A m = 70.0-kg grindstone is a solid disk 0.450 m in diameter. You press an ax do

Question

A m = 70.0-kg grindstone is a solid disk 0.450 m in diameter. You press an ax down on the rim with a normal force of F = 170 N. The coefficient of kinetic friction between the blade and the stone is 0.70, 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 140 rev/min in 11.00 s? N After the grindstone attains an angular speed of 140 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 140 rev/min? N How much time does it take the grindstone to come from 140 rev/min to rest if it is acted on by the axle friction alone?

Explanation / Answer

initial velocity = wo = 0

final velocity = w1 = 140 rev/min = 140*2pi/60 = 14.66 rad/s

time = 11

alpha = (w1-wo)/t = 14.66/11

net torque = I*alpha

(Ftan*r - Tf - u*F*R ) = I*alpha

I = (1/2)*m*R^2

R = 0.45/2 = 0.225 m

(Ftan*0.5 - 6.5 - (0.7*170*0.225)) = (1/2)*70*0.225^2*14.66/11

Ftan = 71.27 N

++++++++

b)

for constant motion

alpha = 0

Ftan = 66.55 N

+++++++++++

torque = 6.5 Nm

torque = I*alpha

6.5 = (1/2)*70*0.225^2*14.66/t

t = 3.99 s

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