ln(<b>m(t)</b>)</i>] is always negative,<br /><br />(d) <i><b>v<sub>e</sub></b></i> is negative since it is a vector <br />directed against the movement of a rocket, which we consider as moving <br />to a positive direction of a coordinate axis,<br /><br />(e) product of two negative values is positive.<br /><br /><br /><br />As is obvious from this equation, unless <i><b>g·</b>d<b>t</b></i> is less than <i><b>v<sub>e</sub>·</b>d</i>[<i>ln(<b>m(t)</b>)</i>], the rocket will not move from the launching pad.<br /><br />We can simplify this launching condition, using the following:<br /><br /><i>d</i>[<i>ln(<b>m(t)</b>)</i>]<i><b> = </b></i>[<i><b>m'(t)/m(t)</b></i>]·<i>d<b>t</b></i><br /><br />This allows to express this condition as<br /><br /><i><b>g</b></i> is less than <i><b>v<sub>e</sub>·</b></i>[<i><b>m'(t)/m(t)</b></i>]<br />or<br /><br /><i><b>v<sub>e</sub>·m'(t)</b></i> is greater than <i><b>g·m(t)</b></i><br /><br /><br /><br />Integrating the differential equation of a rocket in the gravitational field on the interval Δ<b><i>t</i></b> of time from the beginning of engine's work <i><b>t<sub>beg</sub></b></i> to the end of this period <i><b>t<sub>end</sub></b></i>, we obtain the equation for an increment of the rocket's speed during this interval:<br /><br /><i><b>V(t<sub>end</sub>)−V(t<sub>beg</sub>)+g·(t<sub>end</sub>−t<sub>end</sub>)=<br /><br />= v<sub>e</sub>·</b></i>[<i>ln(<b>m(t<sub>end</sub>)</b>)−ln(<b>m(t<sub>beg</sub>)</b>)</i>]<b><i>=<br /><br />= v<sub>e</sub>·</i></b><i>ln</i>[<i><b>m(t<sub>end</sub>)<span style="font-size: medium;">/</span>m(t<sub>beg</sub>)</b></i>]<b><i>=<br /><br />= −v<sub>e</sub>·</i></b><i>ln</i>[<i><b>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</b></i>]<br /><br /><br /><br />That is,<br /><br />Δ<i><b>V(t) = −v<sub>e</sub>·</b>ln</i>[<i><b>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</b></i>]<i><b> − g·(t<sub>end</sub>−t<sub>end</sub>)</b></i><br /><br /><br /><br />The last equation does not take into consideration that the force of <br />gravity decreases with height. It's relatively precise only in the <br />beginning of the rocket's movement. Obviously, taking this factor into <br />consideration will complicate the calculations.<br /><br /><br /><br />The next complication is the drag of the atmosphere, which is not that <br />important in theory, but for practical matters must be taken into <br />consideration.<br /><br /><br /><br />Another important factor of launching is the planet's rotation. If we <br />launch a rocket to the East, the Earth's rotation helps to achieve <br />required speed.<br /><br /><br /><br />All these and other complications make rocket science a rather involved theory.<br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-54065851708879509972018-07-27T13:58:00.001-07:002018-07-27T13:58:59.974-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Calculations<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/Gf3mTxnexGo" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Rocket Calculation 1</u><br /><br /><br /><br />Let's use the rocket equation<br /><br />Δ<b><i>V = −v<sub>e</sub></i></b>·<i>ln</i>[<b><i>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</i></b>]<br /><br />to calculate how much propellant must be taken by a rocket to reach an orbit.<br /><br /><br /><br />Here <b><i>v<sub>e</sub></i></b> is effective exhaust speed, <b><i>m(t<sub>beg</sub>)</i></b> - mass of a rocket in the beginning of a time period during which a rocket's engine is working, <b><i>m(t<sub>end</sub>)</i></b> - mass of a rocket at the end of this period of acceleration or deceleration.<br /><br /><br /><br />Recall that the minus sign in this equation signifies the vector <br />character of the movement: positive direction of the exhausted <br />propellant (that is, the same as the movement of the rocket) causes <br />negative increment in rocket's speed - deceleration, while the negative <br />direction of exhausted propellant (that is, opposite to the movement of a<br /> rocket) causes increase in rocket's speed - acceleration.<br /><br /><br /><br />Contemporary rocket engine can have a very high effective exhaust <br />velocity. The speed of about 4km/sec is mentioned in a few sources we <br />are familiar with. So, we can assume that<br /><br /><b><i>v<sub>e</sub>=4000m/sec.</i></b><br /><br /><br /><br />An international Space Station's speed is about 7.8km/sec, as was calculated in one of the previous lectures on gravity.<br /><br />Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be<br /><br />Δ<b><i>V = 7800m/sec</i></b><br /><br /><br /><br />From this follows that<br /><br /><i>ln</i>[<b><i>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</i></b>]<b><i> =<br /><br />= 7800/4000 = 1.95</i></b><br /><br /><br /><br />Therefore,<br /><br /><b><i>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>) ≅ 7 </i></b><br /><br /><br /><br />So, the mass of a rocket at start is 7 times greater than its mass at <br />the end of its acceleration. For example, to launch 1,000 kg of useful <br />equipment and/or passengers to an International Space Station we will <br />need 6,000 kg of fuel.<br /><br /><br /><br /><br /><br /><u>Rocket Calculation 2</u><br /><br /><br /><br />We still assume that<br /><br /><b><i>v<sub>e</sub>=4000m/sec.</i></b><br /><br /><br /><br />A rocket that goes far from Earth needs about 11.2km/sec speed to escape Earth gravity.<br /><br />Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be<br /><br />Δ<b><i>V = 11200m/sec</i></b><br /><br /><br /><br />From this follows that<br /><br /><i>ln</i>[<b><i>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</i></b>]<b><i> =<br /><br />= 11200/4000 = 2.8</i></b><br /><br /><br /><br />Therefore,<br /><br /><b><i>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>) ≅ 16 </i></b><br /><br /><br /><br />So, the mass of a rocket, that is supposed to leave the Earth's gravity,<br /> at start is 16 times greater than its mass at the end of its <br />acceleration. For example, to launch 1,000 kg of useful equipment and/or<br /> passengers to Mars we will need 15,000 kg of fuel.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-37334166410783422892018-07-25T11:42:00.001-07:002018-07-25T11:42:39.163-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Equation<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/iL_FH8mVNdI" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Ideal Rocket Equation</u><br /><br />(Tsiolkovsky's Equation)<br /><br /><br /><br />The most important difference between the motion of a rocket and motions<br /> analyzed before in this course is that the rocket propels itself by <br />throwing back part of its own mass (propellant), thus becoming lighter. <br />Its mass is variable. Before, in most cases, we were dealing with motion<br /> of objects of some specific mass, not changing during the movement.<br /><br /><br /><br />In this lecture we will analyze the motion of an ideal rocket that <br />throws back propellant with constant (relatively to the rocket) speed. <br />The formula that will be derived was suggested by Russian mathematician <br />K.E.Tsiolkovsky and is historically named after him, though he was not <br />the first to derive it.<br /><br /><br /><br />In particular, we will analyze the dependency between loss of the total <br />mass of a rocket, throwing propellant backward in the absence of any <br />external forces, and its gain in speed caused by this process.<br /><br /><br /><br />The final formula we will derive states that<br /><br />IF<br /><br /><b><i>m(t)</i></b> is the mass of a rocket (including propellant) at time <b><i>t</i></b> and<br /><br /><b><i>V(t)</i></b> is its speed in some inertial reference frame <br />(related to stars, for example, and positioned in such a way that a <br />rocket moves along one axis in a positive direction) and<br /><br /><b><i>m(t+</i></b>Δ<b><i>t)</i></b> is the mass after time interval Δ<b><i>t</i></b>, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed <b><i>v<sub>e</sub></i></b> and <br /><br /><b><i>V(t+</i></b>Δ<b><i>t)</i></b> is its speed after time interval Δ<b><i>t</i></b> in the same inertial reference frame<br /><br />THEN<br /><br />the maximum increment of the rocket's speed<br /><br />Δ<b><i>V=V(t+</i></b>Δ<b><i>t)−V(t)</i></b> during this interval of time Δ<b><i>t</i></b> is<br /><br />Δ<b><i>V = −v<sub>e</sub></i></b>·<i>ln</i>[<b><i>m(t)<span style="font-size: medium;">/</span>m(t+</i></b>Δ<b><i>t)</i></b>]<br /><br /><br /><br />The equation above should be interpreted as the vector equation.<br /><br />If inertial frame of reference is directed in such a way that the rocket<br /> moves along one axis in positive direction and the exhaust is directed <br />backwards relative to a rocket's movement, the <b><i>v<sub>e</sub></i></b> is negative. The mass during this process decreases, so <b><i>m(t)</i></b> is greater than <b><i>m(t+</i></b>Δ<b><i>t)</i></b> and the logarithm is positive. This results in the positive Δ<b><i>V</i></b>, that is a rocket accelerates.<br /><br />If the exhaust is directed forward relative to a rocket's movement, the <b><i>v<sub>e</sub></i></b> is positive, Δ<b><i>V</i></b> is negative and a rocket decelerates.<br /><br /><br /><br />Here is how to derive this formula.<br /><br />First of all, let's recall that in the absence of external forces the <br />combined momentum of motion of a system of objects is constant.<br />This is the Law of Conservation of Momentum.