## Friday, October 12, 2018

### Unizor - Physics4Teens - Mechanics - Work - Golden Rule of Mechanics

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

Golden Rule of Mechanics

Let's summarize what we have learned about mechanical work in the previous lecture.

1. In a simple case of motion along a straight line with a constant force F acting along a trajectory, the most important parameter that quantifies the result of the action is work defined as W=F·S, that fully characterizes and is fully characterized by speed V of an object:

W = F·S = m·V²/2

In particular, it means that increasing the force by a factor of N and decreasing the distance it acts by the same factor of N would result in the same final speed of an object.

So, we can "win" in distance, but we will "lose" in force and vice versa.

This is the first example of the Golden Rule of Mechanics - there
are many ways to achieve the result, you can reduce your distance, but
you will have to increase the force or you can reduce the force, but you
will have to increase the distance.

In short, as we mentioned above, whatever you win in distance you lose in force and vice versa.

2. In case of a constant force acting at an angle to a straight line trajectory, the difference is only a factor cos(φ), where φ is an angle between a force and a direction of a trajectory. In vector form it represents the scalar product (F·S).

So, the Golden Rule works exactly as above.

3. Recall the formula for work of a force F pushing an object of weight P up along an inclined plane of the length S making angle φ with horizon to the height H:

W = F·S = P·H

We've proven this in the previous lecture and, as you see, it's
independent of the slope of an inclined plane. However, the minimum
effort we have to apply as a force to move an object up the slope is F=P·sin(φ), while the distance equals to S=H/sin(φ).

We can reduce the effort (the force F) by using an incline of a smaller slope, but that would lengthen the distance S we have to push an object.

So, again, we see the Golden Rule in action.

4. Consider lifting some heavy object of the weight P using a lever, applying the force F to the opposite to an object end of the lever.

This is a case of equilibrium in rotational motion and the balance can
be achieved by equating the moments of two forces acting against each
other:

F·Lf = P·Lp

If Sf is the distance the force F acts down and Sp is the distance our object moves up,

Sf/Sp = Lf/Lp and

F·Sf = P·Sp

By using a lever with longer arm Lf for the application of force, we can proportionally reduce the force F
achieving the same result - lifting an object to certain height. An
inverse is true as well - we can shorten the arm and proportionally
increase the force.

In any case, the Golden Rule of Mechanics is observed: "winning"
in force - proportionally "losing" in distance or "winning" in distance"
- proportionally "losing" in force.

All the above examples emphasize the importance of the concept of mechanical work as the quantitative measure and characteristic of the purpose and the result of applying a force. The so-called Golden Rule of Mechanics is just a catchy term that underscores the importance of the concept of work.

Acting with the force to achieve certain goal necessitates performing
certain amount of work that depends on the goal, not on how we achieve
it
.

## Wednesday, October 10, 2018

### Unizor - Physics4Teens - Mechanics - Work - Definition

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

Definition of Mechanical Work

Why do we apply a force to an object?

Usually, to achieve certain result, like to accelerate it to a certain speed or to move it from one point to another.

Straight line motion with a force acting along a trajectory

Consider an acceleration of a car by pressing the gas pedal. As a
result, a car will reach certain speed and then, in absence of the
traffic lights, disregarding friction and air resistance, it goes by
inertia, maintaining this constant speed.

A car with a more powerful engine will need a shorter distance to reach
certain speed. A car with a weaker engine accelerates on a longer
distance to reach the same speed.

It seems, there is some relationship between the force, the distance it
is applied to an object and the final speed reached by this object as a
result of the acting force.

Let's consider the final speed of a constant force F acting on an object of mass m, initially at rest, acting on certain distance S.

The acceleration of this object, according to the Second Newton's Law is

a = F/m

The time t to cover distance S with acceleration a is based on the formula of Kinematics

S = a·t²/2 and is equal to

t = √2S/a  = √2S·m/F

From this we derive the final speed at the end of acceleration

V = a·t = (F/m)·√2S·m/F  =

= √2F·S/m

As we see, the final speed of an object of mass m depends on the product of the force and the distance this force is acting.

This product that characterizes the result of applying a force on a certain distance is called the work performed by a given force acting on a given distance:

W = F·S

We can reduce by half the force and double the distance - the resulting speed will be the same.

Using this definition of work, we can represent the final speed as

V = √2W/m

Resolving for work as a function of final speed, we obtain

W = m·V²/2

So, given a final speed, we can determine how much work it takes to
achieve it and, given amount of work, we can calculate the speed this
work will cause.

This concept of work is closely tied with such less precisely
defined concepts of result, purpose, effect etc. While we not always can
compare the results or effect of two physical experiments, we can
always compare the work done by the forces participating in these experiments.

Thus, the work can be used to measure the results or effect of physical experiment.

Straight line motion with a force acting at an angle to a trajectory

Consider a slightly more complicated example with force acting at an angle to a trajectory.

