A flat, solid cylindrical grinding wheel with a diameter of 26.0cm is spinning a

Question

A flat, solid cylindrical grinding wheel with a diameter of 26.0cm is spinning at 2900rpm when its power is suddenly turned off. A workman continues to press his tool bit toward the wheel's center at the wheel's circumference so as to continue to grind as the wheel coasts to a stop.

Part A

If the wheel has a moment of inertia of 5.00kg?m2 , determine the necessary torque that must be exerted by the workman to bring it to rest in 15.0s . Ignore any friction at the axle.?

Part B

If the coefficient of kinetic friction between the tool bit and the wheel surface is 0.900, how hard must the workman push on the bit?

Explanation / Answer

Part A)

Torque = I(alpha)

To find alpha, first, 2900 rpm converts to 303.7 rad/s

Then wf = wo + (alpha)t

0 = 303.7 + (alpha)(15)

alpha = 20.2 rad/s2

Finally, Torque = (5)(20.2) = 101.2 Nm

Part B)

Torque = Fr

101.2 = uFn(r)

101.2 = .9(Fn)(.13)

Fn = 865.2 N

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