<br /><br />Let's apply this law to our situation.<br /><br /><br /><br />Assume, we are comparing the momentum of the system at times <b><i>t</i></b> and infinitesimally incremented <b><i>t+</i></b><i>d<b>t</b></i>.<br /><br /><br /><br />1. At moment <b><i>t</i></b> the momentum of an entire system of a rocket with its propellant was equal to <b><i>m(t)·V(t)</i></b>.<br /><br /><br /><br />2. During the next time interval <b><i></i></b><i>d<b>t</b></i> the rocket has exhausted <b><i>m(t)−m(t+</i></b><i>d<b>t)=−</b>d<b>m(t)</b></i> of propellant with constant relatively to a rocket speed <b><i>v<sub>e</sub></i></b>. Since a rocket moves in some inertial system with speed <b><i>V(t)</i></b> and propellant moves relatively to a rocket with constant speed <b><i>v<sub>e</sub></i></b>, the speed of propellant in the inertial system equals to <b><i>v<sub>e</sub>+V(t)</i></b>. This resulted in the momentum of exhausted propellant at moment <b><i>t+</i></b><i>d<b>t</b></i> to be <b><i>−</i></b><i>d<b>m(t)</b></i>·[<b><i>v<sub>e</sub>+V(t)</i></b>].<br /><br />This equation should be interpreted in the vector form. When a rocket <br />accelerates, velocity vector of its movement and velocity vector of <br />exhausted propellant are opposite in their directions.<br /><br /><br /><br />3. A rocket with remaining propellant at moment <b><i>t+</i></b><i>d<b>t</b></i> has mass <b><i>m(t+</i></b><i>d<b>t)=m(t)+</b>d<b>m(t)</b></i>, velocity <i><b>V(t+</b>d<b>t)=V(t)+</b>d<b>V(t)</b></i> and momentum<br /><br />[<b><i>m(t)+</i></b><i>d<b>m(t)</b></i>]·[<i><b>V(t)+</b>d<b>V(t)</b></i>]<br /><br /><br /><br />Now we are ready to apply the Law of Conservation of Momentum.<br /><br />Item 1 above describes the momentum of the system at time <b><i>t</i></b>.<br /><br />At the moment <i><b>t+</b>d<b>t</b></i> the momentum of the system is a combination of the momentum of the exhausted propellant during time <i>d<b>t</b></i> (see item 2 above) plus the momentum of the remaining rocket mass (see item 3 above).<br /><br /><br /><br />Equalizing these two momentums at time <i><b>t</b></i> and <i><b>t</b>+d<b>t</b></i>, which is the consequence of the Law of Conservation of Momentum, we get the following equation:<br /><br /><b><i>m(t)·V(t) =<br /><br />= −</i></b><i>d<b>m(t)</b></i>·[<b><i>v<sub>e</sub>+V(t)</i></b>] <b>+</b><br /><br /><b>+</b> [<b><i>m(t)+</i></b><i>d<b>m(t)</b></i>]·[<i><b>V(t)+</b>d<b>V(t)</b></i>]<br /><br /><br /><br />This can be simplified. After this the equation looks like<br /><br /><b><i>0 = −v<sub>e</sub>·</i></b><i>d<b>m(t) + m(t)·</b>d<b>V(t) +<br /><br />+ </b>d<b>V(t)·</b>d<b>m(t)</b></i><br /><br /><br /><br />The last member in this equation <i>d<b>V(t)·</b>d<b>m(t)</b></i> is an infinitesimal of the higher order that we can remove from this equation, and the resulting equation looks like<br /><br /><i><b>m(t)·</b>d<b>V(t) = v<sub>e</sub>·</b>d<b>m(t)</b></i><br /><br /><br /><br />Divide both parts by <i><b>m(t)</b></i> and take into consideration that <i>d<b>m(t)<span style="font-size: medium;">/</span>m(t) = </b>d</i>[<i>ln(<b>m(t)</b>)</i>]. Then our equation looks like<br /><br /><i>d<b>V(t) = v<sub>e</sub>·</b>d</i>[<i>ln(<b>m(t)</b>)</i>]<br /><br /><br /><br />Integrating this on the interval Δ<b><i>t</i></b> of time from the beginning of engine's work <i><b>t<sub>beg</sub></b></i> to the end of this period <i><b>t<sub>end</sub></b></i>, we obtain the equation for an increment of the rocket's speed during this interval:<br /><br /><i><b>V(t<sub>end</sub>) − V(t<sub>beg</sub>) = </b></i>Δ<i><b>V(t) =<br /><br />= v<sub>e</sub>·</b></i>[<i>ln(<b>m(t<sub>end</sub>)</b>)−ln(<b>m(t<sub>beg</sub>)</b>)</i>]<b><i>=<br /><br />= v<sub>e</sub>·</i></b><i>ln</i>[<i><b>m(t<sub>end</sub>)<span style="font-size: medium;">/</span>m(t<sub>beg</sub>)</b></i>]<b><i>=<br /><br />= −v<sub>e</sub>·</i></b><i>ln</i>[<i><b>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</b></i>]<br /><br /><br /><br />That is,<br /><br />Δ<i><b>V(t) =−v<sub>e</sub>·</b>ln</i>[<i><b>m(t<sub>beg</sub>)<span style="font-size: medium;">/</span>m(t<sub>end</sub>)</b></i>]<br /><br /><br /><br />So, the increment of a rocket's speed during a period of Δ<i><b>t=t<sub>end</sub>−t<sub>beg</sub></b></i>,<br /> when its engine works, exhausting the propellant, equals to a product <br />of the speed of exhausted propellant times a logarithm of a ratio of the<br /> rocket's mass at the beginning of this period to its mass at the end of<br /> it.<br /><br /><br /><br />The minus in front of a formula is very important. This is a vector <br />equation and, if the exhaust is directed back relatively to rocket's <br />movement (that is, <i><b>v<sub>e</sub></b></i> is negative), the <br />increment of a rocket's speed is positive, a rocket accelerates; if, <br />however, the exhaust is directed towards a tocket movement (that is, <i><b>v<sub>e</sub></b></i> is positive), the increment of a rocket's speed is negative, a rocket slows down.<br /><br /><br /><br />The formula above is the Tsiolkovsky's formula and is called the "ideal rocket equation".Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-79985248977568080242018-07-17T11:38:00.001-07:002018-07-17T11:38:42.576-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Spring Oscillation<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/N4TB2rkwmWY" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Spring</u><br /><br /><br /><br />Consider a point-object of mass <b><i>m</i></b>, hanging vertically at <br />the lower end of a weightless spring, that is fixed at the upper end. <br />Under the weight of this object a spring will stretch a little from its <br />neutral position.<br /><br /><br /><br />The Hook's Law for a spring, which will be used to solve this problem, involves a spring's elasticity constant <b><i>k</i></b>, that we assume is given.<br /><br /><br /><br />Let's stretch this spring even more, so that the distance between an object at its bottom and a spring's neutral level is <b><i>L</i></b> and let it go without any push.<br /><br /><br /><br />Our task is to analyze the oscillation of the object as a function <b><i>x(t)</i></b> of its vertical deviation from a spring's neutral position.<br /><br /><br /><br /><img src="http://www.unizor.com/Pictures/Spring.png" style="height: 270px; width: 200px;" /><br /><br /><br /><br />The obvious initial conditions of the motion of our object are:<br /><br /><b><i>x(0) = L</i></b><br /><br /><b><i>x'(0) = 0</i></b><br /><br /><br /><br />There are two forces acting on our object:<br /><br />(a) its weight <b><i>W</i></b>, directed vertically down and equal in magnitude to <b><i>m·g</i></b>, where <b><i>g</i></b> is the acceleration of free falling<br /><br /><b><i>W = m·g</i></b><br /><br />(b) the spring's elasticity force <b><i>F</i></b>, equal in magnitude to a coefficient of elasticity <b><i>k</i></b><br /> multiplied by a displacement of the spring's bottom end from a neutral <br />level; the direction of this force is always against the direction of <br />the displacement<br /><br /><b><i>F = −k·x(t)</i></b><br /><br /><br /><br />The resultant of the superposition of these two forces can be equated to<br /> mass times acceleration of the object, according to the Newton's Second<br /> Law:<br /><br /><b><i>m·g − k·x(t) = m·x"(t)</i></b><br /><br />This is the differential equation that defines the movement of our object.<br /><br /><br /><br />We don't have to resort to modifying this differential equation with an <br />approximate one to be able to solve it. It is fully solvable and the <br />general solution of this linear differential equation of the second <br />order is<br /><br /><b><i>x(t) = C<sub>1</sub>·cos(t·√<span style="text-decoration-line: overline;">k/m</span>) +<br /><br />+ C<sub>2</sub>·sin(t·√<span style="text-decoration-line: overline;">k/m</span>) + m·g/k</i></b><br /><br /><br /><br />Now we can apply the initial conditions to determine constants <b><i>C<sub>1</sub></i></b> and <b><i>C<sub>2</sub></i></b>.<br /><br />Since <b><i>x(0) = L</i></b>,<br /><br /><b><i>C<sub>1</sub> = L − m·g/k</i></b><br /><br />Since <b><i>x'(0) = 0</i></b>,<br /><br /><b><i>C<sub>2</sub> = 0</i></b><br /><br /><br /><br />This produces the following expression for <b><i>x(t)</i></b>:<br /><br /><b><i>x(t) = (L − m·g/k)·cos(t·√<span style="text-decoration-line: overline;">k/m</span>) +<br /><br />+ m·g/k</i></b><br /><br /><br /><br />Interestingly, if<br /><br /><b><i>L − m·g/k = 0</i></b> or <b><i>L·k = m·g</i></b><br /><br />(which means that the weight <b><i>W=m·g</i></b> is balanced by the force of spring's elasticity <b><i>F=−L·k</i></b> in its initial position with our object at its end) then there are no oscillations, and the object will remain at distance <b><i>L=m·g/k</i></b> from a spring's neutral position.<br /><br /><br /><br />If there are oscillations, their period is<br /><br /><b><i>T = 2π√<span style="text-decoration-line: overline;">m/k</span></i></b>Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-84912794292607306302018-07-16T13:30:00.001-07:002018-07-16T13:30:56.663-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Pendulum<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/8vBYuSfHWKc" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Pendulum</u><br /><br /><br /><br />We will analyze the ideal (<i>mathematical</i>) <i>pendulum</i>, which is a mechanical device placed near the surface of a planet with free fall acceleration <b><i>g</i></b> (to have the gravitational force acting on it) that consists of a point-object of certain mass <b><i>m</i></b>, hanging on a weightless non-stretchable thread of length <b><i>L</i></b>, fixed at the other end, so that the hanging on it object has freedom of motion.