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

In summary: 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.In summary, the question involves two identical cylinders, one above the other with a string wrapped around them. The upper cylinder is free to rotate while the lower one is falling and constrained by the string. The question asks for the relationship between the acceleration of the center of mass and the angular acceleration. After drawing a free body diagram and using the fact that the string does not stretch, it can be proven that the acceleration of the center of mass is 2 times the radius times the angular acceleration.
  • #1
Vriska
138
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
 
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  • #2
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.)
 
  • #3
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?
 
  • #4
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.)
 
  • #5
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 .
 
  • #6
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?
 
  • #7
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
 
  • #8
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?
 
  • #9
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.
 
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  • #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!
 

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

What is the purpose of tying a disk to a stationary rolling disk and letting it go?

The purpose of this experiment is to observe the motion and behavior of the two disks when they are connected and released.

What are the factors that affect the motion of the disks in this experiment?

The motion of the disks can be affected by various factors such as the mass, radius, surface texture, and friction of the disks, as well as the force used to tie them together.

How does the motion of the disks change over time?

As the disks start rolling, they will experience a decrease in speed due to friction and air resistance. The speed will continue to decrease until the disks come to a stop. The direction of motion may also change depending on the initial force and the surfaces of the disks.

What can we learn from this experiment?

By observing the motion of the disks, we can learn about the principles of inertia, friction, and rotational motion. We can also use this experiment to study the effects of different variables on the motion of the disks.

Are there any real-world applications for this experiment?

This experiment can help us understand the behavior of rolling objects, which can be applied to various real-world scenarios such as the motion of wheels on a car or the movement of a ball on a surface. It can also be used in engineering and design processes to optimize the performance of rolling objects.

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