# A disk is tied to a stationary rolling disk and let go

## Homework Statement

The dropped disk rolls down with acceleration of centre of mass. The disks are identical, find accleartion of cm in terms of angular acceleration

[/B]
a = r*alpha

## The Attempt at a Solution

I don't know where to begin, I think the alphas should be equal. but I'm not sure, let's assume that. so the string accelerates at r*alpha, this should be the same as r*alpha + Acm of the other disk... r*alpha = r*alpha + acm...= 0

: (

haruspex
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## Homework Statement

The dropped disk rolls down with acceleration of centre of mass. The disks are identical, find accleartion of cm in terms of angular acceleration

[/B]
a = r*alpha

## The Attempt at a Solution

I don't know where to begin, I think the alphas should be equal. but I'm not sure, let's assume that. so the string accelerates at r*alpha, this should be the same as r*alpha + Acm of the other disk... r*alpha = r*alpha + acm...= 0

: (

You clearly have not stated the whole question. You start mentioning one disc, then imply there is more than one. Later, there is mention of a string.
Please state the whole question as precisely as you can. (I understand that it may be a translation.)

You clearly have not stated the whole question. You start mentioning one disc, then imply there is more than one. Later, there is mention of a string.
Please state the whole question as precisely as you can. (I understand that it may be a translation.)

You have a disk attached like a toilet paper, it can roll, but can't go down. On it a string is wound, the other end of the string is wound on another identical disk, now the second disk is let drop, it accelerates downward without slipping with acceleration of cm as Acm. they're asking the relationship between Acm and angular velocity. does this make sense?

haruspex
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You have a disk attached like a toilet paper, it can roll
Do you mean that it can rotate on an axle through its centre? (That's not rolling, technically.)

## Homework Statement

The dropped disk rolls down with acceleration of centre of mass. The disks are identical, find accleartion of cm in terms of angular acceleration

@Vriska ,

haruspex
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@Vriska ,

The given answer is consistent with:
Two identical cylinders with horizontal axes, one above the other.
A string wraps around both; between the cylinders the string is vertical.
The upper cylinder is free to rotate on an axle through its centre. The lower one is falling, constrained only by the string.

@Vriska, can you confirm that?
Does the question later ask for the acceleration in terms of g?

The given answer is consistent with:
Two identical cylinders with horizontal axes, one above the other.
A string wraps around both; between the cylinders the string is vertical.
The upper cylinder is free to rotate on an axle through its centre. The lower one is falling, constrained only by the string.

@Vriska, can you confirm that?
Does the question later ask for the acceleration in terms of g?

yes that's it, much better put, thanks. it's not in terms of g, the answer is 2*r*alpha

haruspex
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yes that's it, much better put, thanks. it's not in terms of g, the answer is 2*r*alpha
There are two cylinders accelerating. Which has angular acceleration α, or are you told that applies to both?

There are two cylinders accelerating. Which has angular acceleration α, or are you told that applies to both?

There are two cylinders accelerating. Which has angular acceleration α, or are you told that applies to both?

I think they mean lower disk

haruspex
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I think they mean lower disk
Have you drawn a free body diagram showing the forces on each cylinder?
Can you figure out the angular acceleration of the upper one?

Have you drawn a free body diagram showing the forces on each cylinder?
Can you figure out the angular acceleration of the upper one?

I'm getting things with mg which won't go away.

Tr = mr^2/2 *alpha

T = mr/2*alpha

for the lower disk

mg - T = ma

...
Now what?

haruspex
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I'm getting things with mg which won't go away.

Tr = mr^2/2 *alpha

T = mr/2*alpha

for the lower disk

mg - T = ma

...
Now what?
Since α is the angular acceleration of the lower cylinder, I assume your torque equation above is for tgat one. What about the other cylinder?
Another fact to use is that the string does not stretch. That creates a relationship between the accelerations.

Since α is the angular acceleration of the lower cylinder, I assume your torque equation above is for tgat one. What about the other cylinder?
Another fact to use is that the string does not stretch. That creates a relationship between the accelerations.

both alphas are equal right? If the string does not stretch then the acceleration of the string is the accleartion of thr disk. We have the string being removed from top disk at r*alpha, from the bottom at r*alpha, so to keep in touch from all this string-losing, the acceleration of centre of mass should be 2*r*alpha. Thanks!

that's the answer , but I'm super uncomfortable with how informal this is . Is there a better way to put it?

haruspex
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both alphas are equal right?
Provably so, yes.
how informal this is .
As long as you prove the alphas are equal (and opposite) it is rigorous.

• Vriska
Provably so, yes.

As long as you prove the alphas are equal (and opposite) it is rigorous.

Ah thank for your help, really appreciate it. My textbook had given a really confusing solution, that left me scratching my head for a looong time.

thanks again!