Wednesday, August 15, 2018

Unizor - Physics4Teens - Mechanics - Rotational Dynamics





Notes to a video lecture on http://www.unizor.com



Torque



So far we were mostly considering translational motion of point-objects - a motion along a straight line with or without external forces
acting upon this object. We have specified three Newton's Laws of this
motion and derived a lot of interesting facts based on these laws.



Let's recall these laws.



The Newton's First Law is the familiar Law of Inertia that states that
an object at rest stays at rest and an object in uniform motion stays in
this uniform motion, unless acted upon by unbalanced forces.



The Newton's Second Law brings quantitative relationship to vector of force (F), mass (m) and vector of acceleration (a):

F = m·a



The Newton's Third Law states that for every action there is an equal in magnitude and oppositely directed reaction.



Rotational motion obeys the rules in many respects analogous to the laws of translational motion, except we have to change linear movement to rotation, which, in essence, is an angular movement with constant radius.



Consider a point-object of some mass m connected by a weightless rigid rod of the length r to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.

The picture below illustrates such a movement and also indicates the position of angular velocity ω
of a rotation, which was explained earlier, when we studied the
Kinematics of rotation (we assume your familiarity with this topic and
strongly recommend to refresh it before proceeding any further).





Let's discuss the similarities and differences between translational movement along a straight line and rotation around an axis within a plane of rotation perpendicular to this axis.



Obviously, the time concept remains the same in both types of motion.



There is a clear rotational analogy to the Newton's First Law.


An object at rest stays at rest and an object in uniform rotation
stays in this uniform rotation, unless acted upon by unbalanced forces.




The Newton's Third Law is not really specific for a form of motion, so there is no need to address it separately at this moment.



The Newton's Second Law requires certain modification to be applied to rotation.



Let's address the main concept of Dynamics - the force - in connection to rotation.



The very important characteristic of force in Dynamics of translational movement along a straight line is that it can be measured by its effect on objects of certain inertial mass. Thus, a measure of force that gives linear acceleration of 1 m/sec² to an object of inertial mass of 1 kg is 1 newton. The same force applied to an object of 0.5 kg of inertial mass will cause 2 m/sec² linear acceleration.

The same force applied to different objects of the same inertial mass will cause the same linear acceleration etc.



The situation with rotational movement is not the same.



Consider a simple experiment of opening a door. If you apply a force to
open a door at its edge opposite to hinges, where the handle is usually
located, it opens relatively faster than if you apply exactly the same
force in the middle of a door. The closer a point of application of the
same force to hinges - the slower a door opens, if the force applied to
it is the same. In an extreme case, when we apply the pressure where the
hinges are, a door will not open at all.

As a continuation of this experiment, if we want to achieve the same
speed of the opening of a door by applying the force at different
distances from the hinges, we need more efforts for a point of force
application located closer to the hinges.



We can measure the force, the distance from the hinges of a point of application of this force and an angular acceleration of the door and experimentally come up with the fact that for the fixed force the angular acceleration of the door is proportional to the distance of a point of application of the force from the hinges.

Moreover, leaving the point of application of the force the same and changing the force, we can determine that angular acceleration is proportional to the force.



What it means is that in a case of rotation a force by itself does not
determine the final motion of a rotating object. It's a product of a
force F and radius to a point this force is applied r, that determines the final effect. This product τ = F·r is called the torque and it is rotational equivalent of a force in translational movement.

TranslationRotation
Force
F
Torque
τ = F·r


Returning to a picture above, we can apply some force F to
any point on the rigid rod, connecting our point-object to an axis of
rotation, directing this force perpendicularly to the rod, and observe
that the resulting angular acceleration of the object is proportional to
both the force F and the distance r from the axis to a point of application of this force, thus proportional to torque τ = F·r.



Now we have concluded that an angular acceleration α of rotational motion is proportional to a torque τ. This is analogous to linear acceleration a of translational motion being proportional to a force F. The coefficient of proportionality for translational motion is inertial mass m of an object (this is the Newton's Second Law F=m·a).



The obvious question is, what is the coefficient of proportionality between angular acceleration and torque?

Answer to this question will result in rotational equivalent of the Newton's Second Law.



We have experimentally established that equal torques produce equal angular accelerations.

Consider a rotation illustrated on the picture above. Assume that the point of application of force F is exactly at the point-object of mass m rotating around an axis at a distance r
from it on a rigid weightless rod. Assume further that our force acts
within a plane of rotation and directed perpendicularly to the rod.



During an infinitesimal time interval dt the motion of an object can be considered as linear and, therefore, the Newton's Second Law can be applied, giving F=m·a.

Now we can express it in terms of angular acceleration and torque as follows:

a = r·α

τ = F·r

Hence,

F·r = (m·a)·r = m·r²·α

Finally,

τ = (m·r²)·α



One more logical step is needed. We started from a force applied on an object itself at a distance r
from an axis. But we have experimentally established that equal torques
produce equal actions. It means that some other force applied to some
other point will produce the same effect, causing the same angular
acceleration, as long as the torque is the same.

So, the equality τ=(m·r²)·α is universal, regardless of point of application of force since it depends not on force, but on torque.



The above equality represents the rotational analogue of the Newton's Second Law.



Some generalization can be applied to the above.

What if the force, acting within a plane of rotation, is not perpendicular to a rod?

Obvious solution is to replace vector F
with its projection onto a line on the rotation plane that is
perpendicular to a radius and to multiply the product of two scalars F·r by a sine of an angle between corresponding vectors, effectively using a vector product Fr.

So, more general definition of a torque is a vector (or, more precisely, pseudo-vector)

τ = Fr

whose direction, like a direction of an angular acceleration, is along an axis of rotation.

So, for rotational movement the vectors of torque τ and angular acceleration α are collinear, similarly to collinearity of vectors of force F and linear acceleration a for translational movement.



We'd like to note that for purposes of simplicity in this course we will
rarely deal with forces not perpendicular to a radius of rotation.



IMPORTANT TERMINOLOGY POINTS



1. The torque τ is often called the moment of force.



2. Recall the expression tying together a torque τ and an angular acceleration α

τ = (m·r²)·α

The expression m·r² is called moment of inertia, is symbolized by letter I, which allows to specify the formula above as

τ = I·α

where I=m·r² is a moment of inertia, playing the same role in this equation as inertial mass m in the Newton's Second Law and representing resistance to a rotational force.



TranslationRotation
Force
F
Torque
τ = F·r
Acceleration
a
Angular Acc.
α = a/r
Inertial
Mass

m
Moment
of Inertia

I = m·r²
Newton's
Second Law

F=m·
Rotational
Equivalent

τ = I·α

Monday, August 13, 2018

Unizor - Physics4Teens - Mechanics - Rotational Kinematics





Notes to a video lecture on http://www.unizor.com



Rotational Kinematics



So far we were mostly considering translational motion of point-objects - a motion along a straight line with or without external forces acting upon this object.



Rotational motion obeys the rules in many respects analogous to the laws of translational motion, except we have to change linear movement to rotation.



Consider a point-object m connected by a rigid rod of the length r to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.

The picture below illustrates such a movement and also indicates the position of angular velocity ω of a rotation, which we will explain later.





