How Is Torque Generated by Friction on a Rotating Disc Calculated?

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To calculate the torque generated by friction on a rotating disc, the disc can be divided into narrow rings for analysis. The friction force is determined by the equation F = μN, where N is the normal force, which can be expressed as μmg or μg dm for differential mass elements. The torque can be expressed as dτ = μg r dm, where r is the distance from the center of the disc. Integration is necessary to find the total torque, with limits defined by the mass distribution of the disc. Understanding the relationship between dm and dV, along with the symmetry of the disc, is crucial for proper integration.
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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?
 
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Welcome to PF;

You use calculus ... divide the disk into narrow rings.
 
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.
 
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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