# Conservation of Energy and Angular Momentum in a Rotating Train-Disk System

• JD_PM
In summary, the conservation of angular momentum results in that the spin angular momentum of the disk-train system is conserved at all points except when the train is at a distance R > 0 from the disk's centre.In summary, the conservation of angular momentum results in that the spin angular momentum of the disk-train system is conserved at all points except when the train is at a distance R > 0 from the disk's centre.
JD_PM

## Homework Statement

[/B]
A train stands in the middle of a rotating disk with an initial angular velocity of
$\omega_i$. The mass of the train is m and the moment of inertia of the train-disk is I. At one point the train departs on a straight track to a distance R from the disk's centre. (R smaller than the radius of the disk)

a) Find $\omega_f$, the angular velocity of the disk when the train is at a distance R from the centre.

b) Show that if m, R and I are strictly positive, the total energy of the disk-train system is strictly smaller when the train has a distance R > 0 from the disk's centre compared with the total energy when the train is in the middle.

c) The train runs at a constant speed from the centre to
distance R. Show that the work performed by the braking force on the system is equal to
:

$$W = - \int_0^R dr mr (\frac{I_i \omega_i}{I_i + mr^2})^2$$

d) Use the results from questions b) and c) to show that:

$$\int_0^R dr \frac {r}{(1 + r^2)^2} = \frac {1}{2(1 + r^2)}$$

## Homework Equations

- Conservation of angular momentum.

- Rotational kinetic energy.[/B]

## The Attempt at a Solution

[/B]
a) In this system the angular momentum is conserved due to the fact that the net external torque is zero. Therefore the spin angular momentum satisfies that:
$$I_i \omega_i = I_f \omega_f$$
$$\frac{I_i \omega_i}{I_i + mR^2} = \omega_f$$
Note that I applied the parallel axis theorem.

b) The energy of the disk-train-Earth system is conserved, but not if we do not take into consideration the Earth as part of the system.

Actually, the total rotational kinetic energy is:

$$K = \frac{1}{2} I_{CM} \omega^2$$

Note that the higher is the angular velocity the higher is the energy. The angular velocity is higher when the train stands at the centre of the disk and lower when the train is placed at a distance R from the centre because of the conservation of angular momentum. Therefore the energy is higher when the train stands at the centre of the disk and lower when the train is placed at a distance R from the centre.

c) The friction force does a work on the system and the centripetal force is equal to the friction force so:

$$F_c = \frac{m v^2}{r} = mr \omega^2$$

From here get the provided equation.

HERE COMES MY BIG PROBLEM

d) I do not know how to show the equation provided in this section from sections b) and c):

$$\int_0^R dr \frac {r}{(1 + r^2)^2} = \frac {1}{2(1 + r^2)}$$

Any help is appreciated, thank you.

Last edited:
JD_PM said:
The energy of the disk-train-Earth system is conserved, but not if we do not take into consideration the Earth as part of the system.
No, that's not the reason. You are given the reason in part c.
JD_PM said:
Note that the higher is the angular velocity the higher is the energy. The angular velocity is higher when the train stands at the centre of the disk and lower when the train is placed at a distance R from the centre because of the conservation of angular momentum. Therefore the energy is higher when the train stands at the centre of the disk and lower when the train is placed at a distance R from the centre.
That is too hand-waving and should not be trusted. Do the algebra, which you need for part d.
JD_PM said:
The friction force does a work on the system
Yes, that's the reason mechanical work is lost.

JD_PM said:
Note that I applied the parallel axis theorem.
No you did not, but that's irrelevant to your answer in part (a) which is correct.
I am not convinced you have answered part (b) convincingly. I would like to see expressions for the kinetic energy of the train-track system at the centre and at R > 0. Your argument may not be valid because although ##\omega## is decreasing as the train moves out, the moment of inertia of the train-track system increases. Also, when you write the kinetic energy at radius ##K=\frac{1}{2}I_{CM}\omega^2##, what are ##I_{CM}## and ##\omega##? You should be able to write them in terms of the given quantities. As it stands, it is not clear if this expression is correct.

JD_PM
kuruman said:
No you did not
It looked that way to me.

Part b)

$$K_i = \frac{1}{2} I_i (\omega_i)^2$$

$$K_f = \frac{1}{2} I_f (\omega_f)^2 = \frac{1}{2} (I_i + mR^2) (\frac{I_i \omega_i}{I_i + mR^2})^2$$

The factor (I_i + mR^2) can be simplified and we can rearrange the equation like this:

$$K_f = K_i \frac{I}{I + mR^2}$$

As:

$$\frac{I}{I + mR^2} < 1$$

$$K_f < K_i$$

JD_PM said:
Part b)

$$K_i = \frac{1}{2} I_i (\omega_i)^2$$

$$K_f = \frac{1}{2} I_f (\omega_f)^2 = \frac{1}{2} (I_i + mR^2) (\frac{I_i \omega_i}{I_i + mR^2})^2$$

The factor (I_i + mR^2) can be simplified and we can rearrange the equation like this:

$$K_f = K_i \frac{I}{I + mR^2}$$

As:

$$\frac{I}{I + mR^2} < 1$$

$$K_f < K_i$$

Corrections to my mistakes so far:

1) I did not apply the parallel axis theorem; it is just that the final moment of inertia contribution of the train is not zero because R > 0.

