Consider the 9.0 kg motorcycle wheel shown in Figure 9.29. Assume it to be appro

Question

Consider the 9.0 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 = 0.400 m. The motorcycle is on its center stand, so that the wheel can spin freely. Figure 9.29. (a) If the drive chain exerts a force of 2190 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? Incorrect: Your answer is incorrect. rad/s2 (b) What is the tangential acceleration of a point on the outer edge of the tire? Incorrect: Your answer is incorrect. m/s2 (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? Incorrect: Your answer is incorrect. s

Explanation / Answer

Here,

R1 = 0.280 m

R2 = 0.4 m

m = 9 Kg

moment of inertia of wheel , I = 0.5 * m * (R1^2 + R2^2)

I = 0.5 * 9 * (0.280^2 + 0.400^2)

I = 1.073 Kg.m^2

F = 2190 N

a) let the angular acceleration is a

using second law of motion

F * r = I * a

2160 * 0.05 = 1.073 * a

a = 102.1 rad/s^2

the angular acceelration is 102.1 rad/s^2

b)

Tangential acceleration of point at the outer edge = R2 * a

Tangential acceleration of point at the outer edge = 1 102.1 rad/s^2 * 0.4 m

Tangential acceleration of point at the outer edge = 40.8 m/s^2

c)

let the time taken is t

wf = a * t

80 = 102.1 * t

t = 0.783 s

the time taken is 0.783 s

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