A 50.0-kg grindstone is a solid disk 0.550 m in diameter. You press an ax down o

Question

A 50.0-kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a normal force of 150 N (Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60 N.m , and there is a constant friction torque of between the axle of the stone and its bearings.

* I already found out Part A and B, but I'm kinda having diffeculty finding Part C

please help me out solving part C

= 67.8 A 50.0-kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a normal force of 150 N (Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60 N.m , and there is a constant friction torque of 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 9.00 /rms? F = 67.8 /rmN 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? F = 62.5 /rmN Part C 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? t =

Explanation / Answer

(120 rev/min)(2pi rad/1 rev)(1 min/60s)

= 4 rad/s

I=1/2 MR^2

=(1/2)(50 kg)(0.275)^2

= 1.89 kgm^2

torque = Ia = I(w/t)

= (1.89 kgm^2)(4 pi rad/s x 9.0s)

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