70.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on

Question

70.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 180 N. The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50 N.m between the axle of the stone and its bearings. How much force must be applied tangentially at the end of a crank handle 0.500 m long to bring the stone from rest to 120 rev/min in 9.00 s? After the grindstone attains an angular speed of 120 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev/min? How much time does it take the grindstone to come from 120 rev/min to rest if it is acted on by the axle friction alone?

Explanation / Answer

I = ½mr² = ½ * 70kg * (0.26m)² = 2.37 kg·m²
= 120rev/min * 2rad/rev * 1min/60s = 12.6 rad/s
so = / t = 12.6rad/s / 9s = 1.4 rad/s²

A) net torque = I = 2.37kg·m² * 1.4rad/s² = 3.32 N·m
But also = F*r = F*0.5m - 0.6 * 180N * 0.26m - 6.50N·m

F=75.8

B) Now 0 N·m = F*0.5m - 0.6 * 180N * 0.26m - 6.50N·m

F = 69.16 N·m

C) = 6.50 N·m = I = 2.37kg·m² *
= 2.74 rad/s²
t = / = 12.6rad/s / 2.74rad/s² = 4.6 s

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