Why did I get this angular momentum problem wrong?

[dangerboyy]

1. A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it as shown above. Outside edge of the disk is moving at a linear speed of V and the clay is moving at speed v/2. How does angular momentum of system after the clay sticks compare to the angular momentum of system before the clay string?
A) It is the same because there is no external torque acting on the system.
B) It is greater because the rotating mass increases, which increases the rotational inertia.
C) It is less because the speed of the disk decreases when the clay sticks to it.
D) It is less because the angular momentum of the clay opposes that of the disk.

I'm trying to prove my teacher wrong but I can't necessarily figure out how to invalidate his argument. He is trying to argue that the clay creates a torque against the wheel, thus making A not the correct answer. However, my own knowledge and every example of this on the internet chose A as the correct answer (my teacher believes the answer to be C).

2. So I guess my question is... Why doesn't the torque of the clay cause a decrease in momentum?3. I believe it is because the wheel exerts an equal and opposite amount of torque on the clay, thanks to Newton's Third Law, but I am not sure if this is correct. As stated before, I believe the answer is A, but my teacher argued that the answer is C.

 

[phyzguy]

The correct answer is clearly (A). The problem says "the angular momentum of the system", which includes the wheel and the clay. Your teacher is right that the clay exerts a torque on the wheel, thus transferring angular momentum from the clay to the wheel. But there is no net torque on the whole system, so the angular momentum of the whole system is conserved.

 

[dangerboyy]

The correct answer is clearly (A). The problem says "the angular momentum of the system", which includes the wheel and the clay. Your teacher is right that the clay exerts a torque on the wheel, thus transferring angular momentum from the clay to the wheel. But there is no net torque on the whole system, so the angular momentum of the whole system is conserved.

Thank you, I'm waiting for a response from my teacher to see if he'll finally believe me. So, the torque is what transfers angular momentum in this system? Someone on reddit told me that the wheel would create an equal and opposite torque... is this correct?

 

[mpresic3]

1. You can find the angular momentum of the wheel before the clay sticks to it. The angular momentum will be in the direction into the page.
2. You can find the angular momentum of the clay before the clay sticks to it. The angular momentum will be in the direction out of the page.

3. the total system angular momentum before the clay sticks to it: This will be L from 1. minus L from 2. (into the page)

4. Now find the moment of inertia for the clay and the wheel together.

5. Allow the velocity of the flywheel after the sticking together, call it v(final) . Find the angular momentum of the clay and wheel together from the moment of inertia and the radius and the v (final)

6. Now equate the angular momentum before the sticking to the angular momentum after the sticking. You will find the final velocity of the wheel is less than the initial velocity of the wheel.

Upshot: Bottom line: If you follow the steps, you can verify the following conclusions. Even if the angular momentum is the same (conserved), the final velocity of the wheel could be less than the initial velocity of the wheel. (You are showing this) Answer C (even if it is true) does not require that the angular momentum is not the same. Answer C does not nullify answer A, and does not require the angular momentum to change.

 

[haruspex]

The question is deficient in that it fails to specify the axis for measuring the angular momentum. If we were to choose an axis displaced to left or right of the axle then there would be an external torque coming from the axle. Glossing over that, A is clearly the intended answer.

Someone on reddit told me that the wheel would create an equal and opposite torque... is this correct?

Yes, there will be equal and opposite linear impulses at the point of contact, so equal and opposite torques about the axle.

 

[mpresic3]

I would think the conservation of angular momentum holds independent of the axis chosen. It is probably most convenient to choose the axis of rotation of the flywheel as the axis of the problem, but the conclusions should follow for any reasonable axis. Please do not choose an axis traveling near the speed of light or accelerating or something crazy like that.

 

[haruspex]

Answer C does not nullify answer A,

It does. Answer C states that the angular momentum is reduced.

 

[haruspex]

I would think the conservation of angular momentum holds independent of the axis chosen. It is probably most convenient to choose the axis of rotation of the flywheel as the axis of the problem, but the conclusions should follow for any reasonable axis. Please do not choose an axis traveling near the speed of light or accelerating or something crazy like that.

When the clay hits the wheel what stops the wheel rising?

 

[mpresic3]

You are correct. I did not phrase this adequately. I meant given that the consequence of answer c is correct, (the speed of the wheel has decreased), it does not mean the total angular momentum of the system is not conserved (i.e. the same before and after the sticking).

