Consider the 11.0 kg motorcycle wheel shown in the figure below. Assume it to be

Question

Consider the 11.0 kg motorcycle wheel shown in the figure below. Assume it to be approximately an annular ring with an inner radius of 0.330 m and an outer radius of 0.380 m. The motorcycle is on its center stand, so that the wheel can spin freely. If the drive chain exerts a force of 2200 N at a radius of 5.90 cm, what is the angular acceleration of the wheel? Tries 0/10 What is the tangential acceleration of a point on the outer edge of the tire? Tries 0/10 How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

Explanation / Answer

Moment of inertia = 1/2 * M (R22 + R22)

= 0.5 * 11 * (0.332 + 0.382) = 1.39 kg.m2

Torque = F * r = 2200 * 5.90 * 10-2 = 129.8 N.m

T = I * alpha

alpha = 129.8 / 1.39 = 93.38 rad/s2

a) Tangential acceleration = R alpha = 0.38 * 93.38 = 35.48 m/s2

b) wf = wi + alpha t

t = 80 / 93.38

= 0.857 s

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