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Find speed of a rotating disk

  • #1

Homework Statement


A solid uniform disk of mass 21.0 kg and radius 85.0 cm is at rest flat on a frictionless surface A string is wrapped around the rim of the disk and a constant force of 35.0 N is applied to the string. The string does not slip on the rim.
1- When the disk has moved a distance of 5.5 m, determine how fast it is moving.

Homework Equations


Kinetic energy= (mv^2)/2 Rotational kinetic energy = (Iw^2)/2 Work=Fdcos(theta)
I(disk)=(mr^2)/2

The Attempt at a Solution


Since the object is at rest, I know that the initial kinetic energy of the disk is zero and that after applying a force it will have final kinetic energy both rotation and translation.
So my energy of conservation equation would be
Work done by pulling= Kinetic energy Translational + Kinetic energy Rotational
(35)(5.5) = v^2( (m/2) + (m/4)
When I solve for v, I get 3.49. But the answer is 4.3 m/s. Is my conservation of energy equation wrong?
 

Answers and Replies

  • #2
BvU
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Yes.
 
  • #3
Yes.
What is wrong with the equation? Could you please guide me?
 
  • #4
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What is wrong with the equation? Could you please guide me?
Note you used "v" and "w" in your relevant equations. Did you think they are the same?
 
  • #5
Doc Al
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To start with, your calculation of the work done is incorrect. (Reconsider the distance over which the applied force acts.)
 
  • #6
Note you used "v" and "w" in your relevant equations. Did you think they are the same?
No. I wrote the w in terms of v. Isn't that correct?

To start with, your calculation of the work done is incorrect. (Reconsider the distance over which the applied force acts.)
Doesn't the pulling force cause the disk to move a distance of 5.5m? So wouldn't the work formula include 5.5m? I am confused. Could you please elaborate.
 
  • #7
Doc Al
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Doesn't the pulling force cause the disk to move a distance of 5.5m?
True.

So wouldn't the work formula include 5.5m?
Realize that the point of application of the force is on the string. Imagine you are pulling that string. How far will you have to pull it to meet the conditions of this problem?
 
  • #8
True.


Realize that the point of application of the force is on the string. Imagine you are pulling that string. How far will you have to pull it to meet the conditions of this problem?
So is d then (5.5+0.85)m ?
 
  • #9
Doc Al
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So is d then (5.5+0.85)m ?
Where did the 0.85 m come from? (The radius?)

See if you can figure out how much string unwinds from the disk.
 
  • #10
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I suspect that the OP is making the problem more difficult than it has to be (I think you can get 4.3 m/s using a two-step method). Remember that the problem tells you what the net force on the disk is.
 
Last edited:
  • #11
haruspex
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I suspect that the OP is making the problem more difficult than it has to be (I think you can get 4.3 m/s using a two-step method). Remember that the problem tells you what the net force on the disk is.
Quite so, but it is also important that the student come to understand why the attempted method went wrong.
 

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