Consider the 11.0 kg motorcycle wheel shown in Figure 9.29. Assume it to be appr

Question

Consider the 11.0 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 = 0.320 m. The motorcycle is on its center stand, so that the wheel can spin freely.

Consider the 11.0 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 = 0.320 m. The motorcycle is on its center stand, so that the wheel can spin freely. If the drive chain exerts a force of 1930 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)torque provided by driver =Fr = 1930*0.05 =9.65 N

accerelation of the wheell =>torque =I(moment of inertia)*angular acceleration

moment of inertia =0.5*11(0.75*0.04*0.04 +0.28*0.28) =0.4378

therefore angular acceleration = 9.65/(0.4378) = 22.042rad/s^2

b)tangential acceleration =R*angular acceleration =22.042*0.32 = 7.053m?s^2

c)time taken to reach angular velocity of 80 rad/s = 80/22.042 = 3.62sec

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