Let's discuss the similarities and differences between translational movement along a straight line and rotation around an axis within a plane perpendicular to this axis (a plane of rotation).



The first main concept of translational motion is position or distance from the beginning of motion (for a straight line movement) as a function of time. In rotational motion its equivalent is angle of rotation from some original position as a function of time.



TranslationRotation
Distance
s(t)
Angle
φ(t)


The next concept is speed or (better) velocity of translational motion. This is a first derivative of position (or distance) by time:

v(t) = s'(t).

Its equivalent for rotational motion is angular speed, which is a first derivative of angle of rotation by time:

ω(t) = φ'(t)



While vector character of speed of translational motion is obvious and is reflected in the term velocity, vector character of angular speed is less obvious.

The angle of rotation from the first glance is a scalar function of time. But only from the first glance.



In theory, rotational motion always assumes existence of an axis of
rotation and a plane of rotation. To reflect these characteristics and a
magnitude of angular speed, an angular speed is represented by a vector
from a center of rotation along an axis of rotation perpendicularly to a
plane of rotation with a magnitude equal to a value of angular speed.



This allows to represent the rotation in its full spectrum of
characteristics - magnitude, axis, plane of rotation. The picture above
represents angular speed as a vector ω(t), which we may call angular velocity vector.



There is one more characteristic of rotational motion not yet discussed - its direction. It is also reflected in angular velocity
as a vector by its direction. In theory, we can choose two different
directions along the axis of rotation. The direction chosen is such
that, if we look from its end onto a plane of rotation, the rotation is
counterclockwise. Another interpretation of this is the "rule of the
right hand" because if you put you right hand on a plane of rotation
such that your finger go around the axis of rotation pointing to a
direction of rotation, your thumb points to a direction of the angular velocity vector.

So, angular velocity vector represents axis, plane, direction of rotation as well as magnitude of angular speed.



To be more precise, since this vector representation of angular velocity is a little unusual, it is customary to call it "pseudo-vector" instead of "vector".



During infinitesimal time interval dt an object rotating around an axis on a radius r turns by an angle dφ(t), covering the distance ds(t)=r·dφ(t) (this is the length of an arc of radius r and angle , according to a known formula of geometry).

From this follows:

ds(t)/dt = r·dφ(t)/dt or

v(t) = r·ω(t)



TranslationRotation
Speed
v(t)=s'(t)
v(t)=r·ω(t)
Angular Speed
ω(t)=φ'(t)
ω(t)=v(t)/r


The next concept is acceleration that needs its rotational analogue. Obviously, it's the first derivative of angular velocity or the second derivative of an angle of rotation by time.

Using the vector interpretation of angular velocity, we can consider angular acceleration as a vector as well. It is also directed along the axis of rotation.

During infinitesimal time interval dt an angular velocity ω(t) changes by dω(t).

From this follows relationship between linear acceleration a and angular acceleration α:

a(t) = dv(t)/dt = r·dω(t)/dt or

a(t) = r·α(t)



TranslationRotation
Acceleration
a(t)=v'(t)
a(t)=r·α(t)
Angular Acc.
α(t)=ω'(t)
α(t)=a(t)/r


Obviously, integrating the definitions of angular velocity ω and angular acceleration α for constant angular acceleration, we come up with formulas similar to those familiar from translational movement:

ω(t) = ω(0) + α·t

φ(t) = φ(0) + ω(0)·t + α·t²/2



Angular acceleration as a vector (as a pseudo-vector, to be exact), is colinear to the axis of rotation, because angular velocity is.

If rotation goes as on the above picture and the speed of rotation increases, the angular acceleration would be directed upwards, the same way as the angular velocity.

Monday, August 6, 2018

Unizor - Physics4Teens - Mechanics - Statics - Equilibrium





Notes to a video lecture on http://www.unizor.com



Equilibrium



Statics is a part of Mechanics that studies forces not as
quantitative measure of a motion they cause (that is a subject of
Dynamics), but from the more fundamental viewpoint of whether these
forces are or are not balanced, that might or might not cause the motion
of objects these forces act upon.



Historically, studies of static aspects of forces precede quantitative
studies of motion in Dynamics. People, first of all, were concerned with
how to build bridges and buildings, so that they stay in place and not
destroyed by gravity, wind or load. And a concept of equilibrium is a central point of Statics, it's goal and purpose.



Equilibrium, 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.



A person standing on a floor is at rest because two main forces that act
upon him, the gravity and the reaction of the floor are equal in
magnitude and opposite in direction, they balance each other, which
results in a state of equilibrium.



A person standing on the weights to check his weight is in a state of equilibrium
because its weight is balanced by elastic force inside the weights that
is equal in magnitude and opposite in direction to the force of
gravity. The equality in magnitude allows to measure the weight by
measuring the elasticity inside the weights.



A building is in a state of equilibrium because its each part's weight is balanced by equal in magnitude and opposite in direction force of reaction.



An airplane, flying horizontally on its route, is in a state of vertical equilibrium
because its weight is balanced by the lifting power of the air under
its wings, where the air pressure is larger because of the wings' shape.



Let's study forces applied to the same object and balance each other causing this object to be in equilibrium.

As we know, force is a vector. All forces acting on the same point-object can be added as vectors, using the rules of the Vector Algebra.

If the result is a null-vector, we have an equilibrium.



So, if there are N forces Fi (where 1 ≤ i ≤ N) acting simultaneously on the same point-object, the condition of equilibrium is

ΣFi = 0 .



Let's consider a few examples.



1. An object of weight W is hanging on three threads as pictured below.



Thread a is horizontal, thread b is at angle φ to horizon.

The system is in equilibrium.

What is the magnitude of tension forces Ta and Tb on threads a and b?



Solution

There are three forces acting at the point where all threads are connected:

vertical down - the weight W of the object;

tension Ta horizontally to the left along the a thread;

tension Tb at angle φ to horizon along the b thread.

Since the system is in equilibrium, this point where threads come
together does not move in any direction. In particular, it does not move
in a vertical, nor horizontal direction.

This is sufficient to determine magnitude of tension vectors Ta and Tb.

In the horizontal direction the force of gravity is irrelevant, so the
only two forces acting on out object in horizontal direction are tension
Ta acting to the left and a projection of tension Tb on the horizontal line acting to the right.

That gives the first equation about magnitudes of tension forces:

Ta = Tb·cos(φ)

Now consider the vertical direction. The thread a is irrelevant. So, the only two forces acting vertically, are weight W pulling down and a vertical component of Tb that is equal to Tb·sin(φ) pulling upwards.

This gives the second equation

Tb·sin(φ) = W

From these equations we derive:
Tb = W/sin(φ)

After which we can find Ta:

Ta = W·cos(φ)/sin(φ)



2. Two objects of mass M (larger, on an inclined plane) and m (smaller, hanging freely over the edge of an inclined plane) are connected with a thread that goes over a pulley.



What is the angle of an inclined plane φ for this system to be in equilibrium?

Ignore the friction.



Solution

Let's represent the weight of an object on an inclined plane as a sum of two forces:

one is perpendicular to the plane and balanced by a plane's reaction;

another is parallel to the plane and balanced by a tension of a thread.

To be in equilibrium, the force of weight of an object hanging freely
over the edge of an inclined plane must be equal in magnitude to a
tension of a thread (same thread, same tension as before).

