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·L_f = P·L_p

If S_f is the distance the force F acts down and S_p is the distance our object moves up,

S_f/S_p = L_f/L_p and

F·S_f = P·S_p

By using a lever with longer arm L_f 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.

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 = (F ·dS )



The latter represents the most rigorous definition of work.

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?



Answer

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?



Answer



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



Answer

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)



Answer



φ = arctan[1/(2μ)]

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

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×F 



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:

T  + W  + W  = 0 

T  = −2W 



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 = r ×W 

τ _red = −r ×W 

τ _red + τ _blue = 0 



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:

T  + W  + 2W  = 0 

T  = −3W 



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 = 2r ×W 

τ _red = −r ×2W 

τ _red + τ _blue = 0

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:

ΣF_i = 0 .



In three-dimensional world each force F_i
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:

ΣF_i^(x) = 0

ΣF_i^(y) = 0

ΣF_i^(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.

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: V_t perpendicular to a line representing the fully extended long thread (tangential to a future circular trajectory of a radius R) and V_r 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 V_r. To be exact, its moment of inertia M·V_r will be equal to an impulse of the tension force T during interval of time Δt:

T·Δt = M·V_r

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 V_r will be brought down to zero instantaneously.



At the same time the magnitude of tangential vector of speed V_t 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 V_t is the same as between threads φ.



Therefore, V_t = r·ω₁·cos(φ)

But cos(φ) = r/R

from which follows

V_t = r·ω₁·r/R

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

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

I_disk(M, R) =

= ∫[M·2πr·dr/(πR²)]·r² =

= (2M/R²)∫r³dr =

= (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.