Question

A 65.0-kg grindstone is a solid disk 0.550m in diameter. You press an ax down on the rim with a normal force of 150N (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.

Part 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 120 rev/min in 9.00s ?

Part B

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?

Part C

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

Let: m = 65.0 kg be the mass of the grindstone,

r = 0.275 m be its radius,

P = 150 N be the normal force from the axe,

u = 0.60 be the coefficient of friction between the axe and the stone,

F= 6.50 Nm be the friction torque in the bearing, T be the tangential force needed at the end of the crank.

I be the moment of inertia of the grindstone about its centre, a be the angular acceleration of the grindstone,

w = 120 rev/min be its initial angular velocity,

t = 9.00 sec be the stopping time,

L = 0.500 m be the length of the crank handle.

TL - F - uPr = Ia ...(1)

w = at ...(2)

I = mr^2 / 2 ...(3)

Substituting for a from (2) and I from (3) in (1):

TL - F - uPr = mr^2 w / (2t) T

= [ mr^2 w / (2t) + F + uPr ] / L w= 120 * 2pi / 60 = 4pi rad/sec.

T = [ 65.0 * 0.275^2 * 4pi / (2 * 9.00) + 6.50 + 0.60 * 150 * 0.275 ] / 0.500 = 69.36N.

Putting a = 0 in (1):

TL = F + uPr T = (F + uPr) / L

= (6.50 + 0.6 * 150 * 0.275 ) / 0.5 = 62.5 N.

Slowing down with the axle friction alone: 0 = w - at ...(4)

F = Ia = mr^2 a / 2 a = 2F / (mr^2) ...(5)

Substituting for a from (5) in (4): t = mr^2 w / (2F)

= (65 * 0.275^2 * 4pi) / (2 * 6.5) = 4.749 sec.