A m = 60.0 -kg grindstone is a solid disk 0.470 m in diameter. You press an ax d
ID: 2264292 • Letter: A
Question
A m = 60.0-kg grindstone is a solid disk 0.470 m in diameter. You press an ax down on the rim with a normal force of F = 165 N. The coefficient of kinetic friction between the blade and the stone is 0.50, and there is a constant friction torque of 6.50 N
A m = 60.0-kg grindstone is a solid disk 0.470 m in diameter. You press an ax down on the rim with a normal force of F = 165 N. The coefficient of kinetic friction between the blade and the stone is 0.50, and there is a constant friction torque of 6.50 N · m between the axle of the stone and its bearings. 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 10.00 s? N 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? N 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? sExplanation / Answer
(a)
We know that
TL - F - uPr = Ia ----(1)
where T be the tangential force needed at the end of the crank.
L = 0.500 m be the length of the crank handle.
P = 165 N be the normal force from the axe
r = 0.235 m be its radius
I be the moment of inertia of the grindstone about its centre
a be the angular acceleration of the grindstone
F 6.50 Nm be the friction torque in the bearing
u = 0.50 be the coefficient of friction between the axe and the stone,
w = at ----(2)
w = 120 rev/min be its initial angular velocity
where t = 10 sec be the stopping time
I = mr^2 / 2 ----(3)
m = 60 kg be the mass of the grindstone
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 * 2 * pi / 60 = 4*pi rad/sec.
T = [ 60 * 0.235^2 * 4pi / (2 * 10) + 6.50 + 0.50 * 165 * 0.235 ] / 0.500
= [41.62/20 + 6.50 + 19.39] / 0.500
= [2.081 + 6.50 + 19.39] / 0.500
= 27.971 / 0.500
= 55.942 N
(b)
Putting a = 0 in (1):
TL = F + uPr
T = (F + uPr) / L
= (6.50 + 0.5 * 165 * 0.235 ) / 0.5
= 51.775 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)
= (60 * 0.235^2 * 4pi) / (2 * 6.5)
= 3.20 sec.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.