<br /><br /><img src="http://www.unizor.com/Pictures/Pendulum2.png" style="height: 240px; width: 200px;" /><br /><br /><br /><br />Assume that at time <b><i>t=0</i></b> we have tilted a point-object at the end of thread of a pendulum by an angle <b><i>α<sub>0</sub></i></b> from vertical. Then we let it go without any push.<br /><br />Our task is to determine, how an angle of deviation of this pendulum <br />from a vertical changes with time, that is we have to find the function <b><i>α(t)</i></b>.<br /><br /><br /><br />We can say now that initial (at time <b><i>t=0</i></b>) position of a pendulum is<br /><br /><b><i>α(0) = α<sub>0</sub></i></b><br /><br />Considering that linear displacement <b><i>d</i></b> along a circular trajectory of a radius <b><i>L</i></b> and its angular displacement <b><i>α</i></b> are related by a formula<br /><br /><b><i>d = L·α</i></b>,<br /><br />the initial condition of not pushing a pendulum, which means "no initial<br /> linear velocity along its trajectory", means that the first derivative <br />of angular displacement is zero:<br /><br /><b><i>α'(0) = 0</i></b><br /><br /><br /><br />Having these initial conditions, we'll determine the equation that function <b><i>α(t)</i></b> must satisfy, using the Newton's Second Law.<br /><br />The force of gravity <b><i>P=mg</i></b> can be represented as a sum of two forces:<br /><br />- a force along a pendulum's thread, that is completely balanced by a <br />thread's reaction, which results in constant distance of a point-object <br />at the end of a thread from its other (fixed) end; this force constrains<br /> the movement of a point-object within a circular trajectory and is <br />equal to<br /><br /><b><i>mg·cos(α(t))</i></b><br /><br />- a force tangential to a circular trajectory of a point-object at the <br />end of a thread; this force is the source of movement along a trajectory<br /> and is equal to<br /><br /><b><i>F = −mg·sin(α(t))</i></b><br /><br />(negative sign is used because the force is always directed in an opposite direction to the movement)<br /><br /><br /><br />The force tangential to a circular trajectory is the one that <br />accelerates our point-object. Since the displacement along a circular <br />trajectory is, as we indicated, <b><i>d=L·α</i></b>, the linear acceleration along a trajectory is equal to a second derivative of this expression by time<br /><br /><b><i>a = L·α"(t)</i></b><br /><br />The Newton's second law states that<br /><br /><b><i>m·a = F</i></b><br /><br />which results in the following differential equation for function <b><i>α(t)</i></b>:<br /><br /><b><i>m·L·α"(t) = −m·g·sin(α(t))</i></b><br /><br /><br /><br />The good news is that we can reduce this by mass <b><i>m</i></b>, which <br />means that the oscillation of a pendulum does not depend on a mass of a <br />point-object at its end, but only on the length of a thread <b><i>L</i></b> and acceleration of free falling <b><i>g</i></b>.<br /><br />So, we deal with an equation<br /><br /><b><i>L·α"(t) = −g·sin(α(t))</i></b> or<br /><br /><b><i>α"(t) = −(g/L)·sin(α(t))</i></b><br /><br /><br /><br />Another good news is that this is a differential equation of the second <br />order (highest derivative is the second one) and we have two initial <br />conditions for a function <b><i>α(t)</i></b> at <b><i>t=0</i></b> and for its first derivative <b><i>α'(t)</i></b> at <b><i>t=0</i></b>.<br /><br />This fully identifies the function <b><i>α(t)</i></b>.<br /><br /><br /><br />Unfortunately, the bad news is that this differential equation cannot be<br /> solved in terms of simple algebraic functions, but only numerically <br />tabulated using computer.<br /><br /><br /><br />But physicists, in their endless quest for simple solutions to <br />complicated problems of the Universe, have decided that within certain <br />boundaries they can simplify the above equation to approximate its <br />solution, using simple algebraic functions.<br /><br />This simplification is based on the fact that, when an angle is <br />relatively close to zero, its sine is not much different from the value <br />of an angle itself (in radians). This is based on a famous limit<br /><br /><i>lim<sub>x→0</sub></i>[<b><i>sin(x)/x</i></b>]<b><i> = 1</i></b><br /><br />So, for relatively small angles around a vertical, the oscillations of a<br /> pendulum can be approximately expressed by an equation obtained by <br />replacing <b><i>sin(α(t))</i></b> with simple <b><i>α(t)</i></b>.<br /><br /><br /><br />This produces the following equation:<br /><br /><b><i>α"(t) = −(g/L)·α(t)</i></b><br /><br />This is a simple linear differential equation with general solution<br /><br /><b><i>α(t) = C<sub>1</sub>·cos(√<span style="text-decoration-line: overline;">g/L</span>·t) + C<sub>2</sub>·sin(√<span style="text-decoration-line: overline;">g/L</span>·t)</i></b><br /><br />where <b><i>C<sub>1</sub></i></b> and <b><i>C<sub>2</sub></i></b> are some constants.<br /><br />To determine the values of these constants, we will use the initial conditions:<br /><br /><b><i>α(0) = α<sub>0</sub></i></b> and <b><i>α'(0) = 0</i></b><br /><br />This results in the following:<br /><br /><b><i>C<sub>1</sub>·cos(0) + C<sub>2</sub>·sin(0) = α<sub>0</sub></i></b><br /><br />and<br /><br /><b><i>−C<sub>1</sub>·sin(0) + C<sub>2</sub>·cos(0) = 0</i></b><br /><br />from which follows that<br /><br /><b><i>C<sub>1</sub> = α<sub>0</sub></i></b><br /><br />and<br /><br /><b><i>C<sub>2</sub> = 0</i></b><br /><br /><br /><br />Solution to our problem, therefore, is<br /><br /><b><i>α(t) = α<sub>0</sub>·cos(√<span style="text-decoration-line: overline;">g/L</span>·t)</i></b><br /><br /><br /><br />This solution represents <i>harmonic</i> oscillation with an amplitude<br /><br /><b><i>A = α<sub>0</sub></i></b><br /><br />and period<br /><br /><b><i>T = 2π <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">g/L</span> = 2π·√<span style="text-decoration-line: overline;">L/g</span></i></b><br /><br /><br /><br />The above <b>approximate</b> solution satisfies to a certain degree <br />physicists and is accepted as the one describing relatively small <br />harmonic oscillations of pendulum around a vertical.<br /><br />Oscillations on a bigger scale (say, with initial angle of deviation <br />around 45° or so) do not conform to this formula and are not harmonic.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-47183751137859624942018-07-13T13:36:00.001-07:002018-07-13T13:36:13.160-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Weight<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/WWiqtuMYRes" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Weight</u><br /><br /><br /><br /><i>Weight</i> of an object, by definition, is the force of gravity a <br />planet attracts this object with. Usually, the word "weight" implies the<br /> magnitude of this force; its direction is, obviously, always towards a <br />center of a planet.<br /><br /><br /><br />So, <i>weight</i> is not a characteristic of an object itself, it's a <br />characteristic of an object relative to a planet. In most cases, this <br />planet is our Earth, though we sometimes say, for example, that a <br />particular object weighs on the Earth 6 times more than on the Moon. <br />This only means that the force of gravity on the surface of the Moon is 6<br /> times weaker than on the surface of the Earth.<br /><br /><br /><br />Do we feel weight as the force of gravity?<br /><br />Not quite. What we can feel is pressure (<i>reaction force</i>) from the surface we stand on, that equalizes gravitational force to hold us at fixed position on a floor or on a ground.<br /><br />If there is no support (like for a person jumping with a parachute from <br />an airplane before a parachute is open, if we ignore the air <br />resistance), we don't feel weight, we are weightless. We have different <br />senses, but not a sense of gravity.<br /><br />So, feeling weightless is not really an absence of gravity, it's absence<br /> of a reaction force that balances the gravity (equal in magnitude and <br />opposite in direction) and holds us fixed relatively to a planet.<br /><br />This reaction force is not just against our feet, when we stand on the <br />floor, it's everywhere inside our body as well, since the body maintains<br /> its shape. We feel this pressure of a reaction force everywhere inside.<br /> That's why it's very difficult to emulate the gravity with some special<br /> equipment on a spaceship.<br /><br /><br /><br />People on a spaceship with non-working engines flying around the Earth <br />on an orbit feel weightless, because they are constantly falling towards<br /> the Earth together with a spaceship (no support!) from the straight <br />line trajectory tangential to an orbit; planet attracts them with <br />gravitational force, and only because of the speed, they maintain <br />constant distance from the planet.<br /><br /><br /><br />Since weight is a force, it is measured in units of force, like <i>newtons</i> in SI.