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How much friction force must the brake apply to the rim to bring the disk to a halt in 2.30s ?

The 3.10 kg, 31.0-cm-diameter disk is spinning at 320 rpm.

3 Answers

A
Alize Kuhn

Feb 20, 2025

Let F = required frictional force.Then torque = F * (0.31/2) m-N Rate of change in angular momentum = I ω / t= (1/2) m r^2 ω / t= (1/2) (3.1) (.31/2)^2 (320 x 2π / 60) / 2.30= 0.54 m-N Torque = rate of change of angular momentum=> F * (0.31/2) = 0.54=> F = (0.54) / (0.155) = 3.48 N... Show More

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