Finding torque

  • Thread starter kbyws37
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  • #1
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A bicycle wheel, of radius 0.330 m and mass 2.00 kg (concentrated on the rim), is rotating at 4.12 rev/s. After 48.0 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?

First I did
I = (1/2)(mr^2) = (1/2)(2kg)(0.330^2 m) = 0.1089 kg m^2

Then i used the equation
Torque = I*alpha = (I)(omega)/(time)
=((0.1089 kg m^2)(4.12 rev/s x 2pi rad/rev))/(48 s)
=0.0587 N m

however i am getting this question wrong. what am i doing wrong?
 

Answers and Replies

  • #2
just a quick thought that your methods seem to be assuming that the torque force is changing linearly with respect to time. My quick thought says that probably isn't right (usually fiction effects are thought to be proportional to velocity, and probably that isn't linear with respect to time. Does this help you think about the problem differently?
 
  • #3
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sorry, i still don't get it.
i don't know where else i would put time in
 
  • #4
PhanthomJay
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kbyws37 said:
sorry, i still don't get it.
i don't know where else i would put time in
Try using I as mr^2, not mr^2/2. The rotational Inertia of a thin hoop is mr^2, which is the case here since all the mass is concentrated along the rim..
 
  • #5
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thank you so much! i got it!
 

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