Therefore, a component of the weight of an object on a plane that goes
parallel to a plane must be equal in magnitude to a weight of another
object.

M·g·sin(φ) = m·g

From this we derive

sin(φ) = m/M

φ = arcsin(m/M)

Monday, July 30, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rockets and Gravitation





Notes to a video lecture on http://www.unizor.com



Rocket in Gravitational Field



In the previous lectures we examined the motion of a rocket with no
external forces (like gravity or drag) acting on it and came with the
Rocket Equation, stating that

IF

m(t) is the mass of a rocket (including propellant) at time t and

V(t) is its speed in some inertial reference frame
(related to stars, for example, and positioned in such a way that a
rocket moves along one axis in a positive direction) and

m(t+Δt) is the mass after time interval Δt, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed ve and

V(t+Δt) is its speed after time interval Δt in the same inertial reference frame

THEN

the maximum increment of the rocket's speed

ΔV=V(t+Δt)−V(t) during this interval of time Δt is

ΔV = −ve·ln[m(t)/m(t+Δt)]



The equation above should be interpreted as the vector equation.

If inertial frame of reference is directed in such a way that the rocket
moves along one axis in positive direction and the exhaust is directed
backwards relative to a rocket's movement, the ve is negative. The mass during this process decreases, so m(t) is greater than m(t+Δt) and the logarithm is positive. This results in the positive ΔV, that is a rocket accelerates.

If the exhaust is directed forward relative to a rocket's movement, the ve is positive, ΔV is negative and a rocket decelerates.



Now let's add gravity as an external force that acts on a rocket.

There are two cases:

(a) when a rocket is launched from a planet to an orbit, gravity acts
against its movement, thus requiring extra effort by an engine to
overcome the gravity;

(b) when a rocket returns back to a planet for soft landing, gravity
acts in the direction of its movement, but we have to decelerate a
rocket using propellant exhausted also in the same direction, so it also
requires extra effort by an engine.

So, no matter how rocket moves in the gravitational field, an engine
should work harder to either launch it to an orbit or to slow it down
for soft landing.



We will consider the launching from the Earth case only.



Let's follow the same logic as in case of a rocket moving in empty space
with no forces involved and add the effect of gravity in the equation
of conservation of momentum.



1. At moment t the momentum of an entire system of a rocket with its propellant was equal to m(t)·V(t).



2. During the next time interval dt the rocket has exhausted m(t)−m(t+dt)=−dm(t) of propellant with constant relatively to a rocket speed ve. Since a rocket moves in some inertial system with speed V(t) and propellant moves relatively to a rocket with constant speed ve, the speed of propellant in the inertial system equals to ve+V(t). This resulted in the momentum of exhausted propellant at moment t+dt to be dm(t)·[ve+V(t)].

This equation should be interpreted in the vector form. When a rocket
accelerates, velocity vector of its movement and velocity vector of
exhausted propellant are opposite in their directions.



3. A rocket with remaining propellant at moment t+dt has mass m(t+dt)=m(t)+dm(t), velocity V(t+dt)=V(t)+dV(t) and momentum

[m(t)+dm(t)]·[V(t)+dV(t)]



4. When rocket leaves the planet, the force of gravity F=m(t)·g acts against its movement. During time dt it reduces the impulse of a rocket by

dt = m(t)·g·dt.



Now we are ready to apply the Law of Conservation of Momentum.

Item 1 above describes the momentum of the system at time t.

At the moment t+dt the momentum of the system is a combination of the momentum of the exhausted propellant during time dt
(see item 2 above) plus the momentum of the remaining rocket mass (see
item 3 above) plus impulse of the gravitational force (see item 4
above).



Equalizing these two momentums at time t and t+dt, according to the Law of Conservation of Momentum, we get the following equation:

m(t)·V(t) = −dm(t)·[ve+V(t)] + [m(t)+dm(t)]·[V(t)+dV(t)] + m(t)·g·dt.



We would like to express the dependency between rocket's speed and the
way it exhausts its propellant without mentioning the time parameter.
This can be done by using the following:

m'(t) = dm(t)/dt (by definition of the derivative and differential)

From this:

dt = dm(t)/m'(t)



The rocket equation above can be simplified. After this the equation looks like

0 = −ve·dm(t) + m(t)·dV(t) + dV(t)·dm(t) + m(t)·g·dt

Ignoring an infinitesimal of a higher order dV(t)·dm(t), the resulting equation looks like

m(t)·dV(t)+m(t)·g·dt=ve·dm(t)



Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then the differential equation of a rocket in the gravitational field looks like

dV(t) + g·dt = ve·d[ln(m(t))]

Replacing dt with its equivalent dm(t)/m'(t), we obtain an equivalent equation

dV(t)+g·dm(t)/m'(t) = ve·dm(t)



Expression on the right is positive because

(a) m(t) is a decreasing function,

(b) ln(m(t)), therefore, is also a decreasing function,

(c) differential of a decreasing function d[ln(m(t))] is always negative,

(d) ve is negative since it is a vector
directed against the movement of a rocket, which we consider as moving
to a positive direction of a coordinate axis,

(e) product of two negative values is positive.



As is obvious from this equation, unless dt is less than ve·d[ln(m(t))], the rocket will not move from the launching pad.

We can simplify this launching condition, using the following:

d[ln(m(t))] = [m'(t)/m(t)dt

This allows to express this condition as

g is less than ve·[m'(t)/m(t)]
or

ve·m'(t) is greater than g·m(t)



Integrating the differential equation of a rocket in the gravitational field on the interval Δt of time from the beginning of engine's work tbeg to the end of this period tend, we obtain the equation for an increment of the rocket's speed during this interval:

V(tend)−V(tbeg)+g·(tend−tend)=

= ve·
[ln(m(tend))−ln(m(tbeg))]=

= ve·
ln[m(tend)/m(tbeg)]=

= −ve·
ln[m(tbeg)/m(tend)]



That is,

ΔV(t) = −ve·ln[m(tbeg)/m(tend)] − g·(tend−tend)



The last equation does not take into consideration that the force of
gravity decreases with height. It's relatively precise only in the
beginning of the rocket's movement. Obviously, taking this factor into
consideration will complicate the calculations.



The next complication is the drag of the atmosphere, which is not that
important in theory, but for practical matters must be taken into
consideration.



Another important factor of launching is the planet's rotation. If we
launch a rocket to the East, the Earth's rotation helps to achieve
required speed.



All these and other complications make rocket science a rather involved theory.

Friday, July 27, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Calculations





Notes to a video lecture on http://www.unizor.com



Rocket Calculation 1



Let's use the rocket equation

ΔV = −ve·ln[m(tbeg)/m(tend)]

to calculate how much propellant must be taken by a rocket to reach an orbit.



Here ve is effective exhaust speed, m(tbeg) - mass of a rocket in the beginning of a time period during which a rocket's engine is working, m(tend) - mass of a rocket at the end of this period of acceleration or deceleration.



Recall that the minus sign in this equation signifies the vector
character of the movement: positive direction of the exhausted
propellant (that is, the same as the movement of the rocket) causes
negative increment in rocket's speed - deceleration, while the negative
direction of exhausted propellant (that is, opposite to the movement of a
rocket) causes increase in rocket's speed - acceleration.



