A 26inch diameter bicycle wheel supported vertically at its center is spun about

Question

A 26inch diameter bicycle wheel supported vertically at its center is spun about a horizontal axis by a torque of 50 ftlb. If the rim and tire together weigh 3.24 lb determine the approximate angular acceleration of the wheel. Ignore the contribution of the spokes. The units should cancel out. Hint: the net applied torque equals the moment of inertia times the angular acceleration. The wheel is essentially a hoop. A 26inch diameter bicycle wheel supported vertically at its center is spun about a horizontal axis by a torque of 50 ftlb. If the rim and tire together weigh 3.24 lb determine the approximate angular acceleration of the wheel. Ignore the contribution of the spokes. The units should cancel out. Hint: the net applied torque equals the moment of inertia times the angular acceleration. The wheel is essentially a hoop.

Explanation / Answer

I = moment of inertia

w= angular acceleration

t= torque = 50ftlb

mass = 3.24lb

diameter = 26 inch, radius = 13 inch = 1.0833 ft

I = m*r*r = 3.24*1.083*1.083 = 3.8022

t = I*w

w = t/1 = 50/3.8022 = 13.150 (answer)

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