A toy train of mass m is pulled by a child, who stands on a side of a track, with a constant force F, so the force acts at an angle to a straight line trajectory, making an angle φ with it.

Representing this force F (in blue on a picture above) as a
sum of two forces (in red), one acting along a track, forcing the train
to speed up, and another acting perpendicularly to a track (force N), which is balanced by the reaction of the track R (in brown), we see that the force that pushes a train forward equals to F·cos(φ), while the force acting perpendicularly to a track can be simply ignored as being balanced by an opposite reaction force.

Exactly the same considerations as above leads us to a more universal formula for relationship between the final speed and the work:

V = √2W/m ,

where W = F·S·cos(φ)

Obviously, if the force acts along a straight line of a trajectory, angle φ is zero, its cosine is 1 and the formula corresponds to the one derived earlier.

Resolving the formula above for work, we obtain

W = m·V²/2

which indicates that the work depends on the result of an action
only (final speed), not the way how we achieve this result, using a
stronger force on a shorter distance or a weaker force on a longer
distance.

Motion against gravity along an inclined plane

Let's consider a completely different goal of a physical experiment and show the importance of a concept of work.

This time our purpose is to lift an object of mass m to a certain height H above the ground along a frictionless inclined plane making angle φ with horizon.

The force F needed for this must be equal to a component R of the weight P of an object along the inclined because the other component of its weight N, perpendicular to an inclined, is balanced by a reaction of the plane.

F = P·sin(φ)

The distance S this force is acting on equals to

S = H/sin(φ)

From this we can derive the work performed by force F along the distance S:

W = F·S = P·sin(φ)·H/sin(φ) =

= P·H

This is quite a remarkable result. The work does not depend on the angle of an incline, only on the height we lift the object and its weight.

As in the previous cases, the work seems to be a characteristic
of the result and does not depend on how we have achieved it, using an
inclined with bigger or smaller slope.

Rotational motion with a force acting tangentially to a trajectory

Our final example is about rotation.

Consider a person starts rotating a carousel of a radius r, having a moment of inertia I, from the state of rest to some angular speed ω with constant angular acceleration α, applying constant force F tangentially to the carousel's rim.

From the rotational dynamics we can determine angular acceleration, knowing torque τ of the force and a moment of inertia of a carousel I:

τ = F·r = I·α

α = F·r/I

Linear acceleration of the point of application of the force equals to r·α.

To achieve the final angular speed ω with angular acceleration α we need time

t = ω/α = ω·I/(F·r)

During this time the point of application of force travels around a circle for a distance of

S = r·α·t²/2 =

= r·(F·r/I)·[ω·I/(F·r)]²/2 =

= I·ω²/(2F)

Multiplying by F both sides, we obtain the relationship between work W=F·S and final angular speed achieved as a result of application of the force:

W = F·S = I·ω²/2

This is a rotational equivalent of the analogous formula for straight line movement derived in the beginning of this lecture.

So, given a final angular speed, we can determine how much work it takes
to achieve it and, given amount of work, we can calculate the angular
speed this work will cause.

Definition of work in simple cases presented above

All the above examples provide a proper basis for the following definition of the work:

Work of the force F acting at an angle φ to trajectory on the distance S is W = F·S·cos(φ).

General case of a motion along a curved trajectory

To be more precise and to cover a case of a curved trajectory, we have
to require that this definition should be applied to infinitesimal
amount of work dW performed on infinitesimal interval dS of trajectory:

dW = F·dS·cos(φ)

Using a concept of scalar product of vectors and considering force and interval as vectors, the same definition can be written as

dW = (·d)

The latter represents the most rigorous definition of work.

## Wednesday, October 3, 2018

### Unizor - Physics4Teens - Mechanics - Statics - Problems

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

Problems on Equilibrium

Problem 1

Two point-objects of mass M and m are hanging at two opposite ends of a weightless rod of length L.

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?

Does this state of equilibrium imply that a rod is horizontal?

At distance m·L /(M+m) from the end with object of mass M.

Problem 2

An object of weight W is positioned on a flat surface. The coefficient of static friction is μ.

What is the minimum pulling force P to be applied to a
rope, attached to this object, to start pulling it forward, if the angle
between a rope and a horizontal flat surface is φ?

P = μ·W / [cos(φ)+μ·sin(φ)]

Problem 3

Consider the following illustration to this problem.

A person of weight W stands on a platform of weight P. Pulleys and ropes are arranged as on this illustration.

What is the force F a person should apply to a rope to keep the whole system in equilibrium?

Solution

Let N be a reaction of a platform onto a person. It pushes the person upward.

If F is a force of a person pulling a rope down, and the
system is in equilibrium, that is all components are at rest and all
forces are balanced, the rope pulls a person up with the same force F, according to the Third Newton's Law.

