Consider the 12.5-kg motorcycle wheel shown in the figure below. Assume it to be

Question

Consider the 12.5-kg motorcycle wheel shown in the figure below. Assume it to be approximately annular ring with an inner radius of R_1 = 0.280 m and an outer radius of R_2 = 0.300 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2175 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? rad/s^2 (b) What is the tangential acceleration of a point on the outer edge of the tire? m/s^2 (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s? s Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel and after 1.5 revolutions the wheel corners to rest on a space that has a $1500.00 prize, If the initial angular speed of the wheel is 4.10 rad/s, find the angle through which the wheel has turned when the angular speed is 1.80 rad/s. rev

Explanation / Answer

we know

torque = I*alpha

I = moment of inertia

I = 1/2 * m *(r1+r2)^2

r1 = 0.280 m

r2 = 0.300 m

I = 1.0525

torque = F*r

F*r = I*alpha

alpha = 2175 * 0.05 / 1.0525 = 103.325 rad/s^2

part b )

a = alpha*r2 = 30.9976 m/s^2 = 31 m/s^2

part c )

wf = wi + alpha*t

wi = 0

wf = 80 rad/s

t = wf/alpha

t = 0.77425 s

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