Contemporary rocket engine can have a very high effective exhaust
velocity. The speed of about 4km/sec is mentioned in a few sources we
are familiar with. So, we can assume that

ve=4000m/sec.



An international Space Station's speed is about 7.8km/sec, as was calculated in one of the previous lectures on gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 7800m/sec



From this follows that

ln[m(tbeg)/m(tend)] =

= 7800/4000 = 1.95




Therefore,

m(tbeg)/m(tend) ≅ 7



So, the mass of a rocket at start is 7 times greater than its mass at
the end of its acceleration. For example, to launch 1,000 kg of useful
equipment and/or passengers to an International Space Station we will
need 6,000 kg of fuel.





Rocket Calculation 2



We still assume that

ve=4000m/sec.



A rocket that goes far from Earth needs about 11.2km/sec speed to escape Earth gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 11200m/sec



From this follows that

ln[m(tbeg)/m(tend)] =

= 11200/4000 = 2.8




Therefore,

m(tbeg)/m(tend) ≅ 16



So, the mass of a rocket, that is supposed to leave the Earth's gravity,
at start is 16 times greater than its mass at the end of its
acceleration. For example, to launch 1,000 kg of useful equipment and/or
passengers to Mars we will need 15,000 kg of fuel.

Wednesday, July 25, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Equation





Notes to a video lecture on http://www.unizor.com



Ideal Rocket Equation

(Tsiolkovsky's Equation)



The most important difference between the motion of a rocket and motions
analyzed before in this course is that the rocket propels itself by
throwing back part of its own mass (propellant), thus becoming lighter.
Its mass is variable. Before, in most cases, we were dealing with motion
of objects of some specific mass, not changing during the movement.



In this lecture we will analyze the motion of an ideal rocket that
throws back propellant with constant (relatively to the rocket) speed.
The formula that will be derived was suggested by Russian mathematician
K.E.Tsiolkovsky and is historically named after him, though he was not
the first to derive it.



In particular, we will analyze the dependency between loss of the total
mass of a rocket, throwing propellant backward in the absence of any
external forces, and its gain in speed caused by this process.



The final formula we will derive states that

IF

m(t) is the mass of a rocket (including propellant) at time t and

V(t) is its speed in some inertial reference frame
(related to stars, for example, and positioned in such a way that a
rocket moves along one axis in a positive direction) and

m(t+Δt) is the mass after time interval Δt, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed ve and

V(t+Δt) is its speed after time interval Δt in the same inertial reference frame

THEN

the maximum increment of the rocket's speed

ΔV=V(t+Δt)−V(t) during this interval of time Δt is

ΔV = −ve·ln[m(t)/m(t+Δt)]



The equation above should be interpreted as the vector equation.

If inertial frame of reference is directed in such a way that the rocket
moves along one axis in positive direction and the exhaust is directed
backwards relative to a rocket's movement, the ve is negative. The mass during this process decreases, so m(t) is greater than m(t+Δt) and the logarithm is positive. This results in the positive ΔV, that is a rocket accelerates.

If the exhaust is directed forward relative to a rocket's movement, the ve is positive, ΔV is negative and a rocket decelerates.



Here is how to derive this formula.

First of all, let's recall that in the absence of external forces the
combined momentum of motion of a system of objects is constant.
This is the Law of Conservation of Momentum.

Let's apply this law to our situation.



Assume, we are comparing the momentum of the system at times t and infinitesimally incremented t+dt.



1. At moment t the momentum of an entire system of a rocket with its propellant was equal to m(t)·V(t).



2. During the next time interval dt the rocket has exhausted m(t)−m(t+dt)=−dm(t) of propellant with constant relatively to a rocket speed ve. Since a rocket moves in some inertial system with speed V(t) and propellant moves relatively to a rocket with constant speed ve, the speed of propellant in the inertial system equals to ve+V(t). This resulted in the momentum of exhausted propellant at moment t+dt to be dm(t)·[ve+V(t)].

This equation should be interpreted in the vector form. When a rocket
accelerates, velocity vector of its movement and velocity vector of
exhausted propellant are opposite in their directions.



3. A rocket with remaining propellant at moment t+dt has mass m(t+dt)=m(t)+dm(t), velocity V(t+dt)=V(t)+dV(t) and momentum

[m(t)+dm(t)]·[V(t)+dV(t)]



Now we are ready to apply the Law of Conservation of Momentum.

Item 1 above describes the momentum of the system at time t.

At the moment t+dt the momentum of the system is a combination of the momentum of the exhausted propellant during time dt (see item 2 above) plus the momentum of the remaining rocket mass (see item 3 above).



Equalizing these two momentums at time t and t+dt, which is the consequence of the Law of Conservation of Momentum, we get the following equation:

m(t)·V(t) =

= −
dm(t)·[ve+V(t)] +

+ [m(t)+dm(t)]·[V(t)+dV(t)]



This can be simplified. After this the equation looks like

0 = −ve·dm(t) + m(t)·dV(t) +

+
dV(t)·dm(t)




The last member in this equation dV(t)·dm(t) is an infinitesimal of the higher order that we can remove from this equation, and the resulting equation looks like

m(t)·dV(t) = ve·dm(t)



Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then our equation looks like

dV(t) = ve·d[ln(m(t))]



Integrating this on the interval Δt of time from the beginning of engine's work tbeg to the end of this period tend, we obtain the equation for an increment of the rocket's speed during this interval:

V(tend) − V(tbeg) = ΔV(t) =

= ve·
[ln(m(tend))−ln(m(tbeg))]=

= ve·
ln[m(tend)/m(tbeg)]=

= −ve·
ln[m(tbeg)/m(tend)]



That is,

ΔV(t) =−ve·ln[m(tbeg)/m(tend)]



So, the increment of a rocket's speed during a period of Δt=tend−tbeg,
when its engine works, exhausting the propellant, equals to a product
of the speed of exhausted propellant times a logarithm of a ratio of the
rocket's mass at the beginning of this period to its mass at the end of
it.



The minus in front of a formula is very important. This is a vector
equation and, if the exhaust is directed back relatively to rocket's
movement (that is, ve is negative), the
increment of a rocket's speed is positive, a rocket accelerates; if,
however, the exhaust is directed towards a tocket movement (that is, ve is positive), the increment of a rocket's speed is negative, a rocket slows down.



The formula above is the Tsiolkovsky's formula and is called the "ideal rocket equation".

Tuesday, July 17, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Spring Oscillation





Notes to a video lecture on http://www.unizor.com



Spring



Consider a point-object of mass m, hanging vertically at
the lower end of a weightless spring, that is fixed at the upper end.
Under the weight of this object a spring will stretch a little from its
neutral position.



The Hook's Law for a spring, which will be used to solve this problem, involves a spring's elasticity constant k, that we assume is given.



Let's stretch this spring even more, so that the distance between an object at its bottom and a spring's neutral level is L and let it go without any push.



Our task is to analyze the oscillation of the object as a function x(t) of its vertical deviation from a spring's neutral position.