<br /><br />The <i>weight</i> of an object of mass <b><i>m</i></b> on a surface of a planet of mass <b><i>M</i></b> and radius <b><i>R</i></b> is, as we know,<br /><br /><b><i>W = G·M·m <span style="font-size: medium;">/</span>R²</i></b><br /><br />where <b><i>G</i></b> is a universal gravitational constant,<br /><br /><b><i>G = 6.674·10<sup>−11</sup> N·m²/kg²</i></b><br /><br /><br /><br />Since we are talking about weight as a force, which is a subject to the <br />Newton's Second Law, we can determine the acceleration this force causes<br /> to an object of mass <b><i>m</i></b>, if acts alone:<br /><br /><b><i>a = W/m = G·M <span style="font-size: medium;">/</span>R²</i></b><br /><br /><br /><br />Notice that on the surface of Earth this acceleration is constant since <br />all components of this expression (gravitational constant <b><i>G</i></b>,<br />mass of Earth <b><i>M</i></b> and its radius <b><i>R</i></b>) are constants.<br /><br />So, we can calculate this constant once and for all and, knowing the mass of an object <b><i>m</i></b>,<br /> we can determine its weight by multiplying it by this constant, which <br />is, as we determined in the previous lecture, an acceleration of free <br />fall, which on the surface of Earth is traditionally symbolized by <br />letter <b><i>g</i></b>:<br /><br /><b><i>g = G·M <span style="font-size: medium;">/</span>R²</i></b><br /><br /><br /><br />The value of this constant is, approximately, <b><i>9.8 m/sec²</i></b>. <br />But, to be precise, it's not the same at different points on the Earth <br />because the shape of the Earth is not exactly a sphere and its mass is <br />not uniformly distributed within its volume.<br /><br />Moreover, it obviously changes with height (getting smaller) since the higher elevation is equivalent to a greater radius <b><i>R</i></b> (distance to a center of the Earth) of an object.<br /><br /><br /><br />Now we can say that for an object of mass <b><i>m</i></b> the weight on the surface of the Earth is <b><i>W=m·g=9.8·m</i></b>. If mass <b><i>m</i></b> is measured in <i>kilograms</i>, this weight is measured in <i>newtons</i>.<br /><br /><br /><br />Analogous calculation for other planets, based on their mass and radius, show the following values of free falling acceleration:<br /><br />on Sun - 274.1 m/sec²<br /><br />(objects are 28 times heavier on Sun than on Earth),<br /><br />on Jupiter - 25.93 m/sec²<br /><br />(objects are about 2.6 times heavier on Jupiter than on Earth),<br /><br />on Moon - 1.625 m/sec²<br /><br />(objects are about 6 times lighter on Moon than on Earth).<br /><br /><br /><br />Historically, the weight is rarely measured in <i>newtons</i>. More customary units are:<br /><br />1 pound (abbreviated <i>lb</i>) equals to 4.44822 <i>newtons</i> - the weight of an object of mass 0.454 kg on Earth;<br /><br />1 kilogram-force (usually, simply called 1 kilogram, skipping "-force", and abbreviated <i>kgf</i>, but plain <i>kg</i> can also be used, when implication to weight is obvious) equals to weight of an object of mass of 1 kg on Earth, that is 9.8 <i>newtons</i>;<br /><br />and others.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15845874874564706302018-07-10T13:14:00.001-07:002018-07-10T13:14:39.270-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Free Falling<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/wLCbc6Pu1UY" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Free Falling</u><br /><br /><br /><br />Free falling is a movement of an object on a surface of a planet <br />relative to this surface, when the only force acting on this object is <br />the gravitational force of a planet.<br /><br /><br /><br />Our task is to describe this movement in mechanical terms of <i>force</i>, <i>mass</i> and <i>acceleration</i>.<br /><br />In this task we will assume that<br /><br />(a) an object in question is a point-object of mass <b><i>m</i></b>,<br /><br />(b) a planet has a spherical form and its mass <b><i>M</i></b> is uniformly distributed within its volume,<br /><br />(c) a planet has a radius <b><i>R</i></b>,<br /><br />(d) [an important assumption that can be justified by complex <br />calculations] we can model the combined forces of gravitation between <br />all microscopic particles inside a planet and our object in question as a<br /> gravitational force of a point-object of mass <b><i>M</i></b> positioned at the center of a planet.<br /><br /><br /><br />In this case the one and only force of attraction acting on an object <br />and directed towards the center of a planet can be expressed using the <br />Law of Gravitation as follows:<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> is a gravitational constant,<br /><br /><b><i>G = 6.674·10<sup>−11</sup> N·m²/kg²</i></b><br /><br /><br /><br />Knowing the force of gravity <b><i>F</i></b> and mass of an object <b><i>m</i></b>, we can determine the acceleration using the Newton's Second Law:<br /><br /><b><i>a = F/m = G·M <span style="font-size: medium;">/</span>R²</i></b><br /><br /><br /><br />Notice that this acceleration does not depend on <b><i>m</i></b> - mass of an object, which means that all objects fall on the surface of a planet with the same acceleration.<br /><br />An interesting aspect of this formula is that we can imagine how to <br />measure an acceleration (easy) and radius of a planet (more difficult, <br />but possible), while we have no idea how to measure the mass of a <br />planet.<br /><br />So, this formula is used exactly for this purpose - to determine the mass of a planet, resolving the formula above for <b><i>M</i></b>:<br /><br /><b><i>M = a·R² <span style="font-size: medium;">/</span>G</i></b><br /><br /><br /><br />Experiments show that on the surface of our planet Earth the acceleration caused by gravitational force is approximately <b><i>9.8 m/sec²</i></b>.<br /><br />The radius of Earth is approximately <b><i>6.4·10<sup>6</sup> m</i></b>.<br /><br /><br /><br />From this we can calculate the mass of Earth (in kilograms - units of mass in SI):<br /><br /><b><i>M≅9.8· 6.4²·10<sup>12</sup><span style="font-size: medium;">/</span>(6.674·10<sup>−11</sup>)</i></b><br /><br />The result of this calculation is<br /><br /><b><i>M ≅ 6·10<sup>24</sup> kg</i></b><br /><br /><br /><br />Let's solve a different problem now. We'd like to launch a satellite <br />around the Earth that circulates around the planet at height <b><i>H</i></b>. What linear speed should a satellite have to stay on a circular orbit?<br /><br /><br /><br />We know from Kinematics that an object rotating along a circular trajectory of radius <b><i>r</i></b> and angular speed <b><i>ω</i></b> has acceleration <b><i>a=r·ω²</i></b>.<br /><br />In terms of linear speed <b><i>V=r·ω</i></b> along an orbit this formula looks like<br /><br /><b><i>a = V²<span style="font-size: medium;">/</span>r</i></b><br /><br /><br /><br />Since the radius of an orbit is the radius of Earth <b><i>R</i></b> plus the height above its surface <b><i>H</i></b>, we should replace <b><i>r</i></b> in this formula with <b><i>R+H</i></b>.<br /><br /><br /><br />The force of gravity is the only force acting on a satellite and the <br />only source of its acceleration towards the Earth, so the acceleration <br />above must be equal to acceleration of a free fall of a satellite. Here <br />we will take into consideration already known mass of Earth and use <br />distance from the center of the Earth to satellite as <b><i>R+H</i></b>, where <b><i>R</i></b> is the radius of Earth and <b><i>H</i></b> is a height above the Earth's surface.<br /><br /><br /><br />The acceleration of a free fall to Earth at height <b><i>H</i></b> above the surface, using its radius <b><i>R</i></b> and already calculated mass of Earth <b><i>M</i></b>, is:<br /><br /><b><i>a = G·M<span style="font-size: medium;">/</span>(R+H)²</i></b><br /><br /><br /><br />Therefore, equating the acceleration of free fall to acceleration of an <br />object rotating along a circular orbit, we come to the following <br />equation:<br /><br /><b><i>V²<span style="font-size: medium;">/</span>(R+H) = G·M<span style="font-size: medium;">/</span>(R+H)²</i></b><br /><br />from which we derive the value of required linear speed <b><i>V</i></b>:<br /><br /><b><i>V = √<span style="text-decoration-line: overline;">G·M<span style="font-size: medium;">/</span>(R+H)</span></i></b><br /><br /><br /><br />For example, International Space Station rotates around our planet on a height of about 400 kilometers (4·10<sup>5</sup> meters).<br /><br />That means that, to stay on an orbit, it should have linear speed of<br /><br /><b><i>V ≅ 0.78·10<sup>4</sup> m/sec</i></b><br /><br />which is about 28,000 km/hour.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-82597733377840877202018-07-09T11:18:00.001-07:002018-07-09T11:18:03.350-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Gravity<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ZEU81Iy63uQ" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Gravity</u><br /><br /><br /><br />We all know a lot about gravity, weight, weightlessness, rockets flying on orbits calculated based on the laws of gravity etc.<br /><br /><br /><br />But what is <i>gravity</i>?<br /><br /><br /><br />A short answer is: we don't know. It's like most of us can use a smart <br />phone and Global Positioning System (GPS), but don't know how and why it<br /> works. It just works, we know what to do to effectively use it, but <br />have no idea about the real mechanism that allows us to use it.<br />Obviously, designers and engineers who created these technological <br />marvels know, but most of people don't.