A 60.0-kg grindstone is a solid disk 0.530m in diameter. You press an ax down on

Question

A 60.0-kg grindstone is a solid disk 0.530m 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 7.00s ?

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

Explanation / Answer

Given:
m = 60 kg (mass of the grindstone)
r = 0.265m 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 = 7.00 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 = [ 60* 0.265^2 * 4pi / (2 * 7.00) + 6.50 + 0.60 * 150 * 0.265 ] / 0.500
= 68.26 N.

Putting a = 0 in (1):
TL = F + uPr
T = (F + uPr) / L
= (6.50 + 0.6 * 150 * 0.265 ) / 0.5
= 60.7 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)
= (60 * 0.265^2 * 4pi) / (2 * 6.5)
= 4.07 sec.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.