A 55.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down o

Question

A 55.0-kg grindstone is a solid disk 0.540 m in diameter. You press an ax down on the rim with a normal force of 180 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?m between the axle of the stone and its bearings.
1)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 7.00
s ?
2)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?
3)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

1)

frictional force acting due to axe , f = u*N

f = 0.60 * 180

f= 108 N

Now , torque due to friction Tf = (0.540/2)* 108

Tf = 29.16 N.n

wf = 120 rev/min

wf = 120 * 2pi/6-

wf = 12.56 rad/s

Now, angular acceration , a = 12.56/7

a = 1.79 rad/s^2

Now, for the torque ,

F * 0.5 - 29.16 - 6.50 = 55 * 0.5 * (0.54/2)^2 * 1.79

F = 78.52 N

the force needed is 78.52 N

2)

for constant angular speed ,

angular acceleration = 0

F * 0.5 - 29.16 - 6.50 = 0

F = 35.66 N

3)

Here , for leaving the stone to come to rest

29.16 + 6.50 = - 55 * 0.5 * (0.54/2)^2 * a

a = 17.87 rad/s^2

Using first equation of motion

0 = 12.56 - 17.87 * t

t = 0.71 s

the time taken to stop is 0.71 s

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