The obvious initial conditions of the motion of our object are:

x(0) = L

x'(0) = 0



There are two forces acting on our object:

(a) its weight W, directed vertically down and equal in magnitude to m·g, where g is the acceleration of free falling

W = m·g

(b) the spring's elasticity force F, equal in magnitude to a coefficient of elasticity k
multiplied by a displacement of the spring's bottom end from a neutral
level; the direction of this force is always against the direction of
the displacement

F = −k·x(t)



The resultant of the superposition of these two forces can be equated to
mass times acceleration of the object, according to the Newton's Second
Law:

m·g − k·x(t) = m·x"(t)

This is the differential equation that defines the movement of our object.



We don't have to resort to modifying this differential equation with an
approximate one to be able to solve it. It is fully solvable and the
general solution of this linear differential equation of the second
order is

x(t) = C1·cos(t·√k/m) +

+ C2·sin(t·√k/m) + m·g/k




Now we can apply the initial conditions to determine constants C1 and C2.

Since x(0) = L,

C1 = L − m·g/k

Since x'(0) = 0,

C2 = 0



This produces the following expression for x(t):

x(t) = (L − m·g/k)·cos(t·√k/m) +

+ m·g/k




Interestingly, if

L − m·g/k = 0 or L·k = m·g

(which means that the weight W=m·g is balanced by the force of spring's elasticity F=−L·k in its initial position with our object at its end) then there are no oscillations, and the object will remain at distance L=m·g/k from a spring's neutral position.



If there are oscillations, their period is

T = 2π√m/k

Monday, July 16, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Pendulum





Notes to a video lecture on http://www.unizor.com



Pendulum



We will analyze the ideal (mathematical) pendulum, which is a mechanical device placed near the surface of a planet with free fall acceleration g (to have the gravitational force acting on it) that consists of a point-object of certain mass m, hanging on a weightless non-stretchable thread of length L, fixed at the other end, so that the hanging on it object has freedom of motion.





Assume that at time t=0 we have tilted a point-object at the end of thread of a pendulum by an angle α0 from vertical. Then we let it go without any push.

Our task is to determine, how an angle of deviation of this pendulum
from a vertical changes with time, that is we have to find the function α(t).



We can say now that initial (at time t=0) position of a pendulum is

α(0) = α0

Considering that linear displacement d along a circular trajectory of a radius L and its angular displacement α are related by a formula

d = L·α,

the initial condition of not pushing a pendulum, which means "no initial
linear velocity along its trajectory", means that the first derivative
of angular displacement is zero:

α'(0) = 0



Having these initial conditions, we'll determine the equation that function α(t) must satisfy, using the Newton's Second Law.

The force of gravity P=mg can be represented as a sum of two forces:

- a force along a pendulum's thread, that is completely balanced by a
thread's reaction, which results in constant distance of a point-object
at the end of a thread from its other (fixed) end; this force constrains
the movement of a point-object within a circular trajectory and is
equal to

mg·cos(α(t))

- a force tangential to a circular trajectory of a point-object at the
end of a thread; this force is the source of movement along a trajectory
and is equal to

F = −mg·sin(α(t))

(negative sign is used because the force is always directed in an opposite direction to the movement)



The force tangential to a circular trajectory is the one that
accelerates our point-object. Since the displacement along a circular
trajectory is, as we indicated, d=L·α, the linear acceleration along a trajectory is equal to a second derivative of this expression by time

a = L·α"(t)

The Newton's second law states that

m·a = F

which results in the following differential equation for function α(t):

m·L·α"(t) = −m·g·sin(α(t))



The good news is that we can reduce this by mass m, which
means that the oscillation of a pendulum does not depend on a mass of a
point-object at its end, but only on the length of a thread L and acceleration of free falling g.

So, we deal with an equation

L·α"(t) = −g·sin(α(t)) or

α"(t) = −(g/L)·sin(α(t))



Another good news is that this is a differential equation of the second
order (highest derivative is the second one) and we have two initial
conditions for a function α(t) at t=0 and for its first derivative α'(t) at t=0.

This fully identifies the function α(t).



Unfortunately, the bad news is that this differential equation cannot be
solved in terms of simple algebraic functions, but only numerically
tabulated using computer.



But physicists, in their endless quest for simple solutions to
complicated problems of the Universe, have decided that within certain
boundaries they can simplify the above equation to approximate its
solution, using simple algebraic functions.

This simplification is based on the fact that, when an angle is
relatively close to zero, its sine is not much different from the value
of an angle itself (in radians). This is based on a famous limit

limx→0[sin(x)/x] = 1

So, for relatively small angles around a vertical, the oscillations of a
pendulum can be approximately expressed by an equation obtained by
replacing sin(α(t)) with simple α(t).



This produces the following equation:

α"(t) = −(g/L)·α(t)

This is a simple linear differential equation with general solution

α(t) = C1·cos(√g/L·t) + C2·sin(√g/L·t)

where C1 and C2 are some constants.

To determine the values of these constants, we will use the initial conditions:

α(0) = α0 and α'(0) = 0

This results in the following:

C1·cos(0) + C2·sin(0) = α0

and

−C1·sin(0) + C2·cos(0) = 0

from which follows that

C1 = α0

and

C2 = 0



Solution to our problem, therefore, is

α(t) = α0·cos(√g/L·t)



This solution represents harmonic oscillation with an amplitude

A = α0

and period

T = 2π /g/L = 2π·√L/g



The above approximate solution satisfies to a certain degree
physicists and is accepted as the one describing relatively small
harmonic oscillations of pendulum around a vertical.

Oscillations on a bigger scale (say, with initial angle of deviation
around 45° or so) do not conform to this formula and are not harmonic.

Friday, July 13, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Weight





Notes to a video lecture on http://www.unizor.com



Weight



Weight of an object, by definition, is the force of gravity a
planet attracts this object with. Usually, the word "weight" implies the
magnitude of this force; its direction is, obviously, always towards a
center of a planet.



So, weight is not a characteristic of an object itself, it's a
characteristic of an object relative to a planet. In most cases, this
planet is our Earth, though we sometimes say, for example, that a
particular object weighs on the Earth 6 times more than on the Moon.
This only means that the force of gravity on the surface of the Moon is 6
times weaker than on the surface of the Earth.



Do we feel weight as the force of gravity?

Not quite. What we can feel is pressure (reaction force) from the surface we stand on, that equalizes gravitational force to hold us at fixed position on a floor or on a ground.

If there is no support (like for a person jumping with a parachute from
an airplane before a parachute is open, if we ignore the air
resistance), we don't feel weight, we are weightless. We have different
senses, but not a sense of gravity.

So, feeling weightless is not really an absence of gravity, it's absence
of a reaction force that balances the gravity (equal in magnitude and
opposite in direction) and holds us fixed relatively to a planet.

This reaction force is not just against our feet, when we stand on the
floor, it's everywhere inside our body as well, since the body maintains
its shape. We feel this pressure of a reaction force everywhere inside.
That's why it's very difficult to emulate the gravity with some special
equipment on a spaceship.



People on a spaceship with non-working engines flying around the Earth
on an orbit feel weightless, because they are constantly falling towards
the Earth together with a spaceship (no support!) from the straight
line trajectory tangential to an orbit; planet attracts them with
gravitational force, and only because of the speed, they maintain
constant distance from the planet.



Since weight is a force, it is measured in units of force, like newtons in SI.