Since a person is in equilibrium, forces directed upward (N+F) should be equal in magnitude to forces directed downwards (weight of the person W):

N+F = W

The platform is also in equilibrium. The forces that push it down are its weight P and reaction N of a person standing on it. The forces pulling the platform up are tensions of two ropes T on the left and F on the right:

P+N = T+F

Finally, a small pulley is in equilibrium. It's pulled down by two tensions F, from the left and from the right. The tension T pulls it up:

T = 2F

We have three equations with three unknowns: T, N and F, which we have to solve for an unknown force F, with which a person pulls a rope to balance the system:

Substituting T from the last equation into the second one, getting a system of two equations:

N+F = W

P+N = 3F

Solving the first equation for N and substituting into the second:

P+W−F = 3F

Resolving for F:

F = (P+W)/4

F = (P+W)/4

Problem 4

A ladder is standing at some angle to the floor. It's in equilibrium,
that is it's not slipping down to the floor. What holds it is the static friction between it and the floor.

If the ladder stands almost vertically, the static friction is
greater and it easily holds the ladder in the position. As we move the
bottom of a ladder away from the vertical, there will be a point when static friction will no longer could hold the ladder in place and the ladder would slip down to the floor.

What is the smallest angle the ladder would hold if the coefficient of static friction is μ?

Solution

Consider the following illustration to this problem.

While the reaction of the wall N1 is smaller than the static friction F, the ladder would hold its position.

The smallest angle between the ladder and the floor will be when these two forces are equal.

To be in equilibrium, vector sum of all forces must be equal to
null-vector and a sum of all angular momentums also must be equal to
null-vector.

Along the horizontal X-axis the sum of all forces equals to zero if the horizontal reaction of the wall N1 equals to the force of static friction F, which, in turn, equals to a product of the vertical reaction of the floor N2 by a coefficient of static friction μ:

(a) N1 = μ·N2

Along the vertical Y-axis the sum of forces equals to zero if the reaction of the floor N2 equals to weight of a ladder W:

(b) N2 = W

To analyze angular momentums, let's choose an axis of rotation. It's
easier to choose an axis perpendicular to both X- and Y-axis that goes
through a point where a ladder touches the floor. In this case angular
momentums of forces N2 and F are zero because their corresponding radiuses are zero.

Let's assume that the length of the ladder is D and the angle between the ladder and the floor is φ. The magnitude of the angular momentum of the force of weight W will then be W·D·cos(φ)/2. The magnitude of the angular momentum of the reaction of the wall N1 will be N1·D·sin(φ). The condition of the sum of angular momentums to be equal to zero results in the equation:

N1·D·sin(φ) = W·D·cos(φ)/2

The length of the ladder can be canceled:

(c) N1·sin(φ) = W·cos(φ)/2

Equations (a), (b) and (c) form a system of three equations with three unknowns:

(a) N1 = μ·N2

(b) N2 = W

(c) N1·sin(φ) = W·cos(φ)/2

From this

tan(φ) = 1/(2μ)

φ = arctan[1/(2μ)]

Interestingly, the answer does not depend on the weight or the length of the ladder.

## Wednesday, September 26, 2018

### Unizor - Physics4Teens - Mechanics - Statics - Equilibrium of Solids

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

Equilibrium of Solids

As we know, if a vector sum of forces, acting on a point-object, equals
to null-vector, this point-object is at rest or in a state of uniform
motion along a straight line in any inertial frame of reference.
Alternatively, we can say that a frame of reference associated with this
point-object is inertial.

The complexity of a concept of equilibrium of solids lies in the fact
that, even if vector sum of forces, acting on a solid, equals to
null-vector, this solid might not be in the state of rest in any
inertial system. In a simple case of two forces, equal in magnitude and
opposite in direction, applied to two different points of a solid, will
rotate it, making a frame of reference associated with it non-inertial.

Any rotation of a solid, however complex, can be represented as a
combination of three rotations around three coordinate axes. Therefore, a
solid in a state of equilibrium should be in equilibrium relative to
each coordinate axis, and, to study equilibrium of a solid in
three-dimensional space, it is sufficient to study it relative to one
axis.

As we know, in case of rotation around an axis, the main factor replacing a concept of force for translational movement, is torque. Since torque is a cause of rotation, the necessary condition to prevent a rotation is balancing of torques.

Torque is a vector equal to a vector (cross) product of a radius-vector r from the axis of rotation to a point of application of force (perpendicularly to the axis) by a vector of force:

τ = r×

Using this concept of torque, we can formulate the following necessary condition for an equilibrium of a solid.

Sum of torques of all forces acting on a solid relative to each coordinate axis must be equal to null-vector.

In the previous lecture we introduced a concept of a degree of freedom
and, in particular, mentioned six degrees of freedom of a solid: three
degrees of freedom along each coordinate axis and three degrees of
freedom of rotation around each coordinate axis.

