A 60.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down o
ID: 1461385 • Letter: A
Question
A 60.0-kg grindstone is a solid disk 0.520 m in diameter. You press an ax down on the rim with a normal force of 170 N (Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50Nm 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 8.00 s ?
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?
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?
Explanation / Answer
A)
initial velocity = w1 = 0
final angular velocity - w2 = 120 rev/min = (120*2*pi)/60 = 4pi rad/s
angular acceleration = (w2-w1)/t = 4pi/8 = 1.57 rad/s^2
torque = I*alfa
(F*r - Tk - u*N*R = I*alfa = 0.5*m*R^2*alfa
F*0.5 - 6.5 0.6*170*0.26 = 0.5*60*0.26^2*1.57
F = 351.13 N
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at constant speed , alfa = 0
(F*r - Tk - u*N*R = 0
F*0.5 - 6.5 0.6*170*0.26 = 0
F = 344.76 N
++++++++++
net torque = I*alfa
6.5 = 0.5*60*0.26^2*alfa
alfa = 3.2
w2/t = 3.2
t = 4*3.14/3.2
t = 3.925 s
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