A 70.0-kg grindstone is a solid disk 0.550 m in diameter. You press an ax down o

Question

A 70.0-kg grindstone is a solid disk 0.550 m in diameter. You press an ax down on the rim with a normal force of 180 N. 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.

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 8.00 s ?

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: 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

Total frictional torque on the stone, Tf = 6.50 + [0.6 * 180 * (0.55/2)] = 29.7 N-m

Part A) Let F be the required tangential force.

Final angular speed, = 120 rev/min = 120 * 2 / 60 = 12.57 rad/s

Required angular acceleration, = (12.57 - 0)/8 = 1.57 rad/s2

Net torque, T = I = (MR2/2) = [70 * (0.55/2)2 / 2] * 1.57 = 4.16 N-m

Now, T = 0.5F - 29.7 = 4.16

=> F = 67.7 N

Part B) Net torque needs to be zero for constant angular speed.

So, T = 0.5F - 29.7 = 0

=> F = 59.4 N

Part C) Angular deceleration due to axle friction, af = Taf / I = 6.50 / [70 * (0.55/2)2 / 2] = 2.46 rad/s2

Hence required time, t = /af = 12.57/2.46 = 5.1 s

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