<br /><br /><br /><br />With gravity it's similar. We know it exists, we can use it, we feel it,<br /> but we don't know the underlying reason why it is what it is.<br /><br />To be more precise, physicists have certain ideas about the source of <br />gravity, but they are rather vague, on the level of hypothesis.<br /><br />Therefore, we skip this foundational discussion about why gravity <br />exists, what is an underlying mechanism of its work. We will just use it<br /> as we use GPS without getting much deeper.<br /><br /><br /><br />To use a computer game, we just have to know its rules and controls, we don't have to know what software is inside.<br /><br />To use gravity, it's not necessary to know its underlying mechanism, we <br />just need to know its properties, and that's the subject of this <br />lecture.<br /><br /><br /><br />The first fundamental property of gravity is that all objects we deal <br />with attract other objects. This effect of attraction is called <b>gravity</b>.<br /><br /><br /><br />Attraction is a <i>force</i>.<br /><br />Since we usually model physical objects as points, this force is <br />directed along the line connecting these point-objects and pushes them <br />towards each other.<br /><br />It is also important to note that the Newton's Third Law says that the force point-object <i>B</i> attracts point-object <i>A</i> is paired with the same in magnitude and opposite in direction force point-object <i>A</i> attracts point-object <i>B</i>.<br /><br /><br /><br />In more complicated cases of objects that cannot be considered as <br />points, we can assume that every tiny peace of each object, which can be<br /> modeled as a point-object, is attracted to every other tiny peace. Then<br /> some process of integration of all these forces might be used to <br />determine the resultant forces. But we will rarely deal with this type <br />of gravitation, most of cases we will consider will involve <br />point-objects.<br /><br /><br /><br />Forces change the velocity. Therefore, gravity, which is the force <br />observed for any type of object, causes change of motion of objects. If <br />there is only one point-object in the Universe, it will maintain its <br />inertial motion along a straight line with constant velocity. As soon as<br /> another object appears somewhere, the force of gravity will cause a <br />change in the inertial movement of the first object.<br /><br /><br /><br />Our next question is: how exactly forces of gravity change the motion of objects?<br /><br /><br /><br />Different objects attract differently.<br /><br />Consider some probe object <i>A</i> in inertial motion along a straight <br />line with constant velocity. For example, it flies in our <br />three-dimensional space along the X-axis in positive direction, going <br />through point of origin of coordinates <nobr>{<i>0, 0, 0</i>}</nobr> at moment in time <i>t=0</i> towards positive infinity.<br /><br />Let's measure the degree of the change of its motion, when at the later moment of time <i>t=1</i> another object <i>B</i> appears at the origin of coordinates <nobr>{<i>0, 0, 0</i>}</nobr> and stays in this fixed position. This object <i>B</i> possesses the property of gravitational attraction with object <i>A</i> and, therefore, will slow down the velocity of object <i>A</i>, pulling it back to the origin of coordinates, so object <i>A</i><br /> will decelerate. Measuring this deceleration and knowing the mass of <br />objects involved, we can measure the force of attraction between objects<br /> <i>A</i> and <i>B</i> using the Newton's Second Law.<br /><br /><br /><br />Our observations show that different objects <i>B</i> will cause different decelerations of probe object <i>A</i> and the same object <i>B</i> causes different decelerations of probe object <i>A</i> at different distances between them. We conclude then that <i>gravitational force</i> of attraction between objects <i>A</i> and <i>B</i> depends on <u>gravitational properties of objects themselves</u> and on <u>distance between them</u>.<br /><br /><br /><br />Our purpose is to analyze what is the gravitational property of any <br />object, how to measure it and how the force of gravity depends on it and<br /> the distance between objects.<br /><br /><br /><br />The situation with distance is easy.<br /><br />Experiments with the same objects showed that the gravitational force of<br /> attraction between them weakens with distance in inverse proportion to a<br /> square of this distance. In other words, if the distance between any <br />two objects <i>A</i> and <i>B</i> doubles, the gravitational force of attraction weakens by a factor of 4.<br /><br />So, it is sufficient to establish the gravitational force between two <br />objects at a unit length (say, 1 meter), after which the gravitational <br />force between these objects at any distance <i>D</i> will be that force at a unit distance divided by a factor <i>D<sup>2</sup></i>.<br /><br /><br /><br />Let's discuss now the gravitational property of an object, its ability to attract other objects, which in Physics is called <i>gravitational mass</i> of an object.<br /><br /><br /><br />An experimental fact is that two identical objects, <i>B<sub>1</sub></i> and <i>B<sub>2</sub></i> combined together, attract twice as strongly as only one of them, say <i>B<sub>1</sub></i>, providing they attract the same probe object <i>A</i>, and the relative position of participating objects is the same.<br /><br />That means that <i>gravitational mass</i> is additive and the gravitational force is proportional to <i>gravitational mass</i>.<br /><br /><br /><br />Let's choose one particular probe object <i>A</i> and assign it a <i>gravitational mass</i> of a unit and another identical object <i>B</i>. Since they are identical, the <i>gravitational mass</i> of object <i>B</i> is also a unit.<br /><br />Then, comparing the attraction between this unit probe object <i>A</i> and identical unit object <i>B</i> at the unit distance with the attraction of any other object <i>C</i> to the same unit probe object <i>A</i> on the same unit distance, we can assign a <i>gravitational mass</i> to that other object <i>C</i>. Since gravitational mass is additive, the stronger the gravitational force of object <i>C</i> - the proportionally greater is its <i>gravitational mass</i> relative to a unit object <i>B</i>.<br /><br /><br /><br />Notice, that additive property of <i>gravitational mass</i> is similar to a property of <i>inertial mass</i>, which is also additive. This is precisely the reason why both properties are call <i>mass</i>.<br /><br /><br /><br />The analogy goes further. Another experimental fact is that two different objects of the same <i>inertial mass</i> have exactly the same <i>gravitational mass</i>,<br /> that is they attract equal probe objects on equal distance equally. <br />From this follows that the quantitative difference between <i>inertial mass</i> and <i>gravitational mass</i> is just in units of measurement.<br /><br />Based on this, it was decided to measure the <i>gravitational mass</i> in exactly the same units as <i>inertial mass</i> and, by definition, say that an object of 1 kilogram of <i>inertial mass</i> has 1 kilogram of <i>gravitational mass</i>, which, quantitatively, fully characterizes the gravitational properties of an object.<br /><br /><br /><br />When we talk about gravity, 1 kilogram is a measure of gravitational attraction of an object, its <i>gravitational mass</i>. When we discuss inertia, motion, force, 1 kilogram is a measure of an object's <i>inertial mass</i>.<br /><br /><br /><br />Let's derive the formula that expresses the force of gravity between two<br /> point-objects in terms of their gravitational masses and distance <br />between them.<br /><br /><br /><br />We already know that the force of gravity is proportional to a <br />gravitational mass, but, since we always deal with two point-objects, <br />the force must be proportional to a gravitational mass of each of them, <br />that is it is proportional to their product.<br /><br /><br /><br />We also know that the force of gravity is inversely proportional to a square of a distance between objects.<br /><br /><br /><br />These two factors lead to the following formula for the force <i><b>F</b></i> of gravity between two point-objects with gravitational (and inertial, as we defined) masses <b><i>M<sub>1</sub></i></b> and <b><i>M<sub>2</sub></i></b> at distance <b><i>r</i></b> between them:<br /><br /><b><i>F = G·M<sub>1</sub>·M<sub>2</sub> <span style="font-size: medium;">/</span>r²</i></b><br /><br />where <b><i>G</i></b> - a constant of proportionality, since the units <br />of force (N - newtons) have been defined already, and we want to measure<br /> the gravitational force in the same units as any other force.<br /><br /><br /><br />This formula was presented by Sir Isaac Newton in 17th century, though other scientists, like Robert Hooke claimed it as well.<br /><br />Physicists call this formula the Newton's Law of Universal Gravitation.<br /><br /><br /><br />To determine the constant <b><i>G</i></b> in this formula, all we need <br />to do is to place two objects of inertial (and gravitational, as we <br />defined) mass of 1 kilogram each at the distance of 1 meter and measure <br />the force of gravity between them by measuring an acceleration they <br />develop as a result of gravitational force. This force (in newtons) will<br /> be quantitatively equal to a gravitational constant <i><b>G</b></i>.