A m = 70.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax do

Question

A m = 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 F = 150 N. The coefficient of kinetic friction between the blade and the stone is 0.50, and there is a constant friction torque of 6.50N middot m between the axle of the stone and its bearings. (a) 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 150 rev/min in 11.00 s? (b) After the grindstone attains an angular speed of 150 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 150 (c) How much time does it take the grindstone to come from 150 rev/min to rest if it is acted on by the axle friction alone?

Explanation / Answer

mass, m=70 kg

diameter, d=0.52 m

radius, r=0.26 m

normal force, N=150 N

uk= 0.5

frictional torque, Tk=6.5 Nm

a)

w=150 rev/min = 15.71 rad/sec

time, t=11 sec

use,

torque= I*alpa

=1/2*m*r^2*(w/t)

=1/2*70*0.26^2*(15.71/11)

=3.38 N.m

and

net torque =0

F_tan*l - T-Tf-uk*N*r=0

F_tan*0.5 - 3.38 - 6.5 - 0.5*150*0.26=0

===> F_tan = 58.76 N

b)

to maintain constant angular speed,

net torque =0

F_tan*l -Tf-uk*N*r=0

F_tan*0.5 - 6.5 - 0.5*150*0.26=0

===> F_tan = 52 N

c)

by using conservation of energy,

L=I*w

Torque*t=I*w

6.5*t=1/2*m*r^2*w

6.5*t=1/2*70*0.26^2*15.71

===> t=5.72 sec

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