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

  • Thread starter Thread starter Vriska
  • Start date Start date
  • Tags Tags
    Disk Rolling
Click For Summary

Homework Help Overview

The discussion revolves around the dynamics of two identical disks connected by a string, where one disk is allowed to drop while the other remains stationary. The participants are tasked with finding the acceleration of the center of mass in terms of angular acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between angular acceleration and linear acceleration, with some suggesting that the angular accelerations of both disks may be equal. There are attempts to clarify the setup and the forces acting on each disk, including discussions about free body diagrams and the implications of a non-stretching string.

Discussion Status

The conversation includes various interpretations of the problem setup, with some participants seeking clarification on the initial conditions and assumptions. Guidance has been offered regarding the relationship between the accelerations of the disks and the implications of the string's behavior, although no consensus has been reached on a final solution.

Contextual Notes

There is mention of potential confusion due to the wording of the problem and the need for a clearer statement of the scenario. Some participants express discomfort with the informality of their reasoning, indicating a desire for a more rigorous approach.

Vriska
Messages
138
Reaction score
2

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

Homework Equations


[/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

: (

ps: answer is 2 r*alpha
 
Physics news on Phys.org
Vriska said:

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

Homework Equations


[/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

: (

ps: answer is 2 r*alpha
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.)
 
haruspex said:
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?
 
Vriska said:
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.)
 
Vriska said:

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 ,

Please attach a picture of your setup . That would be quite helpful .
 
conscience said:
@Vriska ,

Please attach a picture of your setup . That would be quite helpful .
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?
 
haruspex said:
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
 
Vriska said:
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?
 
haruspex said:
There are two cylinders accelerating. Which has angular acceleration α, or are you told that applies to both?

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

I think they mean lower disk
 
  • #10
Vriska said:
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?
 
  • #11
haruspex said:
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?
 
  • #12
Vriska said:
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.
 
  • #13
haruspex said:
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?
 
  • #14
Vriska said:
both alphas are equal right?
Provably so, yes.
Vriska said:
how informal this is .
As long as you prove the alphas are equal (and opposite) it is rigorous.
 
  • Like
Likes   Reactions: Vriska
  • #15
haruspex said:
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!
 

Similar threads

Replies
11
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 97 ·
4
Replies
97
Views
6K
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
30
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K