<br /><br />This measurement shows a very weak force, and the gravitational constant equals to<br /><br /><b><i>G = 6.674·10<sup>−11</sup> N·m²/kg²</i></b><br /><br /><br /><br />Finally, let's attempt to explain the phenomenon of weakening of the <br />gravitational force inversely proportional to a square of a distance <br />from the gravitating object.<br /><br />This is not really a theoretical proof, but a reasonable explanation of this fact.<br /><br /><br /><br />Assume that the source of gravitational force around an object is <br />something similar to tentacles of an octopus with objects of larger <br />gravitational mass corresponding to greater number of tentacles. The <br />gravitational grip, presumably. depends on the density of tentacles per <br />unit of covered area.<br /><br />To propel gravity on a longer distance the tentacles should be longer, while their quantity remains the same.<br /><br />Now, the longer these tentacles - the more is the area they have to spread around. This area for tentacle of the length <b><i>r</i></b> is a surface of the sphere of this radius, that is <b><i>4r²</i></b>.<br /> So, area these tentacles are supposed to cover is increasing as a <br />square of their length, which causes a gravitational grip to be weaker <br />in exactly the same proportion.<br /><br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-35927379667312776732018-06-07T10:01:00.001-07:002018-06-07T10:01:56.653-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 3<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/kyuOCZdwOCw" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Friction Problems 3</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />An object of mass <b><i>M</i></b> can slide on a horizontal table. Two weights of masses <b><i>M<sub>1</sub></i></b> and <b><i>M<sub>2</sub></i></b> are attached to it on both sides of a table with weightless thread as on this picture.<br /><br /><img src="http://www.unizor.com/Pictures/Friction2weights.png" style="height: 130px; width: 210px;" /><br /><br />Assuming mass <b><i>M<sub>1</sub></i></b> is greater than <b><i>M<sub>2</sub></i></b>, the object will start sliding to the left under a pull of a bigger weight.<br /><br />Tensions <b><i>T<sub>1</sub></i></b> and <b><i>T<sub>2</sub></i></b>, as well as friction act on an object on a table.<br /><br />Assume that this object, as a result of actions of these forces, moves with acceleration <b><i>a</i></b>.<br /><br />What is the friction coefficient <b><i>μ</i></b> and magnitudes of tensions <b><i>T<sub>1</sub></i></b> and <b><i>T<sub>2</sub></i></b>?<br /><br /><br /><br /><i>Hint</i>:<br /><br />Use the Second Newton's Law for each of three objects (one on a table and two weights) participating in the motion.<br /><br /><br /><br /><i>Solution</i>:<br /><br /><br /><br />For object on a table:<br /><br /><b><i>T<sub>1</sub> − T<sub>2</sub> − μ·M·g = M·a</i></b><br /><br />For weight on the left:<br /><br /><b><i>M<sub>1</sub>·g − T<sub>1</sub> = M<sub>1</sub>·a</i></b><br /><br />For weight on the right:<br /><br /><b><i>T<sub>2</sub> − M<sub>2</sub>·g = M<sub>2</sub>·a</i></b><br /><br />Got linear system of three equations with three unknowns: friction coefficient <b><i>μ</i></b> and magnitudes of tensions <b><i>T<sub>1</sub></i></b> and <b><i>T<sub>2</sub></i></b>.<br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com1tag:blogger.com,1999:blog-3741410418096716827.post-39327047079020788292018-06-01T11:56:00.001-07:002018-06-01T11:56:33.867-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 2<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/z1VQKlyzMPQ" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Friction Problems 2</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />There is a ramp of mass <i><b>M</b></i> and angle <i><b>φ</b></i> to horizon, lying, but not fixed, on a horizontal surface (it's like a triangular prism lying on a side).<br /><br />An object of mass <i><b>m</b></i> slides down this ramp. As it slides, the ramp also moves along the horizontal surface it's lying on.<br /><br />What is the acceleration of a ramp relative to a horizontal surface, as <br />object slides down, if the coefficient of friction between an object and<br /> a ramp is <i><b>μ<sub>0</sub></b></i>, and the coefficient of friction between a ramp and a horizontal surface is <i><b>μ<sub>1</sub></b></i>?<br /><br />The free fall acceleration is <i><b>g</b></i>, so the weight of an object of mass <i><b>m</b></i> is <i><b>m·g</b></i>.<br /><br /><br /><br /><i>Hint</i>:<br /><br />(1) Review Problem B from Mechanics - Dynamics - Superposition of Forces - Inclined Plane of this course.<br /><br />(2) Take into account that, according to the Third Newton's Law, <br />friction between an object and a ramp is, on one hand, a force exhorted <br />by a ramp that pulls object uphill against its sliding downhill and, on <br />the other hand, is an action of an object on the ramp in an opposite <br />direction, slowing its horizontal movement.Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-56451067829003582062018-05-31T12:08:00.001-07:002018-05-31T12:08:56.739-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 1<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ktWCreDOQ40" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Friction Problems 1</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />There is an object of mass <i><b>m</b></i> on an inclined plane that makes angle <i><b>φ</b></i> with horizon.<br /><br />To move this object up on this inclined plane with constant speed you need to apply a force <i><b>F<sub>up</sub></b></i> directed uphill.<br /><br />The coefficient of friction is unknown.<br /><br />Let's assume that, left by itself, the object would slide down the <br />inclined plane under its own weight. The free fall acceleration is <i><b>g</b></i>, so the weight of an object of mass <i><b>m</b></i> is <i><b>m·g</b></i>.<br /><br />What is the acceleration <i><b>a</b></i> of an object when it moves downhill under its own weight?<br /><br /><br /><br /><i>Hint</i><br /><br /><i><b>F<sub>up</sub> = m·g·sin(φ)+μ·m·g·cos(φ)</b></i><br /><br /><i><b>a = g·sin(φ)−μ·g·cos(φ)</b></i><br /><br /><br /><br /><i>Answer</i>:<br /><br /><i><b>a = 2·g·sin(φ) − F<sub>up</sub><span style="font-size: medium;">/</span>m</b></i><br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />An object is lying on a horizontal platform that moves with acceleration <i><b>a=10 m/sec²</b></i>.<br /><br />The <i>coefficient of kinetic friction</i> between an object and a surface of a platform is <i><b>μ=0.3</b></i>, while <i>coefficient of static friction</i> is <i><b>μ<sub>s</sub>=0.4</b></i>.<br /><br />The free fall acceleration is <i><b>g=9.8m/sec²</b></i>, so the weight of an object of mass <i><b>m</b></i> is <i><b>m·g</b></i>, but mass <i><b>m</b></i> is unknown.<br /><br />(a) How the object will behave?<br /><br />(b) Why was coefficient of static friction <i><b>μ<sub>s</sub></b></i> given?<br /><br />(c) What is the acceleration of the object relative to the ground and relative to the platform?<br /><br /><br /><br /><i>Answer</i>:<br /><br />(a) The object will slide back along the platform's surface, but forward relative to the ground.<br /><br />(b) If the coefficient of static friction is too high, the object will <br />not change its position relatively to a platform, and the next question <br />would make no sense.<br /><br />(c) Relative to the ground acceleration is <i><b>a<sub>0</sub>=g·μ=2.94m/sec²</b></i>.<br /><br />Relative to the platform acceleration is <i><b>a<sub>1</sub>=a<sub>0</sub>−a=−7.06m/sec²</b></i><br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-38678497806961201832018-05-29T13:47:00.001-07:002018-05-29T13:47:01.208-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Superposition of Forces ...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/uhPbKX4Qf20" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Motion on Inclined Plane</u><br /><br /><br /><br /><i>Problem A</i><br /><br /><br /><br />Consider a slope fixed on the ground that makes an angle <b><i>φ</i></b><br /> with horizon. Let's analyze the motion of an object, as it slides down a<br /> slope under its own weight. Our task is to determine its acceleration <br />along the slope.<br /><br /><img src="http://www.unizor.com/Pictures/Inclined1.png" style="height: 100px; width: 200px;" /><br /><br />First of all, we have to choose a reference frame - a system of coordinates suitable for this problem.<br /><br />The main force acting on this object is its <i>weight</i> <b><i><span style="text-decoration: overline;">W </span></i></b> - a vector of <i>gravity</i><br /> directed vertically down to the ground. That prompts us to choose a <br />frame of reference with horizontal X-axis and vertical Y-axis. So, the <br />vector of weight has non-zero Y-component and zero X-component: <nobr><b><i><span style="text-decoration: overline;">W </span></i></b> = {<b><i>0</i></b>,<b><i>W</i></b>},</nobr> where <b><i>W</i></b> is a magnitude of vector of weight <b><i><span style="text-decoration: overline;">W</span></i></b>. However, this obvious choice is not the best in a case like this.<br /><br /><br /><br />There are usually numerous forces involved in an experiment and only few<br /> objects. In this case there is only one object. So, instead of catering<br /> to one particular force, like <i>weight</i>, it's better to simplify the motion of a single object.