The weight of an object of mass m on a surface of a planet of mass M and radius R is, as we know,

W = G·M·m /

where G is a universal gravitational constant,

G = 6.674·10−11 N·m²/kg²



Since we are talking about weight as a force, which is a subject to the
Newton's Second Law, we can determine the acceleration this force causes
to an object of mass m, if acts alone:

a = W/m = G·M /



Notice that on the surface of Earth this acceleration is constant since
all components of this expression (gravitational constant G,
mass of Earth M and its radius R) are constants.

So, we can calculate this constant once and for all and, knowing the mass of an object m,
we can determine its weight by multiplying it by this constant, which
is, as we determined in the previous lecture, an acceleration of free
fall, which on the surface of Earth is traditionally symbolized by
letter g:

g = G·M /



The value of this constant is, approximately, 9.8 m/sec².
But, to be precise, it's not the same at different points on the Earth
because the shape of the Earth is not exactly a sphere and its mass is
not uniformly distributed within its volume.

Moreover, it obviously changes with height (getting smaller) since the higher elevation is equivalent to a greater radius R (distance to a center of the Earth) of an object.



Now we can say that for an object of mass m the weight on the surface of the Earth is W=m·g=9.8·m. If mass m is measured in kilograms, this weight is measured in newtons.



Analogous calculation for other planets, based on their mass and radius, show the following values of free falling acceleration:

on Sun - 274.1 m/sec²

(objects are 28 times heavier on Sun than on Earth),

on Jupiter - 25.93 m/sec²

(objects are about 2.6 times heavier on Jupiter than on Earth),

on Moon - 1.625 m/sec²

(objects are about 6 times lighter on Moon than on Earth).



Historically, the weight is rarely measured in newtons. More customary units are:

1 pound (abbreviated lb) equals to 4.44822 newtons - the weight of an object of mass 0.454 kg on Earth;

1 kilogram-force (usually, simply called 1 kilogram, skipping "-force", and abbreviated kgf, but plain kg 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 newtons;

and others.

Tuesday, July 10, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Free Falling





Notes to a video lecture on http://www.unizor.com



Free Falling



Free falling is a movement of an object on a surface of a planet
relative to this surface, when the only force acting on this object is
the gravitational force of a planet.



Our task is to describe this movement in mechanical terms of force, mass and acceleration.

In this task we will assume that

(a) an object in question is a point-object of mass m,

(b) a planet has a spherical form and its mass M is uniformly distributed within its volume,

(c) a planet has a radius R,

(d) [an important assumption that can be justified by complex
calculations] we can model the combined forces of gravitation between
all microscopic particles inside a planet and our object in question as a
gravitational force of a point-object of mass M positioned at the center of a planet.



In this case the one and only force of attraction acting on an object
and directed towards the center of a planet can be expressed using the
Law of Gravitation as follows:

F = G·M·m /

where G is a gravitational constant,

G = 6.674·10−11 N·m²/kg²



Knowing the force of gravity F and mass of an object m, we can determine the acceleration using the Newton's Second Law:

a = F/m = G·M /



Notice that this acceleration does not depend on m - mass of an object, which means that all objects fall on the surface of a planet with the same acceleration.

An interesting aspect of this formula is that we can imagine how to
measure an acceleration (easy) and radius of a planet (more difficult,
but possible), while we have no idea how to measure the mass of a
planet.

So, this formula is used exactly for this purpose - to determine the mass of a planet, resolving the formula above for M:

M = a·R² /G



Experiments show that on the surface of our planet Earth the acceleration caused by gravitational force is approximately 9.8 m/sec².

The radius of Earth is approximately 6.4·106 m.



From this we can calculate the mass of Earth (in kilograms - units of mass in SI):

M≅9.8· 6.4²·1012/(6.674·10−11)

The result of this calculation is

M ≅ 6·1024 kg



Let's solve a different problem now. We'd like to launch a satellite
around the Earth that circulates around the planet at height H. What linear speed should a satellite have to stay on a circular orbit?



We know from Kinematics that an object rotating along a circular trajectory of radius r and angular speed ω has acceleration a=r·ω².

In terms of linear speed V=r·ω along an orbit this formula looks like

a = V²/r



Since the radius of an orbit is the radius of Earth R plus the height above its surface H, we should replace r in this formula with R+H.



The force of gravity is the only force acting on a satellite and the
only source of its acceleration towards the Earth, so the acceleration
above must be equal to acceleration of a free fall of a satellite. Here
we will take into consideration already known mass of Earth and use
distance from the center of the Earth to satellite as R+H, where R is the radius of Earth and H is a height above the Earth's surface.



The acceleration of a free fall to Earth at height H above the surface, using its radius R and already calculated mass of Earth M, is:

a = G·M/(R+H)²



Therefore, equating the acceleration of free fall to acceleration of an
object rotating along a circular orbit, we come to the following
equation:

/(R+H) = G·M/(R+H)²

from which we derive the value of required linear speed V:

V = √G·M/(R+H)



For example, International Space Station rotates around our planet on a height of about 400 kilometers (4·105 meters).

That means that, to stay on an orbit, it should have linear speed of

V ≅ 0.78·104 m/sec

which is about 28,000 km/hour.

Monday, July 9, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Gravity





Notes to a video lecture on http://www.unizor.com



Gravity



We all know a lot about gravity, weight, weightlessness, rockets flying on orbits calculated based on the laws of gravity etc.



But what is gravity?



A short answer is: we don't know. It's like most of us can use a smart
phone and Global Positioning System (GPS), but don't know how and why it
works. It just works, we know what to do to effectively use it, but
have no idea about the real mechanism that allows us to use it.
Obviously, designers and engineers who created these technological
marvels know, but most of people don't.



With gravity it's similar. We know it exists, we can use it, we feel it,
but we don't know the underlying reason why it is what it is.

To be more precise, physicists have certain ideas about the source of
gravity, but they are rather vague, on the level of hypothesis.

Therefore, we skip this foundational discussion about why gravity
exists, what is an underlying mechanism of its work. We will just use it
as we use GPS without getting much deeper.



To use a computer game, we just have to know its rules and controls, we don't have to know what software is inside.

To use gravity, it's not necessary to know its underlying mechanism, we
just need to know its properties, and that's the subject of this
lecture.



The first fundamental property of gravity is that all objects we deal
with attract other objects. This effect of attraction is called gravity.



Attraction is a force.

Since we usually model physical objects as points, this force is
directed along the line connecting these point-objects and pushes them
towards each other.

It is also important to note that the Newton's Third Law says that the force point-object B attracts point-object A is paired with the same in magnitude and opposite in direction force point-object A attracts point-object B.



In more complicated cases of objects that cannot be considered as
points, we can assume that every tiny peace of each object, which can be
modeled as a point-object, is attracted to every other tiny peace. Then
some process of integration of all these forces might be used to
determine the resultant forces. But we will rarely deal with this type
of gravitation, most of cases we will consider will involve
point-objects.



Forces change the velocity. Therefore, gravity, which is the force
observed for any type of object, causes change of motion of objects. If
there is only one point-object in the Universe, it will maintain its
inertial motion along a straight line with constant velocity. As soon as
another object appears somewhere, the force of gravity will cause a
change in the inertial movement of the first object.



Our next question is: how exactly forces of gravity change the motion of objects?



