(a) How much force must be applied tangentially at the end of a crank handle 0.5

Question

(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 130 rev/min in 10.00 s?
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(b) After the grindstone attains an angular velocity of 130 rev/min, what tangential force at the end of the handle is needed to maintain a constant angular velocity of 130 rev/min?
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(c) How much time does it take the grindstone to come from 130 rev/min to rest if it is acted on by the axle friction alone?
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Explanation / Answer

(130 rev/min) (2pi rad/1 rev) (1 min/60s) = 4.364 pi rad/s

I = 1/2 MR2
I = (1/2)(50 kg)(0.275)2
I = 1.89 kgm2

torque = Ia = I(w/t)
Torque = (1.89 kgm2)(4.364 pi rad/s / 10.0s) = 2.59 Nm

(a)
F = (1/R)(torque + frictional torque + coeff. friction x normal F x R)
F = 1/0.5 (2.59 + 6.50 + (0.60 x 170 x 0.275)
F = 74.28 N

(b)
F' = 1/0.5 (6.50 + (0.60 x 170 x 0.275))
F' = 69.1 N

(c)
t = L/frictional torque
t = wI/frictional torque
t = ((4.364 pi rad/s)(1.89 kgm^2))/6.50 Nm
t = 3.98 s

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