# Man rotating in a merry-go-round and grabbing a pendulum

## Homework Statement:

A horizontal disk rotates at an angular speed ##\omega_1## clockwise around its center of mass (which is where the origin of the non-inertial frame lies). A person with a rope is turning counterclockwise at an angular speed ##\omega_2## on that disk. The rope has length ##l_2##. The man himself is at a distance ##l_1## from the origin of the disk and the rope also moves at a height ##h## above the horizontal disk. A stone is attached to the end of that rope, which becomes a pendulum. At ##t=0##, the pendulum is maximally far away from the origin (thus, the origin, the man and the stone stand on a straight line).

a) Find an expression for the speed ##v(t)## of the stone at a given moment ##t## in terms of the Cartesian coordinate system at rest. Choose the origin of such a system to be located at the center of the disk. Choose the ##z## axis to be vertically upwards, the ##x## axis according to the straight line that goes through the stone at ##t=0## and the ##y## axis perpendicular to the previous ones.

b) Suppose the man releases the stone at ##t=0##. Where does the stone fall on the disk?

NOTE:This exercise has been translated from Dutch by Google Translator. Let me know if there's something unclear and we can all discuss about it.

## Relevant Equations:

Where:

1) ##A## is the translational acceleration, ##\Omega## the angular velocity and ##\dot \Omega## the angular acceleration (all relative to the inertial frame attached to the ground ##F##).

2) ##r'##, ##v'## and ##a'## are the position, velocity and acceleration vectors, all relative to the frame attached to the roundabout ##F'## (and thus ' has nothing to do with derivatives on above equation).

- Picture of the problem (imagine the man has a pendulum):

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a) I think the exercise is just asking us to find ##v(t)## of the stone with respect to the inertial frame ##F##.

If the pendulum is not released, then its velocity will be same that the velocity of the man. Thus:

Expression for ##v(t)## with respect to the inertial frame F:

$$\omega_s=\omega_m−\omega_d$$

$$\frac{v_s}{l_1 + l_2} = \frac{v_m}{l_1} - \frac{v_d}{l_2}$$​

Thus the velocity of the stone is:

$$v_s = (l_1 + l_2) \frac{Rv_m + l_1 v_d}{R l_1}$$​

But this velocity is not in function of ##t##, so I guess this is not what the exercise is asking for.

Besides I am not convinced that the assumption I made above is correct (If the pendulum is not released, then its velocity will be same that the velocity of the man)...

b) Here I guess the exercise is asking for a qualitative answer. I expect the pendulum when released to have the direction of the black arrow:

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haruspex
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If the pendulum is not released, then its velocity will be same that the velocity of the man.
Why would you think that?
But I don't see where you used that, so maybe I don't understand what you mean.
However, the equation you did write is only true at t=0.
I would approach it by finding the velocity of the stone relative to the man at time t.
b) Here I guess the exercise is asking for a qualitative answer.
Again, why? There seems to be enough information.

JD_PM

The exercise statement is not clear, I agree. I will edit it.

haruspex
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The exercise statement is not clear, I agree. I will edit it.
No, I think I understand the problem statement.

No, I think I understand the problem statement.
Alright so I will not edit it. I was a little worried about it because I used a translator for all the text.

If the pendulum is not released, then its velocity will be same that the velocity of the man. Thus:
Why would you think that?
Because I thought that the man on the merry-go-round would be holding the pendulum at a small angle while rotating.

I would approach it by finding the velocity of the stone relative to the man at time t.
So you are suggesting we approach the problem with respect to the rotating frame, aren't you?

Again, why? There seems to be enough information.
What kind of information do you think is missing?

haruspex
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Because I thought that the man on the merry-go-round would be holding the pendulum at a small angle while rotating.
It says the "rope moves at height h". This implies the stone is moving in a horizontal circle about the man. Now, strictly speaking I don't think that would happen, but if the man is rotating much faster than the disc then this would be a reasonable approximation.
The problem is gravity. Without the disc rotation, we could work out the angle the rope makes to the horizontal; with the disc rotation, it is not obvious that this is constant.
I need to think about that some more.
So you are suggesting we approach the problem with respect to the rotating frame, aren't you?
Yes.
What kind of information do you think is missing?
I said there is no information missing. You wrote that you thought only a qualitative answer was wanted. That could be true if information were missing.

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haruspex
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It says the "rope moves at height h". This implies the stone is moving in a horizontal circle about the man.
On second thoughts, that is before the stone is attached. After the stone is attached it says it becomes a pendulum.
My problem is that unless the man is pulling on the rope in a varying way, I don't think the stone will maintain a constant angular speed about the man.
@Orodruin , any suggestions here?

Orodruin
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On second thoughts, that is before the stone is attached. After the stone is attached it says it becomes a pendulum.
My problem is that unless the man is pulling on the rope in a varying way, I don't think the stone will maintain a constant angular speed about the man.
@Orodruin , any suggestions here?
I think it is not very clear exactly what is being asked. This may be a problem of the translation. Unfortunately, I do not know Dutch so there is also no way for me to have a look at the statement in the original language.

It may be of relevance to know what level this question is posed at to try to understand the intentions of the problem writer.

JD_PM
haruspex
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I think it is not very clear exactly what is being asked. This may be a problem of the translation. Unfortunately, I do not know Dutch so there is also no way for me to have a look at the statement in the original language.

It may be of relevance to know what level this question is posed at to try to understand the intentions of the problem writer.
Thanks.

@JD_PM , I suggest you treat the rope as horizontal for part a. Or, you could take it that the rate of rotation of the disk is so low that it does not significantly affect the angle the rope makes to the horizontal. If that is not the intent then I don't think all the given conditions can be met.

For part b, you don't need to make any such assumptions, but if you don't then the answer will be inconsistent with that for part a.

JD_PM
Unfortunately, I do not know Dutch so there is also no way for me to have a look at the statement in the original language.
Me neither, so I had no option but use an online translator.

It may be of relevance to know what level this question is posed at to try to understand the intentions of the problem writer.
This is a problem coming from a Classical mechanics course (2nd year of the Physics' Bachelor). Our textbook is Gregory's Classical mechanics.

We are expected to master general Physics already.

I suggest you treat the rope as horizontal for part a.
Alright, I will make such assumption.

you could take it that the rate of rotation of the disk is so low that it does not significantly affect the angle the rope makes to the horizontal.
If we were to take the angular velocity of the disk to be so low, the centrifugal and Coriolis forces would become negligible.

I think this was not the intention of the exercise, as our teacher insisted we should be able to work out exercises in rotating frames where both centrifugal and Coriolis forces become a factor.

Orodruin
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I believe the problem is too complex for the level if you do not make some very basic assumptions that generally are not going to be satisfied:
- The rope indeed moves in the horizontal plane. This is essentially saying that it is not affected by gravity.
- The rope rotates with the man. In general, the mention of a pendulum draws the mind to some sort of oscillations that would be very complex in this scenario otherwise.

In many respects, it would then have been better to replace the rod with a solid stick beigh held horizontally and rotating with the man.

With these assumptions, the problem becomes relatively easily solvable using basic vector algebra. (With the slight ambiguity of whether the man’s angular frequency is relative to the disc or the non-rotating frame).

In many respects, it would then have been better to replace the rod with a solid stick beigh held horizontally and rotating with the man.
So you suggest approaching the problem as if the man were to hold a rod instead of a pendulum?