Question

A 70.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 170 N (Figure 1). The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N⋅mbetween 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?

Answer 1

given
Torque due to friction, Tf = -6.5 N.m

w1 = 0

w2 = 120 rev/mi = 120*2*pi/60 = 12.57 rad/s

t = 9.00 s

angular acceleration during speeding up, alfa = (w2 - w1)/t

= (12.57 - 0)/9

= 1.397 rad/s^2

moment of inertia of grindstone, I = 0.5*M*R^2

= 0.5*70*(0.54/2)^2

= 2.55 kg.m^2

now net torque, Tnet = I*alfa1

Tf + T_applied = I*alfa1

Tf + F_applied*R*sin(90) = I*alfa

F_applied*R = I*alfa1 - Tf

F_applied = (I*alfa1 - Tf)/R

= (2.55*1.397 - (-6.5))/0.27

= 37.3 N <<<<<<<<<<<<-------------------Answer

b) Apply, Tnet = 0

Tf + T_appled = 0

T_applied = -Tf

F_applied*R*sin(90) = -(-6.5))

F_applied = 6.5/R

= 6.5/0.27

= 24.1 N <<<<<<<<<<<<-------------------Answer

c) angular acceleration while stoping, alfa2 = Tf/I

= -6.5/2.55

= -2.55 rad/s^2

time taken to stop, t = (wf - wi)a/lfa2

= (0 - 12.57)/(-2.55)

= 4.93 s <<<<<<<<<<<<-------------------Answer