Different objects attract differently.

Consider some probe object A in inertial motion along a straight
line with constant velocity. For example, it flies in our
three-dimensional space along the X-axis in positive direction, going
through point of origin of coordinates {0, 0, 0} at moment in time t=0 towards positive infinity.

Let's measure the degree of the change of its motion, when at the later moment of time t=1 another object B appears at the origin of coordinates {0, 0, 0} and stays in this fixed position. This object B possesses the property of gravitational attraction with object A and, therefore, will slow down the velocity of object A, pulling it back to the origin of coordinates, so object A
will decelerate. Measuring this deceleration and knowing the mass of
objects involved, we can measure the force of attraction between objects
A and B using the Newton's Second Law.



Our observations show that different objects B will cause different decelerations of probe object A and the same object B causes different decelerations of probe object A at different distances between them. We conclude then that gravitational force of attraction between objects A and B depends on gravitational properties of objects themselves and on distance between them.



Our purpose is to analyze what is the gravitational property of any
object, how to measure it and how the force of gravity depends on it and
the distance between objects.



The situation with distance is easy.

Experiments with the same objects showed that the gravitational force of
attraction between them weakens with distance in inverse proportion to a
square of this distance. In other words, if the distance between any
two objects A and B doubles, the gravitational force of attraction weakens by a factor of 4.

So, it is sufficient to establish the gravitational force between two
objects at a unit length (say, 1 meter), after which the gravitational
force between these objects at any distance D will be that force at a unit distance divided by a factor D2.



Let's discuss now the gravitational property of an object, its ability to attract other objects, which in Physics is called gravitational mass of an object.



An experimental fact is that two identical objects, B1 and B2 combined together, attract twice as strongly as only one of them, say B1, providing they attract the same probe object A, and the relative position of participating objects is the same.

That means that gravitational mass is additive and the gravitational force is proportional to gravitational mass.



Let's choose one particular probe object A and assign it a gravitational mass of a unit and another identical object B. Since they are identical, the gravitational mass of object B is also a unit.

Then, comparing the attraction between this unit probe object A and identical unit object B at the unit distance with the attraction of any other object C to the same unit probe object A on the same unit distance, we can assign a gravitational mass to that other object C. Since gravitational mass is additive, the stronger the gravitational force of object C - the proportionally greater is its gravitational mass relative to a unit object B.



Notice, that additive property of gravitational mass is similar to a property of inertial mass, which is also additive. This is precisely the reason why both properties are call mass.



The analogy goes further. Another experimental fact is that two different objects of the same inertial mass have exactly the same gravitational mass,
that is they attract equal probe objects on equal distance equally.
From this follows that the quantitative difference between inertial mass and gravitational mass is just in units of measurement.

Based on this, it was decided to measure the gravitational mass in exactly the same units as inertial mass and, by definition, say that an object of 1 kilogram of inertial mass has 1 kilogram of gravitational mass, which, quantitatively, fully characterizes the gravitational properties of an object.



When we talk about gravity, 1 kilogram is a measure of gravitational attraction of an object, its gravitational mass. When we discuss inertia, motion, force, 1 kilogram is a measure of an object's inertial mass.



Let's derive the formula that expresses the force of gravity between two
point-objects in terms of their gravitational masses and distance
between them.



We already know that the force of gravity is proportional to a
gravitational mass, but, since we always deal with two point-objects,
the force must be proportional to a gravitational mass of each of them,
that is it is proportional to their product.



We also know that the force of gravity is inversely proportional to a square of a distance between objects.



These two factors lead to the following formula for the force F of gravity between two point-objects with gravitational (and inertial, as we defined) masses M1 and M2 at distance r between them:

F = G·M1·M2 /

where G - a constant of proportionality, since the units
of force (N - newtons) have been defined already, and we want to measure
the gravitational force in the same units as any other force.



This formula was presented by Sir Isaac Newton in 17th century, though other scientists, like Robert Hooke claimed it as well.

Physicists call this formula the Newton's Law of Universal Gravitation.



To determine the constant G in this formula, all we need
to do is to place two objects of inertial (and gravitational, as we
defined) mass of 1 kilogram each at the distance of 1 meter and measure
the force of gravity between them by measuring an acceleration they
develop as a result of gravitational force. This force (in newtons) will
be quantitatively equal to a gravitational constant G.

This measurement shows a very weak force, and the gravitational constant equals to

G = 6.674·10−11 N·m²/kg²



Finally, let's attempt to explain the phenomenon of weakening of the
gravitational force inversely proportional to a square of a distance
from the gravitating object.

This is not really a theoretical proof, but a reasonable explanation of this fact.



Assume that the source of gravitational force around an object is
something similar to tentacles of an octopus with objects of larger
gravitational mass corresponding to greater number of tentacles. The
gravitational grip, presumably. depends on the density of tentacles per
unit of covered area.

To propel gravity on a longer distance the tentacles should be longer, while their quantity remains the same.

Now, the longer these tentacles - the more is the area they have to spread around. This area for tentacle of the length r is a surface of the sphere of this radius, that is 4r².
So, area these tentacles are supposed to cover is increasing as a
square of their length, which causes a gravitational grip to be weaker
in exactly the same proportion.

Thursday, June 7, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 3





Notes to a video lecture on http://www.unizor.com



Friction Problems 3



Problem A



An object of mass M can slide on a horizontal table. Two weights of masses M1 and M2 are attached to it on both sides of a table with weightless thread as on this picture.



Assuming mass M1 is greater than M2, the object will start sliding to the left under a pull of a bigger weight.

Tensions T1 and T2, as well as friction act on an object on a table.

Assume that this object, as a result of actions of these forces, moves with acceleration a.

What is the friction coefficient μ and magnitudes of tensions T1 and T2?



Hint:

Use the Second Newton's Law for each of three objects (one on a table and two weights) participating in the motion.



Solution:



For object on a table:

T1 − T2 − μ·M·g = M·a

For weight on the left:

M1·g − T1 = M1·a

For weight on the right:

T2 − M2·g = M2·a

Got linear system of three equations with three unknowns: friction coefficient μ and magnitudes of tensions T1 and T2.

Friday, June 1, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 2





Notes to a video lecture on http://www.unizor.com



Friction Problems 2



Problem A



There is a ramp of mass M and angle φ to horizon, lying, but not fixed, on a horizontal surface (it's like a triangular prism lying on a side).

An object of mass m slides down this ramp. As it slides, the ramp also moves along the horizontal surface it's lying on.

What is the acceleration of a ramp relative to a horizontal surface, as
object slides down, if the coefficient of friction between an object and
a ramp is μ0, and the coefficient of friction between a ramp and a horizontal surface is μ1?

The free fall acceleration is g, so the weight of an object of mass m is m·g.



Hint:

(1) Review Problem B from Mechanics - Dynamics - Superposition of Forces - Inclined Plane of this course.

(2) Take into account that, according to the Third Newton's Law,
friction between an object and a ramp is, on one hand, a force exhorted
by a ramp that pulls object uphill against its sliding downhill and, on
the other hand, is an action of an object on the ramp in an opposite
direction, slowing its horizontal movement.