2) The energy of the system is not conserved just due to the fact that the friction force does a work on the system.

JD_PM said:
1) I did not apply the parallel axis theorem; it is just that the final moment of inertia contribution of the train is not zero because R > 0.
Then I do not understand how you arrived at that exact formula.

haruspex said:
Then I do not understand how you arrived at that exact formula.

The final moment of inertia is the sum of the moment of inertia of the disk + the moment of inertia of the train:

$$I_f = I_d + I_t = I_i + mR^2$$

Note that at the beginning the train stands at the centre so R = 0 and:

$$I_i = I_d = I_{CM}$$

JD_PM said:
Part b)

$$K_i = \frac{1}{2} I_i (\omega_i)^2$$

$$K_f = \frac{1}{2} I_f (\omega_f)^2 = \frac{1}{2} (I_i + mR^2) (\frac{I_i \omega_i}{I_i + mR^2})^2$$

The factor (I_i + mR^2) can be simplified and we can rearrange the equation like this:

$$K_f = K_i \frac{I}{I + mR^2}$$

As:

$$\frac{I}{I + mR^2} < 1$$

$$K_f < K_i$$
Ok, so using that, write an expression for the KE lost and equate it to the expression you got for that in c.

JD_PM said:
The final moment of inertia is the sum of the moment of inertia of the disk + the moment of inertia of the train:

$$I_f = I_d + I_t = I_i + mR^2$$
The initial MoI you were given, I, should be taken to include a component from the train. It is not a point mass.
Assume R is the displacement of its mass centre.

haruspex said:
The initial MoI you were given, I, should be taken to include a component from the train. It is not a point mass.
Assume R is the displacement of its mass centre.

What I mean is that at the beginning we have:

$$I = I_d + I_t$$

But we know that R = 0 so:

$$I = I_d$$

Do you agree?

haruspex said:
Ok, so using that, write an expression for the KE lost and equate it to the expression you got for that in c.
You mean I equate it to the work expression?

JD_PM said:
You mean I equate it to the work expression?
Equate the work done against friction to the KE lost. This is what it implies when it says to use the results from b and c.

JD_PM said:
d) Use the results from questions b) and c) to show that:
$$\int_0^R dr \frac {r}{(1 + r^2)^2} = \frac {1}{2(1 + r^2)}$$
The equation in (d) should be
$$\int_0^R dr \frac {r}{(1 + r^2)^2} = \frac {1}{2(1 + R^2)}$$After you integrate there is no ##r## dependence.

JD_PM and haruspex
JD_PM said:
What I mean is that at the beginning we have:

$$I = I_d + I_t$$

But we know that R = 0 so:

$$I = I_d$$

Do you agree?
No.
Allow the train to have some length. That means it has a moment about its own centre. Call that It. We are given I=Id+It as the initial total moment.
What is the train's moment when it is displaced by R?

JD_PM
haruspex said:
No.
Allow the train to have some length. That means it has a moment about its own centre. Call that It. We are given I=Id+It as the initial total moment.
What is the train's moment when it is displaced by R?

OK So let's give the train some length r. If that is the case, we get:

$$I_i = I_d + I_t = I_{CM} + mr^2$$

The work equation is:

$$W = - \int_0^R dr mr (\frac{I_i \omega_i}{I_{CM} + mr^2})^2$$

If we plugg ##I_i## into it we get:

$$W = - \int_0^R dr mr (\omega_i)^2$$

But this is not the provided integral. What am I missing here?

JD_PM said:
The work equation is:$$W = - \int_0^R dr mr (\frac{I_i \omega_i}{I_{CM} + mr^2})^2$$
It's not that until you have shown that it is.
JD_PM said:
If we plugg ##I_i## into it we get: ...
Why would you do that? The work energy theorem in this case says ##dW=dK.## It follows that ##dW=\frac{dK}{dr}dr##. You found ##K(r)## in #5, so how should you proceed to get ##W##?

kuruman said:
It's not that until you have shown that it is.

Why would you do that? The work energy theorem in this case says ##dW=dK.## It follows that ##dW=\frac{dK}{dr}dr##. You found ##K(r)## in #5, so how should you proceed to get ##W##?

