A m = 50.0-kg grindstone is a solid disk 0.510 m in diameter. You press an ax do

Question

A m = 50.0-kg grindstone is a solid disk 0.510 m in diameter. You press an ax down on the rim with a normal force of F = 155 N. The coefficient of kinetic friction between the blade and the stone is 0.70, and there is a constant friction torque of 6.50 N · 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 110 rev/min in 11.00 s? (b) After the grindstone attains an angular speed of 110 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular speed of 110 rev/min? c) How much time does it take the grindstone to come from 110 rev/min to rest if it is acted on by the axle friction alone?

Explanation / Answer

(a)

Let:
m = 50.0 kg be the mass of the grindstone,
r = 0.255 m be its radius,
P = 155 N be the normal force from the axe,
u = 0.70 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 = 110 rev/min be its initial angular velocity,
t = 11.00 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 = 110 * 2pi / 60 = 11.52 rad/sec.

T = [ 50.0 * 0.255^2 * 11.52 / (2 * 11.00) + 6.50 + 0.70 * 155 * 0.255 ] / 0.500
= 71.74 N.

(b)
Putting a = 0 in (1):
TL = F + uPr
T = (F + uPr) / L
= (6.50 + 0.70 * 155 * 0.255 ) / 0.5
= 68.335 N.

(c)

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)
= (50 * 0.255^2 * 11.52) / (2 * 6.5)
= 2.88 sec

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