Acceleration of the center of mass of this cylinder

  • #1
A 2.81 kg hollow cylinder with inner radius 0.29 m and outer radius 0.5 m rolls without slipping when it is pulled by a horizontal string with a force of 47.7 N, as shown in the diagram below.

Its moment of inertia about the center of mass is .5m(r(out)^2 + r(in)^2).

What is the accelereation of the cylinder's center of mass? Answer in units of m/s^2.


What am I doing wrong? I found the Torque of the hollow cylinder by T = F(r). Then I found the angular acceleration by Torque = Interia * Alpha. Inertia was found using the supplied forumula. After finding the angular acceleration I found the Tangential Acceleration by TangentialAcceleration = radius * AngularAcceleration. What am I doing wrong?
 

Answers and Replies

  • #2
siddharth
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Isn't the moment of inertia of a hollow cylinder
[tex] \frac{1}{2} M (R_1^2 + R_2^2) [/tex]
So, your value of M/2 is not 0.5 but 2.81/2
 
  • #3
siddharth said:
Isn't the moment of inertia of a hollow cylinder
[tex] \frac{1}{2} M (R_1^2 + R_2^2) [/tex]
So, your value of M/2 is not 0.5 but 2.81/2
Hence, .5M which is the same as 2.81/2.
 
  • #4
siddharth
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Oh, you mean 0.5 * 2.81 . Didn't see that, sorry.

There wil be a torque due to friction, the value of which is not known. So, I don't think you can use the above equations alone to get the answer.

Have you applied Newton's second law in the horizontal direction? (ie, F-f = ma). Then eliminate f using all the equations and solve for a. That should give you the correct answer.
 

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