<br /><br />Much more convenient frame of reference would be the one, where our object in motion has only one non-zero coordinate.<br /><br /><br /><br />Recall that there must be another force acting on an object - <i>reaction force</i> of the slope, that prevents an object to go vertically down to the ground through a slope and forces it, in cooperation with <i>gravity force</i>, to slide along a slope. This reaction force <b><i>R</i></b><br /> is always perpendicular to the surface, where an object is on (in this <br />case, perpendicular to a slope), and its magnitude is such that the <i>resultant</i> of the weight <b><i>W</i></b> and reaction <b><i>R</i></b> is directed along a slope.<br /><br /><br /><br />Consider a frame of reference with X-axis going along the slope, where the object slides down and Y-axis perpendicular to it.<br /><br /><img src="http://www.unizor.com/Pictures/Inclined2.png" style="height: 100px; width: 200px;" /><br /><br />Granted, the weight now has both coordinates non-zero:<br /><br />(a) perpendicular to a surface of a slope and (on the picture above) directed along negative Y-coordinates vector <b><i><span style="text-decoration: overline;">W<sub>R</sub></span></i></b> with magnitude <b><i>W<sub>R</sub>=W·cos(φ)</i></b>, that causes reaction force <i><b>R</b></i>, that is equal in magnitude and opposite in direction to <b><i><span style="text-decoration: overline;">W<sub>R</sub></span></i></b>, and<br /><br />(b) parallel to a slope, directed (on the picture) towards negative X-coordinate, vector <b><i><span style="text-decoration: overline;">W<sub>F</sub></span></i></b> with magnitude <b><i>W<sub>F</sub>=W·sin(φ)</i></b>, that is the cause of motion of our object down a slope.<br /><br /><br /><br />As it is pictured, both components of weight are negative since they are<br /> directed towards negative direction of the X- and Y-axes:<br /><br /><b><i><span style="text-decoration: overline;">W </span></i></b> = {<b><i>−W·sin(φ)</i></b>,<b><i>−W·cos(φ)</i></b>}<br /><br />Force <b><i><span style="text-decoration: overline;">R </span></i></b>, as opposite and equal in magnitude to <b><i><span style="text-decoration: overline;">W<sub>R</sub></span></i></b>, is<br /><br /><b><i><span style="text-decoration: overline;">R</span></i></b> = {<b><i>0</i></b>,<b><i>W·cos(φ)</i></b>}.<br /><br /><br /><br />The resultant of three vectors <b><i><span style="text-decoration: overline;">W<sub>R</sub></span></i></b>, <b><i><span style="text-decoration: overline;">W<sub>F</sub></span></i></b> and <b><i><span style="text-decoration: overline;">R </span></i></b> is <b><i><span style="text-decoration: overline;">W<sub>F</sub></span></i></b>, directed down a slope, equaled in magnitude to <b><i>W·sin(φ)</i></b>.<br /><br /><br /><br />Therefore, an object of mass <i><b>M</b></i> will slide down a slope with acceleration equaled in magnitude to<br /><br /><i><b>a = W<sub>F</sub><span style="font-size: medium;">/</span>M = W·cos(φ)<span style="font-size: medium;">/</span>M</b></i><br /><br />Since weight and mass of an object are related as <i><b>W=M·g</b></i>, where <i><b>g</b></i> is a known acceleration of free falling (9.8 m/sec² on the Earth ground), the resulting acceleration equals in magnitude to<br /><br /><i><b>a = M·g·sin(φ)<span style="font-size: medium;">/</span>M = g·sin(φ)</b></i><br /><br />In vector form in the chosen reference frame:<br /><br /><b><i><span style="text-decoration: overline;">a </span></i></b> = {<i><b>−g·sin(φ), 0</b></i>}<br /><br /><br /><br /><br /><br /><i>Problem B</i><br /><br /><br /><br />Consider an object A of mass <i><b>m</b></i>, sliding without friction <br />on a slope of slide B, which itself lies on a horizontal surface and can<br /> slide on it without friction. An angle of a slope of slide B is <b><i>φ</i></b> with horizon, its mass is <i><b>M</b></i>.<br /><br />Our task is to determine acceleration <i><b>a</b></i> of slide B.<br /><br />Let's analyze the motion of object A on a slope and the motion of a <br />slide B, as it moves horizontally, as a result of the weight of object A<br /> on it.<br /><br /><img src="http://www.unizor.com/Pictures/Inclined3.png" style="height: 100px; width: 200px;" /><br /><br /><br /><br />As object A presses down with its weight, it slides downhill along a <br />slope of a slide B. At the same time slide B moves to the right (on the <br />picture above).<br /><br />The horizontal component of the pressure <i><b>W<sub>R</sub></b></i> of <br />an object A on slide B perpendicularly to its slope is the cause of the <br />motion of a slide B. However, this pressure is not the same as in the <br />problem A above. It will be less. Its opposite reaction force <i><b>R</b></i>, that is equal in magnitude to <i><b>W<sub>R</sub></b></i>, but acting on the object A, will also be less than in the problem A above.<br /><br />The <i>resultant</i> of weight <i><b>W</b></i> and reaction force <i><b>R</b></i> is force <i><b>W<sub>F</sub></b></i> that is not parallel to a slope, but tilted downwards.<br /><br />So, the combination of object A sliding downhill on a slope and slide B <br />movement to the right produces the resultant move of object A that is <br />not parallel to a slope, neither it is directed vertically down, but <br />will be somewhere in-between.<br /><br /><br /><br />Let's consider the same reference frame as in the problem A above. Now <br />both object A and slide B, as they move, have both X- and Y-components <br />not equal to zero.<br /><br /><br /><br />Consider only Y-coordinate of the A object now and Y-components of forces acting on it.<br /><br />In the direction of Y-axis the force <nobr><b><i><span style="text-decoration: overline;">W<sub>F</sub></span></i></b> = <b><i><span style="text-decoration: overline;">W</span></i></b> + <b><i><span style="text-decoration: overline;">R</span></i></b></nobr>, acting against object A, has value<br /><br /><b><i>W<sub>Fy</sub></i></b> = <b><i>W<sub>y</sub></i></b> + <b><i>R<sub>y</sub></i></b> = <i><b>R − W·cos(φ)</b></i><br /><br /><br /><br />This force <nobr><i><b>R − W·cos(φ)</b></i></nobr>, according to the Newton's Second Law, should be equal to object A's mass <i><b>m</b></i>, multiplied by a Y-component of its acceleration, which so far is unknown.<br /><br /><br /><br />Let <i><b>a</b></i> be an acceleration of slide B in the direction of <br />its horizontal movement. Since displacement of object A in the direction<br /> of the Y-axis (perpendicularly to a slope) equals to horizontal <br />displacement of a slide B multiplied by <i><b>sin(φ)</b></i>, the <br />acceleration of slide B in the horizontal direction and acceleration of <br />the object A in a direction perpendicular to a slope, maintain the same <br />factor.<br /><br />Therefore, the acceleration of object A perpendicularly to a slope of slide B equals to <i><b>a·sin(φ)</b></i>, where <i><b>a</b></i> is the acceleration of the slide B that we have to determine in this problem.<br /><br /><br /><br />The Newton's Second Law for object A in the direction of the Y-axis (perpendicular to a slope of slide B) is<br /><br /><i><b>R − W·cos(φ) = −m·a·sin(φ)</b></i><br /><br />(minus on the right because the acceleration of object A relative to Y-axis is negative).<br /><br />This equation is the first in a system of two equations that include two unknowns <i><b>R</b></i> and <i><b>a</b></i>.<br /><br /><br /><br />On the other hand, a horizontal component of vector <b><i><span style="text-decoration: overline;">−R</span></i></b> is the cause of horizontal acceleration <i><b>a</b></i> of slide B that has mass <i><b>M</b></i>.<br /><br />Therefore,<br /><br /><i><b>R·sin(φ) = M·a</b></i><br /><br />This is the second equation in a system of two equations that include two unknowns <i><b>R</b></i> and <i><b>a</b></i>.<br /><br /><br /><br />Solving this system as follows.<br /><br />From the second equation:<br /><br /><i><b>R = M·a/sin(φ)</b></i><br /><br />Substitute it in the first equation:<br /><br /><i><b>M·a/sin(φ) − W·cos(φ) =<br /><br />= −a·m·sin(φ)</b></i><br /><br /><br /><br />The solution for horizontal acceleration of slide B is<br /><br /><i><b>a = W·cos(φ) <span style="font-size: medium;">/<br /><br />/</span> </b></i>[<i><b>m·sin(φ) + M/sin(φ)</b></i>]<i><b> =<br /><br />= W·sin(φ)·cos(φ) <span style="font-size: medium;">/<br /><br />/</span> </b></i>[<i><b>m·sin²(φ)+M</b></i>]<br /><br /><br /><br />So, our final result for an acceleration <i><b>a</b></i> of slide B, as it moves horizontally, is<br /><i><b>W·sin(φ)·cos(φ)<span style="font-size: medium;">/</span></b></i>[<i><b>m·sin²(φ)+M</b></i>]<br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-65067031925144951152018-05-23T18:50:00.001-07:002018-05-23T18:50:27.572-07:00Unizor - Physics4Teens - Mechanics - Dynamics - Friction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/wiVJJkgn9Tw" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Friction</u><br /><br /><br /><br />What's good and bad about friction?<br /><br /><br /><br />Friction prevents us to easily move furniture around. We can buy special<br /> devices to make it easier, but friction is definitely prevents to move <br />furniture easily.<br /><br />Friction is the reason why parts in many machines wear out and need replacement.<br /><br />Friction between the air and a spaceship, when it enters the atmosphere,<br /> causes tremendous heating of the spaceship's body and requires special <br />layer of heat-resistant material to cover the spaceship.