If a vector sum of all forces acting on a solid is a null-vector, there
will be a point that is not moving anywhere or, more precisely, a frame
reference associated with this point is inertial. This condition assures
an equilibrium by three degrees of freedom associated with
translational motion (a motion along coordinate axes).

If a vector sum of all torques relatively to each of the three
coordinate axes is a null-vector, there will be no rotation and the
system will be in equilibrium by three other degrees of freedom
associated with rotational motion (a motion around coordinate axes).

So, for a solid to be in equilibrium, the necessary conditions are
all of the above - sum of all the forces and sum of all the torques must
be null-vectors
.

Consider the following example.

A weightless rod of a length 2r is hanging horizontally on a thread fixed at its midpoint.

On both ends of a rod there are equal weights W hanging down.

To balance the translational movement of a system vertically down the force of tension T on a thread that holds the rod must balance a sum of forces of weight acting in opposite direction:

+ + =

= −2

To balance the rotational movement of a system around a midpoint of a
rod the torque of one weight should be equal in magnitude to the torque
of another weight since the directions of these two torques are
opposite:

τ blue = ×

τ red = −×

τ red + τ blue =

Consider a less symmetrical case below.

Here a rod of the length 3r is hanging by a thread fixed not in the middle, but at a distance r from the left edge and, therefore, at a distance 2r from the right edge.

At the same time, the weight on the left is twice as heavy as the one on the right.

Tension on the thread holding a rod must be greater to hold more weight:

+ + 2 =

= −3

From the rotational viewpoint, this system is in equilibrium because the
torques on the left of a rod and on the right are equal in magnitude,
while opposite in direction:

τ blue = 2×

τ red = −×2

τ red + τ blue = 0

## Thursday, September 20, 2018

### Unizor - Physics4Teens - Mechanics - Statics - Point-Objects

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

Equilibrium of Point-Objects

The state of equilibrium in a case of possibility of a
point-object to move in some direction implies that the position of this
object does not change, which, in turn, implies that the vector sum of
all forces acting on it is a null-vector:

ΣFi = 0 .

In three-dimensional world each force Fi
has three components - the projections of this force on X-, Y- and
Z-axis. In order for a sum of vectors to be equal to a null-vector, it
is necessary and sufficient for a sum of their X-components to be equal
to 0, same for Y- and Z-components.

Therefore, we can construct three equations, one for each axis, to determine forces participating in the state of equilibrium:

ΣFi(x) = 0

ΣFi(y) = 0

ΣFi(z) = 0

The fact that only three equations determine the conditions on forces to
be in equilibrium in a three-dimensional world means that unique
determination of these forces is possible only if the number of unknowns
does not exceed three. If four or more unknown variables participate in
the state of equilibrium, we will have infinite number of solution of
the system of three linear equations with four or more unknowns.

Most likely, for our purposes we will consider situations with only unique solutions.

As a simple example of an equilibrium in three-dimensional space,
consider a chandelier of some weight hanging from a ceiling on three
threads attached to a ceiling at three different points. For simplicity,
assume that a chandelier is a point-object and three points of
attachment of the threads to a ceiling are making an equilateral
triangle and the length of all threads is the same.

The weight of a chandelier W is balanced by three equal in magnitude T,
but acting at different angles, tension forces along each thread.
Knowing the weight of a chandelier and an angle of each thread with a
vertical φ (the same for each thread in our simple case), we can easily determine the magnitude of the tension force of each thread:

3·T·cos(φ) = W

T = W / [3·cos(φ)]

Here we only constructed the equation along the vertical Z-axis because
in a horizontal plane (along X- and Y-axis) the tension forces balance
each other.

The example above is a perfect illustration to our next discussion.

Stable Equilibrium

We call an equilibrium stable if, after a small movement of an object from its position of equilibrium, it returns back to this position.

A small push of a chandelier will result in its light swinging and eventual return to the original position of equilibrium.

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 stable equilibrium. If we push it off this position, it will eventually return back because the sum F of reaction force R of the cup and the weight W of a ball always act in a direction of the lowest point of this semi-spherical cup.

Let's formulate the definition of a stable equilibrium in more precise terms.

For this we have to consider a position of an object in equilibrium and
its immediate neighborhood. Since we consider the equilibrium to be
stable, we have to consider the deviation of an object from a position
of equilibrium.

Basically, we would like to say that the equilibrium is stable, if an
object returns to it from any neighboring point. How big is the
neighborhood, from which this return takes place is, obviously, a
subject of a definition.

The rigorous definition of a stable equilibrium defines the neighborhood as any, however small, one.

So, if exists a real positive value ε such that from any point within ε-neighborhood
of a position of equilibrium an object, placed at this point, returns
to a position of equilibrium, the equilibrium is stable.

Unstable Equilibrium

The unstable equilibrium is the opposite of a stable one. That
means that any, however small, deviation of the object's position from
the position of equilibrium results in moving of this object away from
the equilibrium point.