Oh so I should equate these equations:

$$W = - \int_0^R dr mr (\frac{I_i \omega_i}{I_{CM} + mr^2})^2 = \frac{1}{2} I_{i} (\omega_i)^2[\frac{I_{i}}{I_{CM} + mr^2} -1]$$

Where the lost kinetic energy is:

$$\Delta K= \frac{1}{2} I_{i} (\omega_i)^2[\frac{I_{i}}{I_{CM} + mr^2} -1]$$

You have correctly calculated the change in kinetic energy. That is part of what you need to do to address part (d). You still have not derived the equation for ##W## shown in part (c). Please reread #17 and answer the last question. Hint: Think in terms of what ##dW## means, not in terms of "equating" and "plugging in".

JD_PM
kuruman said:
You have correctly calculated the change in kinetic energy. That is part of what you need to do to address part (d). You still have not derived the equation for ##W## shown in part (c). Please reread #17 and answer the last question. Hint: Think in terms of what ##dW## means, not in terms of "equating" and "plugging in".

OK, let's derive the work's equation:

We know that the work has to be done by a force. What force? Well, that force in this problem is the kinetic friction force. Besides, in this problem the kinetic friction force is equal to the centripetal force, which makes the train go through a circular path instead of undergoing a straight line path. Therefore:

$$dW = F_c dr = ma_c dr$$

$$a_c = \frac{v^2}{r}$$

$$v= r \omega$$

$$dW = F_c dr = ma_c dr = mr \omega^2 dr$$

And we know that the angular velocity when the train is at a distance r from the centre is:

$$\omega = \frac{I_{i}}{I_{CM} + mr^2}$$

So we end up getting:

$$dW = mr (\frac{I_{i}}{I_{CM} + mr^2})^2 dr$$

Please let me know if you disagree.

JD_PM said:
So we end up getting:$$dW = mr (\frac{I_{i}}{I_{CM} + mr^2})^2 dr$$
You could have obtained this result more directly after noting that ##dW=\frac{dK}{dr}dr## and taking the derivative of ##K## with respect to ##r##, but OK. Now what? How should you proceed to get ##W##?

JD_PM
kuruman said:
You could have obtained this result more directly after noting that ##dW=\frac{dK}{dr}dr## and taking the derivative of ##K## with respect to ##r##, but OK. Now what? How should you proceed to get ##W##?

Now I would integrate. But notice that we have r both in the numerator and in one term of the denominator so this seems a rational integral. But I am thinking that now I should be able to use the lost kinetic energy expression before integrating... Am I right?

JD_PM said:
Now I would integrate. But notice that we have r both in the numerator and in one term of the denominator so this seems a rational integral.
Just write down the integral. That's the answer to part (c).
JD_PM said:
But I am thinking that now I should be able to use the lost kinetic energy expression before integrating... Am I right?
Use the lost kinetic energy expression to what purpose?

JD_PM said:
OK So let's give the train some length r. If that is the case, we get:

$$I_i = I_d + I_t = I_{CM} + mr^2$$
That's not what I meant.
You do not need to assign the train a length variable, and certainly not r, which is already in use in the integral as a variable part of the displacement R.
At the start, yes, I=Ii=Id+It. When the train is displaced by r, the parallel axis theorem says its new moment is It+mr2, so the new moment of the system is Id+It+mr2=I+mr2.

JD_PM
kuruman said:
Just write down the integral. That's the answer to part (c).

Use the lost kinetic energy expression to what purpose?

I wanted to use it to get expression d) but did not succeed. I got something like #16, which seems to be wrong. May you please provide a hint?

JD_PM said:
I wanted to use it to get expression d) but did not succeed. I got something like #16, which seems to be wrong. May you please provide a hint?
Please post your attempt.

JD_PM
haruspex said:
Please post your attempt.

Sorry for the delay.

OK, what I think is that we are dealing with a nonisolated system in terms of energy; the change in the total amount of energy in the system is due to the work done by the kinetic frictional force. Therefore we have:

$$W = \Delta K$$

$$- \int_0^R dr mr (\frac{I_i \omega_i}{I_{CM} + mr^2})^2 = \frac{1}{2} I_{i} (\omega_i)^2[\frac{I_{i}}{I_{CM} + mr^2} -1]$$

(Please see #18 and #20 for more details on how the work and ##\Delta K## were obtained).

But I am still not getting d)...

What is missing is the Algebra. If you think that my starting point for obtaining d) :

$$W = \Delta K$$

is correct please let me know and I will include all the algebra in order to see my mistake. Otherwise I will change my approach.

Thanks.

JD_PM said:
Sorry for the delay.