Thursday, May 31, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Friction - Problems 1





Notes to a video lecture on http://www.unizor.com



Friction Problems 1



Problem A



There is an object of mass m on an inclined plane that makes angle φ with horizon.

To move this object up on this inclined plane with constant speed you need to apply a force Fup directed uphill.

The coefficient of friction is unknown.

Let's assume that, left by itself, the object would slide down the
inclined plane under its own weight. The free fall acceleration is g, so the weight of an object of mass m is m·g.

What is the acceleration a of an object when it moves downhill under its own weight?



Hint

Fup = m·g·sin(φ)+μ·m·g·cos(φ)

a = g·sin(φ)−μ·g·cos(φ)



Answer:

a = 2·g·sin(φ) − Fup/m





Problem B



An object is lying on a horizontal platform that moves with acceleration a=10 m/sec².

The coefficient of kinetic friction between an object and a surface of a platform is μ=0.3, while coefficient of static friction is μs=0.4.

The free fall acceleration is g=9.8m/sec², so the weight of an object of mass m is m·g, but mass m is unknown.

(a) How the object will behave?

(b) Why was coefficient of static friction μs given?

(c) What is the acceleration of the object relative to the ground and relative to the platform?



Answer:

(a) The object will slide back along the platform's surface, but forward relative to the ground.

(b) If the coefficient of static friction is too high, the object will
not change its position relatively to a platform, and the next question
would make no sense.

(c) Relative to the ground acceleration is a0=g·μ=2.94m/sec².

Relative to the platform acceleration is a1=a0−a=−7.06m/sec²

Tuesday, May 29, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Superposition of Forces ...





Notes to a video lecture on http://www.unizor.com



Motion on Inclined Plane



Problem A



Consider a slope fixed on the ground that makes an angle φ
with horizon. Let's analyze the motion of an object, as it slides down a
slope under its own weight. Our task is to determine its acceleration
along the slope.



First of all, we have to choose a reference frame - a system of coordinates suitable for this problem.

The main force acting on this object is its weight W - a vector of gravity
directed vertically down to the ground. That prompts us to choose a
frame of reference with horizontal X-axis and vertical Y-axis. So, the
vector of weight has non-zero Y-component and zero X-component: W = {0,W}, where W is a magnitude of vector of weight W. However, this obvious choice is not the best in a case like this.



There are usually numerous forces involved in an experiment and only few
objects. In this case there is only one object. So, instead of catering
to one particular force, like weight, it's better to simplify the motion of a single object.

Much more convenient frame of reference would be the one, where our object in motion has only one non-zero coordinate.



Recall that there must be another force acting on an object - reaction force of the slope, that prevents an object to go vertically down to the ground through a slope and forces it, in cooperation with gravity force, to slide along a slope. This reaction force R
is always perpendicular to the surface, where an object is on (in this
case, perpendicular to a slope), and its magnitude is such that the resultant of the weight W and reaction R is directed along a slope.



Consider a frame of reference with X-axis going along the slope, where the object slides down and Y-axis perpendicular to it.



Granted, the weight now has both coordinates non-zero:

(a) perpendicular to a surface of a slope and (on the picture above) directed along negative Y-coordinates vector WR with magnitude WR=W·cos(φ), that causes reaction force R, that is equal in magnitude and opposite in direction to WR, and

(b) parallel to a slope, directed (on the picture) towards negative X-coordinate, vector WF with magnitude WF=W·sin(φ), that is the cause of motion of our object down a slope.



As it is pictured, both components of weight are negative since they are
directed towards negative direction of the X- and Y-axes:

W = {−W·sin(φ),−W·cos(φ)}

Force R , as opposite and equal in magnitude to WR, is

R = {0,W·cos(φ)}.



The resultant of three vectors WR, WF and R is WF, directed down a slope, equaled in magnitude to W·sin(φ).



Therefore, an object of mass M will slide down a slope with acceleration equaled in magnitude to

a = WF/M = W·cos(φ)/M

Since weight and mass of an object are related as W=M·g, where g is a known acceleration of free falling (9.8 m/sec² on the Earth ground), the resulting acceleration equals in magnitude to

a = M·g·sin(φ)/M = g·sin(φ)

In vector form in the chosen reference frame:

a = {−g·sin(φ), 0}





Problem B



Consider an object A of mass m, sliding without friction
on a slope of slide B, which itself lies on a horizontal surface and can
slide on it without friction. An angle of a slope of slide B is φ with horizon, its mass is M.

Our task is to determine acceleration a of slide B.

Let's analyze the motion of object A on a slope and the motion of a
slide B, as it moves horizontally, as a result of the weight of object A
on it.





As object A presses down with its weight, it slides downhill along a
slope of a slide B. At the same time slide B moves to the right (on the
picture above).

The horizontal component of the pressure WR of
an object A on slide B perpendicularly to its slope is the cause of the
motion of a slide B. However, this pressure is not the same as in the
problem A above. It will be less. Its opposite reaction force R, that is equal in magnitude to WR, but acting on the object A, will also be less than in the problem A above.

The resultant of weight W and reaction force R is force WF that is not parallel to a slope, but tilted downwards.

So, the combination of object A sliding downhill on a slope and slide B
movement to the right produces the resultant move of object A that is
not parallel to a slope, neither it is directed vertically down, but
will be somewhere in-between.



Let's consider the same reference frame as in the problem A above. Now
both object A and slide B, as they move, have both X- and Y-components
not equal to zero.



Consider only Y-coordinate of the A object now and Y-components of forces acting on it.

In the direction of Y-axis the force WF = W + R, acting against object A, has value

WFy = Wy + Ry = R − W·cos(φ)



This force R − W·cos(φ), according to the Newton's Second Law, should be equal to object A's mass m, multiplied by a Y-component of its acceleration, which so far is unknown.



Let a be an acceleration of slide B in the direction of
its horizontal movement. Since displacement of object A in the direction
of the Y-axis (perpendicularly to a slope) equals to horizontal
displacement of a slide B multiplied by sin(φ), the
acceleration of slide B in the horizontal direction and acceleration of
the object A in a direction perpendicular to a slope, maintain the same
factor.

Therefore, the acceleration of object A perpendicularly to a slope of slide B equals to a·sin(φ), where a is the acceleration of the slide B that we have to determine in this problem.



The Newton's Second Law for object A in the direction of the Y-axis (perpendicular to a slope of slide B) is

R − W·cos(φ) = −m·a·sin(φ)

(minus on the right because the acceleration of object A relative to Y-axis is negative).

This equation is the first in a system of two equations that include two unknowns R and a.



On the other hand, a horizontal component of vector −R is the cause of horizontal acceleration a of slide B that has mass M.

Therefore,

R·sin(φ) = M·a

This is the second equation in a system of two equations that include two unknowns R and a.



Solving this system as follows.

From the second equation:

R = M·a/sin(φ)

Substitute it in the first equation:

M·a/sin(φ) − W·cos(φ) =

= −a·m·sin(φ)




The solution for horizontal acceleration of slide B is

a = W·cos(φ) /

/
[m·sin(φ) + M/sin(φ)] =

= W·sin(φ)·cos(φ) /

/
[m·sin²(φ)+M]



So, our final result for an acceleration a of slide B, as it moves horizontally, is
W·sin(φ)·cos(φ)/[m·sin²(φ)+M]