Obvious example is a small ball on the top of a sphere. If its position
is exactly at the North Pole, it's in equilibrium. But any, however
small, deviation from the NorthPole results in complete departure from
the position of equilibrium.

Neutral Equilibrium

Consider a ball on a horizontal plane. There are two forces acting on it
- gravity vertically down and reaction of the plane vertically up and
equal in magnitude to the ball's weight. These two forces neutralize
each other, no matter where on the plane our ball is positioned. So, any
position is a position of equilibrium. This is exactly what neutral equilibrium is.

Any deviation from the position of neutral equilibrium results in the new equilibrium.

Strictly speaking, neutral equilibrium implies existence of some
neighborhood around the position of equilibrium such as any position
within this neighborhood is a position of equilibrium.

Solids and Six Degrees of Freedom

Before we talked about equilibrium of a point-object, that is a state of
rest, when all forces applied to it are in balance and their sum is
null-vector.

Let's expand our concept of equilibrium to solids. Obviously, they not
only can move in three different directions (forward-backward along
X-axis, left-right along Y-axis, up-down along Z-axis), but also can
rotate around any of the three coordinate axis.

That means that a position of a solid is determined by six parameters - so called six degrees of freedom. Equilibrium of this solid object implies its state of rest along each of these six degrees of freedom.

We know how to deal with the first three degrees of freedom associated with translational motion.

Conditions of equilibrium related to rotation - the other three degrees of freedom - will be analyzed in the next section.

## Thursday, September 6, 2018

### Unizor - Physics4Teens - Mechanics - Rotational Dynamics - Conservation ...

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

Conservation of
Angular Momentum

We ended up the previous lecture deriving the Law of Conservation of
Angular Momentum. In this lecture we will continue discussing this
particular conservation law.

We will consider a thought experiment suggested by Arkady Kokish in our private discussion of this topic.

The results of this thought experiment will confirm the Angular Momentum Conservation Law.

Consider a point-object of mass M with no external forces (like gravity) present and two threads, short of length r and long of length R tied to it.

A person holds both threads and starts spinning an object with angular velocity ω1. It will rotate along a circular trajectory of a radius r (the length of a shorter thread).

Let's analyze what happens if this person lets a shorter thread go, while holding the long one.

On the picture above the short thread is let go when an object was at
the top. Then, since nothing holds it on a circular trajectory, it will
fly along a tangential line maintaining the same linear speed it had
when it was rotating, that is r·ω1.

But our object will not fly too far. The longer thread will stop it, when it will be on a distance R from a center of rotation.

Here is a picture of this particular moment, when the long thread is fully extended.

At this moment the linear speed of an object flying (on the picture) left to right equals to V=r·ω1.

Let's represent this vector as a sum of two vectors: Vt perpendicular to a line representing the fully extended long thread (tangential to a future circular trajectory of a radius R) and Vr going along that thread line (radial).

Assuming an angle between a short thread at a moment it was let go and a
long one at the moment it is fully extended, stopping tangential
movement of an object, is φ, it is obvious that r=R·cos(φ).

At the moment when the long thread is fully extended the tension of the
thread instantly (actually, during a very short interval of time Δt) nullifies the radial speed Vr. To be exact, its moment of inertia M·Vr will be equal to an impulse of the tension force T during interval of time Δt:

Δt = M·Vr

This tension force decelerates the radial speed to zero during the interval of time Δt. The shorter this interval - the more ideal our experiment is.

In an ideal case of non-stretchable thread that can withstand infinite tension the interval Δt is infinitely small, and deceleration is infinitely large in magnitude and short in time.

In this case the radial speed Vr will be brought down to zero instantaneously.

At the same time the magnitude of tangential vector of speed Vt is not affected by the tension since this vector is perpendicular to a tension force.

We see that the angle between vectors V=r·ω1 and Vt is the same as between threads φ.

Therefore, Vt = r·ω1·cos(φ)

But cos(φ) = r/R

from which follows

Vt = r·ω1·r/R

Tangential linear speed Vt is related to angular speed of rotation ω2 of our object on a new radius R as Vt=R·ω2.

Hence,

R·ω2 = r·ω1·r/R

R²·ω2 = r²·ω1

M·R²·ω2 = M·r²·ω1

Expression M·r² represents the moment of inertia I1 of an object of mass M rotating on a radius r.

Analogously, expression M·R² represents the moment of inertia I2 of the same object of mass M rotating on a radius R.

Therefore, we have derived the Law of Conservation of the Angular Momentum:

I1·ω1 = I2·ω2

## Monday, August 27, 2018

### Unizor - Physics4Teens - Mechanics - Rotational Dynamics - Problems

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

Problems on
Rotational Dynamics

Problem 1

Consider a device that consists of a wheel of radius r and mass M, freely rotating on some fixed axis, a thread rolled around it a few times and a point-object of mass m hanging off that thread.

