Consider the 14.0 kg motorcycle wheel shown in Figure 9.29. Assume it to be appr

Question

Consider the 14.0 kg motorcycle wheel shown in Figure 9.29. Assume it to be approximately an annular ring with an inner radius of R1 = 0.280 m and an outer radius of R2 =0.300 m. The motorcycle is on its center stand, so that the wheel can spin freely.

Consider the 14.0 kg motorcycle wheel shown in Figure 9.29. Assume it to be approximately an annular ring with an inner radius of R1 = 0.280 m and an outer radius of R2 =0.300 m. The motorcycle is on its center stand, so that the wheel can spin freely. If the drive chain exerts a force of 2060 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? What is the tangential acceleration of a point on the outer edge of the tire? How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

Explanation / Answer

(a) the moment of inertia of an annular ring is

I = m*(R1^2 + R2^2) / 2

So for the wheel,

I = 14.0*(0.28^2 + 0.30^2) / 2 = 1.179 kg.m^2

torque T = 2060 N * 0.05m = 103 N.m

angular accel alpha = T / I = 103 / 1.179 rad/s^2 = 87.36 rad/s^2

(b) a = r*alpha = 0.30 * 87.36 rad/s^2 = 26.21 m/s^2

(c) t = w / alpha = 80 / 26.21 s = 3.05 s

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