Question

A 70.0-kg grindstone is a solid disk 0.530m in diameter. You press an ax down on the rim with a normal force of 160N (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.

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 8.00s ?

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?

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 = 70.0 kg be the mass of the grindstone,
r = 0.265 m be its radius,
P = 160 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 = 8.00 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 = mr² / 2 ---------------(3)

Substituting for a from (2) and I from (3) in (1):
TL - F - uPr = mr² w / (2t)

T = [ mr² w / (2t) + F + uPr ] / L

w = 120 * 2pi / 60 = 4pi rad/sec.

T = [ 70.0 * 0.265² * 4pi / (2 * 8.00) + 6.50 + 0.60 * 160 * 0.265 ] / 0.500
= 71.59 N.

Putting a = 0 in (1):
TL = F + uPr

T = (F + uPr) / L
= (6.50 + 0.6 * 160 * 0.265 ) / 0.5
= 63.88 N.

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

F = Ia
= mr² a / 2
a = 2F / (mr²) ---------(5)

Substituting for a from (5) in (4):
t = mr² w / (2F)
= (70.0* 0.265² * 4pi) / (2 * 6.5)
= 4.74 sec.