Assume that all mass of a wheel is concentrated in its rim, while spokes are weightless.

There is no friction in the rotating wheel.

A thread is weightless and unstrechable.

The force of gravity pulls the object down with free fall acceleration g.

What will be the acceleration of an object?

Solution

The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.

The only force acting tangentially on a wheel is the tension of a tread T trying to rotate it.

Therefore, we can equate its momentum τ=T·r (torque) to a product of its moment of inertia I and angular acceleration α:

τ = T·r = I·α

The fact that all mass of a wheel is concentrated in its rim allows us
to easily calculate its moment of inertia. We can imagine a rim
consisting of a large number N of point-objects of mass M/N each. The moment of inertia of each is (M/N)·r² and combined moment of inertia of a wheel is N·(M/N)·r²=M·r². If the mass is not concentrated in the rim, analogous logic would lead us to more involved calculations of inertial mass.

Considering, I=M·r² and r·α=a (where a is a linear acceleration of a thread) we can write the following equation:

T·r = M·r²·a/r = M·r·a

Radius cancels out and we get

T = M·a

Notice that this is exactly the same equation as the Newton's Second Law, as if an object of mass M is pulled along a straight line by a force T with linear acceleration a.

Another important detail is that this formula is independent of a radius
of a wheel - a direct consequence of the fact that the mass of a wheel
is concentrated in its rim.

Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration a down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,

m·g − T = m·a

Now we have a system of two equations with two unknowns T and a, which is easy to solve.

Substitute T from the first equation to the second:

m·g − M·a = m·a

m·g = (M + m) · a

a = m·g / (M + m)

Problem 2

Calculate a moment of inertia of disk of mass M and radius R rotating around an axis going through its center perpendicularly to its surface.

Solution

Divide a disk into a set of concentric rings of infinitesimal width dr. Let the inner radius of a particular ring be r and it outer radius be r+dr.

Moment of inertia of each ring is

I(r) = m·r²

where m is its mass (see the previous Problem 1 for explanation).

Mass of a ring m is a mass of a disk M multiplied by a ratio of a ring's area 2πr·dr to the area of an entire disk πR².

So, the moment of inertia of our ring of radius r and infinitesimal width dr is

I(r) = [M·2πr·dr/(πR²)]·r²

Integrating this from r=0 to r=R, we get

Idisk(M, R) =

=
[M·2πr·dr/(πR2)]·r2 =

= (2M/R2)r3
dr =

= (2M/R2)·(r4/4) =

= M·R2/2

Problem 3

Consider Problem 1, but change a wheel with all mass concentrated in the
rim with a wheel with mass evenly distributed inside the circumference,
making it a disk.

Solution

We follow the same logic as in Problem 1.

The weight of a wheel is balanced by the reaction of fixed axis and, therefore, can be ignored.

The only force acting tangentially on a wheel is the tension of a tread T trying to rotate it.

Therefore, we can equate its momentum τ=T·r (torque) to a product of its moment of inertia I and angular acceleration α:

τ = T·r = I·α

The fact that all mass of a wheel is evenly distributed within its
circumference allows us to easily calculate its moment of inertia using
the Problem 2 above:
I=M·r²/2
Linear acceleration a and angular acceleration α are related:

r·α=a

Therefore,

T·r = [M·r²/2]·(a/r) = M·r·a/2

Radius cancels out and we get

T = M·a/2

An important detail is that this formula is independent of a radius of a wheel.

Analyzing the movement of an object hanging on a thread, we conclude that it moves with linear acceleration a down by the forces of gravity (directed down) and tension of a thread (directed up). Therefore, using the Newton's Second Law,

m·g − T = m·a

Now we have a system of two equations with two unknowns T and a, which is easy to solve.

Substitute T from the first equation to the second:

m·g − M·a/2 = m·a

m·g = (m + M/2) · a

a = m·g / (m + M/2)

If we compare this formula with the one in Problem 1, we see that final acceleration is greater because denominator is smaller.

So, the disk in this problem will rotate faster than a wheel with empty middle part from Problem 1.

Problem 4

Let's rotate a small ball of mass M within a horizontal plane on a thread of length D. The thread will make certain angle φ with the horizon. Experiments show that the angle will be smaller if a ball rotates faster.

Determine the relationship between angular speed of rotation ω, mass of a ball M, length of a thread D and angle φ, ignoring friction and air resistance.

Solution

The tension of a thread T keeps a ball on its orbit. Radius of an orbit is R=D·cos(φ).

The tension of a thread T serves dual purpose - its vertical component T·sin(φ) acts against gravity M·g, its horizontal component T·cos(φ) acts as a centripetal force and should be equal to M·V2/R, where V is a linear speed of a ball, which is equal, in turn, to R·ω, where ω is angular speed of rotation of a ball.

