Friction on a rotating disc

  • #1
If I have a spinning circular disc of uniform density, how would I find the torque generated by friction, if the disc is lying flat against a table with coefficient of friction μ? τ=Fxr, but what is F, and what is r in this case?
 

Answers and Replies

  • #2
Simon Bridge
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Welcome to PF;

You use calculus ... divide the disk into narrow rings.
 
  • #3
Thank you Mr. Bridge, for your reply. I am still confused however, as towhat the force on a differential element would be. The friction force would be μ N=μ m g(or would it be μ g dm?) The friction force would always act perpendicular to position, so torque would be upward with magnitude equal to the product of the distance from the center and the force. However, I am not sure what I would be integrating with respect to, and I am also unsure as to what the bounds on the integration would be. Please clarify? Thank you in advance.
 
  • #4
Simon Bridge
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That would be:
##d\tau = \mu g r \text{d}m##

The limits of the integration depend on how you define dm ...
hint: relate dm to dV (volume) and exploit the symmetry.

Friction force always acts opposite to the velocity vector.
Friction torque always twists the opposite way to the angular velocity.
 

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