How Is the Retarding Torque Calculated in a Rotating System with Friction?

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SUMMARY

The discussion focuses on calculating the retarding torque in a rotating system with friction, specifically involving a thin steel rod with attached balls. The system consists of a rod of mass 6.4 kg and length 1.2 m, rotating at an initial angular velocity of 30 revolutions per second, which converts to 188.4 rad/s. The angular acceleration is determined to be -5.89 rad/s², leading to the conclusion that the net torque can be calculated using the formula torque = I * angular acceleration, where I is the moment of inertia of the entire system, including the rod and the attached balls.

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1. "A small ball of mass 1.06 is attached to each end of a thin steel rod of length 1.2 m and mass 6.4kg. The rod is constrained to rotate in the horizontal plane about a vertical axis through its midpoint. At some instant, it is observed to rotate with an angular velocity of 30 rev / s. Due to friction, it comes to rest 32 seconds later. Assuming a constant frictional torque, compute

the retarding torque exerted by friction.



2.

First I converted the initial angular velocity to rad/s which is 188.4 rad/s

I computed the angular acceleration to be -5.89 rad/s2

we know the final angular velocity is 0

I THOUGHT i could do

torque,net = I*angular acceleration

and assume that friction is the only torque that's making it slow down ??

I'm so confused..
I think I is (1/3)M,rod*L^2

any help or suggestions would be greatly appreciated.
 
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I is not that, since a ball is attached to the end, I,total would be I of the rod plus M(ball)(L/2)^2
hope it helps
 

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