So, we have the following equations:
T·sin(φ) = M·g

T·cos(φ) = M·(V2)/R = M·R·ω2 = M·D·cos(φ)·ω2

The second equation is simplified to
T = M·D·ω2

Substituting it to the first equation,
D·ω2·sin(φ) = g

Therefore,
sin(φ) = g/[D·ω2]

For a given angular speed we can find an angle of a thread to horizon φ. It in inversely proportional to a length of a thread D and to a square of angular speed ω, which seems to be reasonable.

Interestingly, this angle is independent of the mass M.

## Tuesday, August 21, 2018

### Unizor - Physics4Teens - Mechanics - Dynamics - Angular Momentum

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

Angular Momentum

Recall the rotational equivalent of the Newton's Second Law:

τ = I·α

where

τ = F·r - a torque, a product of tangentially applied force F and the distance from the point of application of force to the axis of rotation r;

I = m·R² - a moment of inertia, a product of inertial mass of a point-object by a square of its distance from the axis of rotation R (might be different from the distance r above);

α - angular acceleration of rotation.

It is very important to correspond straight line translational elements of motion with their rotational counterparts:

 Translation Rotation ForceF Torqueτ = F·r Linear Accelerationa Angular Accelerationα = a/r InertialMassm Momentof InertiaI = m·r² Newton'sSecond LawF=m·a RotationalEquivalentτ = I·α

Now we will consider a rotational equivalent of a familiar equation
between impulse and momentum in translational motion along a straight
line.

As is known from the previous material, there is a correspondence between these quantities: impulse of the force F exhorted during an infinitesimal time period dt equals to an increment of the momentum d(m·v), where m is mass and v - velocity of an object:

dt = d(m·v)

Let's see if the corresponding rotational characteristics of motion have similar dependency.

We know the rotational equivalent of the Newton's Second Law τ=I·α.

Multiplying it by dt, we obtain

τ·dt = I·α·dt

Since a product of angular acceleration α and infinitesimal time interval dt is an infinitesimal increment of angular speed dω, there is an equality

τ·dt = I·dω = d(I·ω)

The latter fully corresponds to an equation between impulse and momentum
of translational movement. It represents the relationship between rotational impulse τ·dt of torque τ during time interval dt and increment of rotational (angular) momentum d(I·ω) of an object with a moment of inertia I and angular speed ω.

We can continue our table of correspondence between translational motion
along a straight line and rotation along a circular trajectory.

 Translation Rotation ImpulseF·dt Rotational Impulseτ·dt Momentumd(m·v)=F·dt Rotational Momentumd(I·ω)=τ·dt

Trivial consequence of an equation

τ·dt = d(I·ω)

is that

τ = d(I·ω)/dt = (I·ω)'

From this follows that if the balance of all torques acting on an object is zero, the rotational momentum remains constant.

This is the Law of Conservation of the rotational (angular) momentum.

## Monday, August 20, 2018

### Unizor - Physics4Teens - Mechanics - Dynamics - Moment of Inertia

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

Moment of Inertia

Recall the rotational equivalent of the Newton's Second Law:

τ = I·α

where

τ = F·r - a torque, a product of tangentially applied force F and the distance from the point of application of force to the axis of rotation r;

I = m·R² - a moment of inertia, a product of inertial mass of a point-object by a square of its distance from the axis of rotation R (might be different from the distance r above);

α - angular acceleration of rotation.

It is very important to correspond straight line translational elements of motion with their rotational counterparts:

 Translation Rotation ForceF Torqueτ = F·r Linear Accelerationa Angular Accelerationα = a/r InertialMassm Momentof InertiaI = m·r² Newton'sSecond LawF=m·a RotationalEquivalentτ = I·α

In this lecture we will study the moment of inertia and its properties. We will also calculate the moment of inertia in a few simple cases.

The first very important property of the moment of inertia is that it is additive.
It means that the moment of inertia of two rigidly connected
point-objects rotating around the same axis of rotation equals to sum of
their individual moments of inertia.

Let's explain why.

Consider a rigid weightless disk rotating around an axis going through
its center perpendicularly to it and two point-objects of mass m1 and m2 fixed on its surface at distances, correspondingly, r1 and r2.

Assume that this disk with both fixed on it objects is rotating with angular acceleration α caused by some force F acting on a distance R from its center tangentially to a circle of a radius R within a plane of this disk.

The picture below illustrates this.

Our purpose is to explain that the combined moment of inertia of both objects equals to a sum of their corresponding individual moments of inertia.

The fact that object of mass m1 on a distance r1 from the axis of rotation has angular acceleration α implies that there is a force f1 acting on it tangentially to its trajectory such that (from rotational equivalent of the Newton's Second Law):

f1·r1 = I1·α

where I1=m1·r1² - moment of inertia of this object.