OK, what I think is that we are dealing with a nonisolated system in terms of energy; the change in the total amount of energy in the system is due to the work done by the kinetic frictional force. Therefore we have:

$$W = \Delta K$$

$$- \int_0^R dr mr (\frac{I_i \omega_i}{I_{CM} + mr^2})^2 = \frac{1}{2} I_{i} (\omega_i)^2[\frac{I_{i}}{I_{CM} + mr^2} -1]$$

(Please see #18 and #20 for more details on how the work and ##\Delta K## were obtained).

But I am still not getting d)...

What is missing is the Algebra. If you think that my starting point for obtaining d) :

$$W = \Delta K$$

is correct please let me know and I will include all the algebra in order to see my mistake. Otherwise I will change my approach.

Thanks.
Why are you still writing ICM? You liked my post #24 but don't seem to have understood it. Just use Ii and mr2. When you have that you will get some simplification.

Compared to the expression you have to prove, you have extra unkowns: m, Ii etc. The equality you have found must be true for all values of these, so you can replace those with constants as you wish. It is a matter of finding a combination that gives the required answer.

JD_PM
haruspex said:
Why are you still writing ICM? You liked my post #24 but don't seem to have understood it. Just use Ii and mr2. When you have that you will get some simplification.

Compared to the expression you have to prove, you have extra unkowns: m, Ii etc. The equality you have found must be true for all values of these, so you can replace those with constants as you wish. It is a matter of finding a combination that gives the required answer.

Touché but my outcome does not change. I am pretty close though. The algebraic steps are:

$$- \int_0^R dr mr (\frac{I_i \omega_i}{I_{i} + mr^2})^2 = \frac{1}{2} I_{i} (\omega_i)^2[\frac{I_{i}}{I_{i} + mr^2} -1]$$

$$- \int_0^R dr \frac{mr I_i}{(I_{i} + mr^2)^2} = \frac{1}{2}(\frac{I_{i}-I_{i}-mr^2}{I_{i}+mr^2})$$

At this point I assumed that:

$$I_{i} = m = 1$$

To get:

$$\int_0^R dr \frac{r}{(1 + r^2)^2} = \frac{r}{2(1+r^2)}$$

These are my objections to the final result:

1) The assumption ##I_{i} = m = 1## simply does not make sense for me. I just did it with the objective of getting the desired result.

2) It does not meet the provided answer. I got an extra r on the right hand side of the equation (see #1).

What am I missing here?

JD_PM said:
The assumption Ii=m=1Ii=m=1I_{i} = m = 1 simply does not make sense for me.
It is not an assumption; they are possible values of those variables.
You obtained a general equation, so it is valid for all values. You can plug in other constant values and obtain different equations. They will all be true, but won't give the particular result you are asked for.
JD_PM said:
It does not meet the provided answer.
That is partly because the equation to be proved, as stated in post #1, is wrong. Yours is also wrong, but because you dropped a factor r (or rather, R). It should be R2 in the numerator on the right.
Also, as has been pointed out to you, "r" should only appear inside the integral. In the expressions for initial and final KE it should be "R".
So the RHS at the end should be ##\frac{R^2}{2(1+R^2)}##.

JD_PM
Thank you all for your help! My first experience on PF has been really positive! I will be back soon.

## 1. What is the conservation of energy and angular momentum in a rotating train-disk system?

The conservation of energy and angular momentum in a rotating train-disk system refers to the principle that the total energy and angular momentum of the system remain constant, even as the system undergoes rotational motion. This means that the energy and angular momentum cannot be created or destroyed, but can only be transferred between different forms within the system.

## 2. Why is conservation of energy and angular momentum important in a rotating train-disk system?

The conservation of energy and angular momentum is important in a rotating train-disk system because it helps us understand and predict the behavior of the system. It allows us to make calculations and predictions about the motion of the train and disk, and also helps us to design and optimize the system for maximum efficiency.

## 3. How does the conservation of energy and angular momentum affect the motion of a rotating train-disk system?

The conservation of energy and angular momentum affects the motion of a rotating train-disk system by ensuring that the total energy and angular momentum of the system remain constant. This means that any changes in the system's motion must be accompanied by corresponding changes in its energy and angular momentum.

## 4. Can the conservation of energy and angular momentum be violated in a rotating train-disk system?

No, the conservation of energy and angular momentum cannot be violated in a rotating train-disk system. This is a fundamental principle in physics and has been extensively tested and proven through experiments. Any apparent violations of this principle can be explained by factors such as external forces or energy losses due to friction.

## 5. How can the conservation of energy and angular momentum be applied to real-life situations involving rotating train-disk systems?

The conservation of energy and angular momentum can be applied to real-life situations involving rotating train-disk systems by using mathematical equations and principles to analyze and predict the behavior of the system. This can be useful in designing efficient and safe train-disk systems for transportation, energy production, or other applications.

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