I feel answer C is terrible. For one problem, the answer suggests the angular momentum of the system is solely a function of the final speed of the combined system, independent of the (changed) moment of inertia. Telling us the final speed of the wheel is less than the initial speed (alone) does not even address the question as to whether the angular momentum has changed. In that respect, answer D, while incorrect at least addresses the question as to the final angular momentum. Even answer D is a better answer than answer C.

 

[dangerboyy]

Glossing over that, A is clearly the intended answer

Thank you so much, but now I think the conversation has gone beyond my understanding haha, but I'll try to show this to my teacher and see if he'll decide to change his mind.

Thank you mpresic3 as well for the help.

 

[mpresic3]

As to why the wheel rises to any extent. It is impossible to say to what degree the wheel rises. If the mass off the wheel is much greater than the clay, it will not rise much. Your suggestion that the question is deficient has merit. I do believe, outside of these deficiencies, Choice A is the most correct answer, as you have noted

 

[haruspex]

As to why the wheel rises to any extent. It is impossible to say

You miss my point. The wheel will not rise at all since it is mounted, securely one trusts, on an axle. The axle will exert a downwards impulse on the wheel equal and opposite to to the upward impulse from the clay.
If you choose an axis left or right of the axle then that impulse will create an impulsive torque about the axis.

 

[mpresic3]

Fair enough. I did not consider reaction forces the mounting would have to apply to the wheel to keep it stationary after the clay attached.

 

[Friend of Kalina]

The linear velocity of the clay is opposed to the tangential velocity of the wheel at the point of impact. Consider the case where the relative mass of the clay is such that the wheel, on its fixed axis, stops...

It is not a closed system. Neither linear nor angular momentum are conserved.

 

[haruspex]

nor angular momentum are conserved.

Why would it not be conserved about the wheel's axle? What external torque is there?

 

[PeroK]

Thank you so much, but now I think the conversation has gone beyond my understanding haha, but I'll try to show this to my teacher and see if he'll decide to change his mind.

Just to add that @haruspex is totally reliable on these matters. The other posts are misguided.

My only contribution is to ask this:

If you specify the mass of of the disc and the clay, how would your teacher go about calculating the rotation of the system after the collision?

You would do it by conservation of AM about the axis: you'd get an answer, and that answer would be correct.

But, if AM about the axis is not conserved, then how could the problem be solved?

 

[CWatters]

I guess there are better sources for the definition but..

https://en.wikipedia.org/wiki/Angular_momentum#Conservation_of_angular_momentum

A rotational analog of Newton's first law of motion might be written, "A body continues in a state of rest or of uniform rotation unless acted by an external torque."[18] Thus with no external influence to act upon it, the original angular momentum of the system is conserved

The problem defines the system as both the disc and the clay..

A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it as shown above.

There are no external torques acting on the system so whatever net angular momentum the system has before impact is conserved. Answer A.

Edit: My guess is the teacher was looking for a problem that required you to calculating the effect of the clay impacting the wheel and overlooked the exact wording of the problem. No calculation is required.

 

[PeroK]

There are no external torques acting on the system so whatever net angular momentum the system has before impact is conserved. Answer A.

Although, as @haruspex has already mentioned, there is an external force; and, whether that external force represents a torque depends on the point about which we are considering the AM.

AM is conserved about the centre of the disc; but not, for example, about the centre of mass of the system.

 

[Friend of Kalina]

Why would it not be conserved about the wheel's axle? What external torque is there?

The clay moving vertically upwards has zero angular momentum. It is moving in a straight line. If you try to impose an arbitrary axis about which it "rotates," you need to explain away an angular momentum under acceleration in a system with no friction and no external forces.

If the wheel is free to move (i.e., rotating about its axis on a frictionless surface), angular and linear momentum of the system would be conserved, although the axis of rotation would shift. But it isn't. The wheel is fixed on its axle, which, as you pointed out, does represent an (instantaneous) external force on the system. When the clay hits the wheel and sticks, it imposes a torque on the wheel. Because it is moving in a direction opposed to the tangential velocity of the wheel at the point of impact, the wheel slows. In the limit mass_clay >> mass_wheel, the wheel ends up rotating with a tangential velocity of -v/2 relative to it's initial velocity.

Imagine a mass of clay exactly sized to bring the wheel to a halt. Initially, the system has the angular momentum of the wheel alone, and linear momentum of the clay alone. After, the system is static with zero angular momentum and zero linear momentum. Neither angular nor linear momentum are conserved.

