How much friction force must the brake apply to the rim to bring the disk to a halt in 2.30s ?
How much frictional force must the brake apply to the rim to bring the disk to a complete stop in 2.30 seconds? The disk has a mass of 3.10 kg and a diameter of 31.0 cm, and it is initially spinning at 320 revolutions per minute (rpm).
3 Answers
Feb 06, 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
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