Similarly, the fact that object of mass m2 on a distance r2 from the axis of rotation has angular acceleration α implies that there is a force f2 acting on it tangentially to its trajectory such that (from rotational equivalent of the Newton's Second Law):

f2·r2 = I2·α

where I2=m2·r2² - moment of inertia of this object.

Behavior of this system depends only on parameters specified above -
masses, distances from the center of a disk and forces. So, if we change
the initial position of masses without changing their distances from
the center, the system will work exactly the same.

Our first modification then will be to shift the position of mass m2 along its circular trajectory of radius r2 to a position, where it is on the same radius as m1. So, both masses and a center of a disk are on one line.

Similarly, let's shift the point of application of force F to a point on the same line where our two objects are now located without changing the distance R
from the point of application of the force to a center. Now both masses
and a point of application of main force that rotates the disk are on
the same line with a center of the disk, while their distances from a
center are still the same as before.

As we know, the angular acceleration of a rotating object depends on its
mass, its distance from an axis of rotation and a torque of a force
rotating this object.

If force f1 applied directly to an object of mass m1 located on a distance r1 from the axis of rotation caused the angular acceleration α, any other force having the same torque τ1=f1·r1 would result in the same angular acceleration.

Let's replace then the force f1 acting on a distance r1 from the axis with force g1=f1·r1/R, acting on a distance R at the point where our main force F was applied. There will be no change in angular acceleration since the torque acting on our object is the same:

τ'1 = (f1·r1/R)·R = f1·r1 = τ1

Same trick with the second object results in using force g2=f2·r2/R, acting on a distance R at the point where our main force F was applied, with no change in the behavior of the system.

τ'2 = (f2·r2/R)·R = f2·r2 = τ2

We have brought all forces (g1, g2 and F)
to the same point of application. They act in the same direction -
tangentially to a circular trajectory. We know that the effect of main
force F applied to this point is equivalent to a combined effect of forces g1=f1·r1/R and g2=f2·r2/R applied to the same point in the same direction. Therefore, sum of forces g1 and g2 must be equal to force F:

g1 + g2 = F

Now what remains is simple algebra.

f1·r1/R + f2·r2/R = F

f1·r1 + f2·r2 = F·R

τ1 + τ2 = F·R

I1·α + I2·α = F·R

(I1 + I2)·α = F·R

The expression on the right is the torque of the main force F acting tangentially to a circle of rotation on a distance R from the axis:

τ = F·R

Therefore, the expression on the left, according to rotational equivalent of the Newton's Second Law, must be a moment of inertia of the entire system I multiplied by the system's angular acceleration α:

I·α = τ = F·R

from which follows the equality

I·α = (I1 + I2)·α

and, finally,

I = I1 + I2

That means that the combined moment of inertia of two objects equals to a sum of their individual moments of inertia.

In short, moment of inertia is additive.

The most important consequence of this is that, to calculate a moment of
inertia of a rotating solid object of some complicated geometric shape,
we can divide it in small (strictly speaking, infinitesimally small)
pieces, calculate a moment of inertia of each piece and add all these
moments of inertia together (strictly speaking, integrating them) to get
a moment of inertia of the whole rotating solid object.

Let's calculate a moment of inertia of some solids, using this approach.

1. Rod rotating around its edge

Consider a solid thin rod of a length L and mass m rotating around an axis that is perpendicular to it and going through its one end.

To calculate its moment of inertia, we divide it in small pieces of the length dr, located on a distance r from the axis and having mass dm=(m/L)·dr.

Each such piece has a moment of inertia

dI = (dm)·r² = (m/L)·r²·dr

Since moment of inertia is additive, all we have to do now is to integrate this by r from 0 to L.

The indefinite integral of function is r³/3. So, by Newton-Leibniz formula the result of integration on an interval [0,L] is

L³/3 − 0³/3 = L³/3.

So, the total moment of inertia of a rod is

Irod = (m/L)·L³/3 = m·L²/3

2. Rod rotating around its center

Let's calculate the moment of inertia, if the rod rotates around its center point.

We can consider now this rod as a system of two small half-rods, each being a half of the original one, with length L/2 and mass m/2.

Each half-rod rotates around its end, so the formula above is applicable to each half-rod.

Its moment of inertia is

Ihalf = (m/2)·(L/2)²/3 = m·L²/24

The whole rod, being a system of two half-rods, has a momentum of
inertia equal to a sum of momentums of its two halves, so its value is
double of the above moment of inertia for each half-rod:

Iwhole = m·L²/12

## 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.

 Translation Rotation ForceF Torqueτ = 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.

 Translation Rotation ForceF Torqueτ = F·r Accelerationa Angular Acc.α = a/r InertialMassm Momentof InertiaI = m·r² Newton'sSecond LawF=m· RotationalEquivalentτ = 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.

 Translation Rotation Distances(t) Angleφ(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)

 Translation Rotation Speedv(t)=s'(t)v(t)=r·ω(t) Angular Speedω(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)

 Translation Rotation Accelerationa(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.