Answer C is correct. In the limit mass_clay >> mass_wheel, the absolute value of the angular momentum may be larger, but (given the original direction of rotation to be positive), the new angular momentum will be negative.

 

[haruspex]

The clay moving vertically upwards has zero angular momentum. It is moving in a straight line.

That is a basic misunderstanding. A moving object has angular momentum about any axis offset from its instantaneous line of velocity.
Let the clay's mass be m_c. It has velocity v/2 and its path misses the axle by distance r. So before the impact it has anticlockwise angular momentum rm_cv/2 about the wheel’s axle.

 

[Friend of Kalina]

That is a basic misunderstanding. A moving object has angular momentum about any axis offset from its instantaneous line of velocity.
Let the clay's mass be m_c. It has velocity v/2 and its path misses the axle by distance r. So before the impact it has anticlockwise angular momentum rm_cv/2 about the wheel’s axle.

Okay. And what is its angular momentum at distance 2r from the axle? Or its angular momentum at t = -infinity, r = -infinity when it's moving in an essentially radial direction? Your definition requires that for any fixed axis, the angular momentum of the clay is constantly varying, unless the axis of "rotation" is also moving at v/2. Where is the force accomplishing this acceleration?

Your suggestion is that the act of stickiness converts linear momentum to angular momentum, which I will accept, but that's not the same as "conserved." Unless you constrain the problem to the instant of impact and after, and decide to ignore the linear momentum of the system as irrelevant. In other words, you are redefining the problem from "A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it" to "A system at the time of impact..."

Unless the angular momentum of the clay is rm_cv sin(theta)/2, where theta is measured from a line parallel to the line of motion of the clay, with the axis of the wheel as the origin?

 

[haruspex]

what is its angular momentum at distance 2r from the axle?

It depends what you mean by that. If you mean it is moving at v/2 on a straight path that would miss the axle by 2r then its angular momentum about the axle would be m(v/2)(2r)=mvr.

when it's moving in an essentially radial direction?

Zero. For a body at $\vec r$ from an an axis and having linear momentum $\vec L$ its angular momentum about the axis is $\vec r\times\vec L$.

Your definition requires that for any fixed axis, the angular momentum of the clay is constantly varying

If it is moving in a straight line then the distance of that line from the axis is not varying.

See http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html.

 

[Friend of Kalina]

If it is moving in a straight line then the distance of that line from the axis is not varying.

See http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html.

What I come up with is that the angular momentum of a mass $m$ moving in a straight line at velocity $\frac v 2$ at an angle $\Theta$ between the direction of motion and an offset fixed axis, at a distance $d$ from that axis, has (constant) angular momentum $\omega = {mvd} {\sin(\theta)}$.

At the point of closest approach, $d = r$, $\Theta$ = $\frac \Pi 2$, and

$\omega = \frac {mvr} 2$

At least, I think that's what you're trying to say. With an appropriate mass, you can get the wheel to stop, which means the net angular momentum of the system prior to impact was zero.

 

[haruspex]

at an angle Θ between the direction of motion and an offset fixed axis, at a distance d from that axis, has (constant) angular momentum $\omega = {mvd} {\sin(\theta)}$.

Yes, but that’s a somewhat complicated way of saying it. For a straight line path, $d\sin(\theta)$ is constant, being the distance between the path and the point.
(L is the usual symbol for an angular momentum. And you specified a velocity of v/2, so you mean m(v/2)d sin(θ).)

So do you agree now that the angular momentum of the system about the axle as axis is conserved?

 

[Friend of Kalina]

Now I do. Because I can see how a mass with a changing angular velocity and changing distance reduces to the constant angular momentum. Otherwise, I was having trouble seeing how a mass at a distance of, say, 100,000r, on a "nearly" radial approach to the axis and with "near zero" angular velocity, had the same angular momentum, or how the angular momentum with respect to a series of axes arranged parallel to the linear motion would be identical at all times for all axes.

If I start with $d \sin \theta$, the effect of distance and angle is apparent.

 

[CWatters]

Although, as @haruspex has already mentioned, there is an external force; and, whether that external force represents a torque depends on the point about which we are considering the AM.

AM is conserved about the centre of the disc; but not, for example, about the centre of mass of the system.

Ah ok. The axle is included in the system but the bearings and it's mounting are not. So an external force crosses the boundary