<br /><br />Friction prevents a rollerblader to roll indefinitely along the smooth straight road.<br /><br /><b>In a word, friction is bad</b>.<br /><br /><br /><br />But is it really?..<br /><br /><br /><br />Friction is needed to walk on a road, to drive a car, to wear clothes, <br />to have furniture in place where we want it, to have plates on a table <br />in a steady position without slipping over the edge, to fork a piece of <br />meat, to hold a glass of water in a hand.<br /><br />Enough! As we see, <b>we cannot live without friction</b>, but sometimes would like to reduce it.<br /><br /><br /><br />So, what is <i>friction</i>?<br /><br />Here is what causes friction:<br /><br /><img src="http://www.unizor.com/Pictures/Friction.png" style="height: 110px; width: 200px;" /><br /><br />On a micro level the surfaces of two objects touching each other are not<br /> perfect. Each little bump of one is catching a little bump of another, <br />thus preventing smooth gliding of one surface over another.<br /><br />As was mentioned above, there are good and bad sides of this story, but <br />it is what it is. We have to use this friction, when needed, and reduce,<br /> when desired.<br /><br /><br /><br /><b>Static Friction</b><br /><br /><br /><br /><i>Static friction</i> is the friction between an object at rest and surface it is positioned on.<br /><br />A chair stands at a fixed place on a floor, even if the floor is slightly tilted, because <i>static friction</i> holds it in place.<br /><br />Micro-bumps on the floor are caught in-between micro-bumps on the chair's legs and fix the chair's position on a floor.<br /><br /><br /><br />Friction is the source of the force preventing the motion. If there is <br />no force that attempts to move an object from the position of rest, <br />friction results in no force preventing this move, but, if there is a <br />force that attempts to move an object, friction exhorts a force against <br />it, thus preventing the move.<br /><br /><br /><br />If the force that attempts to move an object is not too strong, bumps on<br /> the surfaces of object and its support are still caught between each <br />other, an object stays in place, which means that the force of friction,<br /> directed against the force that attempts to move an object, is equal to<br /> a force of moving by magnitude and opposite in direction.<br /><br /><br /><br />Obviously, the force of friction has its limits, so, if a very strong <br />force attempts to move an object, the bumps on the surface will no <br />longer be able to hold an object, and an object moves.<br /><br />So, as the force attempting to move an object from the state of rest <br />grows in magnitude, so does a force of static friction that prevents <br />this move, but only up to a maximum value, after which the force of <br />friction cannot grow any more, and the object moves. This <b>maximum value</b> of the <i>friction force</i> is the one when it is specified explicitly.<br /><br /><br /><br />To move an object from the position of rest we have to apply force. The <br />heavier an object - the deeper bumps on its surface are caught <br />in-between bumps of the surface it stands on. Therefore, the force <br />needed to move it from the position of rest depends on an object's <br />weight. At the same time, surfaces can be more or less smooth, that is <br />they can have deeper or shallower bumps. Obviously, the degree of <br />smoothness must be a factor in the amount of force needed to move an <br />object from the position of rest.<br /><br /><br /><br />So, two major factors contribute to the amount of force needed to shift <br />an object from the position of rest: the pressure between an object and a<br /> surface it stands on (for example, the component of its weight) and the<br /> quality of surfaces of an object and its support.<br /><br /><br /><br />According to numerous experiments, for any two surfaces (surface of an object and surface of a support it rests on) there is a <i>coefficient of static friction</i> <i><b>μ</b></i>, such that the force needed to shift an object from the position of rest <i><b>F</b></i> depends on this coefficient and the normal pressure <i><b>N</b></i> between an object and its supporting surface as<br /><br /><i><b>F = μ·N</b></i><br /><br /><br /><br />If the experiment is conducted on Earth and the supporting surface is horizontal, normal pressure of an object is its weight.<br /><br />If the surface is an incline, the vector of weight should be represented<br /> as a combination of a normal to a supporting surface and tangential to <br />it components. The normal component contributes to friction, while the <br />tangential component represents the force that attempts to move object <br />from the position of rest.<br /><br /><br /><br />Consider an incline of angle <i><b>φ</b></i> and an object of weight <i><b>W</b></i> on it. Will it slide down?<br /><br /><img src="http://www.unizor.com/Pictures/FrictionInclined.png" style="height: 145px; width: 210px;" /><br /><br />The following forces are acting upon this object:<br /><br />(a) vector of gravity of Earth <i><b><span style="text-decoration: overline;">W</span></b></i> directed vertically down;<br /><br />(b) reaction of the surface <i><b><span style="text-decoration: overline;">R</span></b></i> directed perpendicularly to a surface of an inclined;<br /><br />(c) force of static friction <i><b><span style="text-decoration: overline;">F<sub>f</sub></span></b></i> directed against potential movement down the slope.<br /><br />The right approach to analysis of this experiment is to represent a vector of weight <i><b><span style="text-decoration: overline;">W</span></b></i> as a combination of a vector perpendicular to an inclined <i><b><span style="text-decoration: overline;">W<sub>R</sub></span></b></i> and the vector of force attempting to move an object down a slope <i><b><span style="text-decoration: overline;">W<sub>F</sub></span></b></i>.<br /><br />Now the only force that attempts to move an object is a component of its weight tangential to an inclined <i><b><span style="text-decoration: overline;">W<sub>F</sub></span></b></i>.<br /><br />The reaction force <i><b>R</b></i> equals to a component <i><b>W<sub>R</sub></b></i> of weight.<br /><br /><br />Obviously, the magnitudes of these vectors are:<br /><br /><i><b>W<sub>F</sub> = W·sin(φ)</b></i><br /><br /><i><b>W<sub>R</sub> = W·cos(φ)</b></i><br /><br />Assuming, we know that the <i>coefficient of static friction</i> equals <i><b>μ</b></i>. Then the maximum force of static friction equals to<br /><br /><i><b>F<sub>f</sub> = W·cos(φ)·μ</b></i><br /><br /><br /><br />Therefore, if <i><b>W<sub>F</sub></b></i> is greater than <i><b>F<sub>f</sub></b></i>, the object will slide down a slope, otherwise the static friction will be able to hold it in place.<br /><br />The minimum angle when the movement occurs is a solution of the following equation for angle <i><b>φ</b></i>:<br /><br /><i><b>W·sin(φ) = W·cos(φ)·μ</b></i> or<br /><br /><i><b>tan(φ) = μ</b></i> or<br /><br /><i><b>φ = arctan(μ)</b></i><br /><br />This is an angle, when the sliding of an object under its own weight begins.<br /><br /><br /><br /><b>Kinetic Friction</b><br /><br /><br /><br /><i>Kinetic friction</i> is the friction between a moving object and <br />surface it moves on or between surfaces of two touching each other <br />objects moving relative to each other.<br /><br /><br /><br />Generally speaking, we all know that it's easier to move a sofa on the <br />floor after the movement has started than to initiate a movement from <br />the state of rest.<br /><br /><br /><br />The theory behind the friction, based on tiny bumps on the surfaces of <br />touching each other objects, explains this as a result of deeper <br />penetration of bumps of one surface between the bumps of another, when <br />objects are at rest, than when objects are in motion and bumps have <br />little time to deeply penetrate each other, they slide across each <br />other, causing <i>kinetic friction</i>, which is less than <i>static</i>.<br /><br />These bumps are so small that relative speed of one object against another plays little role in the strength of <i>kinetic friction</i>, but the pressure between the objects and material their surfaces are made of are very important.<br /><br /><br /><br />Quantitatively, kinetic friction is very much like static one with the only difference in the <i>coefficient of friction</i>.<br /><br />According to numerous experiments, for any two surfaces there is a <i>coefficient of kinetic friction</i> <i><b>μ</b></i>, such that any moving object experiences the force <i><b>F</b></i>, acting against its movement, which depends on this coefficient and the normal pressure <i><b>N</b></i> between an object and its supporting surface as<br /><br /><i><b>F = μ·N</b></i><br /><br /><br /><br />This is an experimental law. Coefficients of friction are measured for <br />many pairs of surfaces, which allows to solve such problems as to find <br />the distance of breaking from some speed to full stop, determine forces <br />needed to overcome the friction to move an object with constant speed <br />etc.<br /><br />These are the problems we will solve in the next lectures.<br />Zor Shekhtmanhttps://plus.google.com/112487435710670506083noreply@blogger.com0