Consider the 10.5 kg motorcycle wheel shown in the figure. Assume it to be appro

Question

Consider the 10.5 kg motorcycle wheel shown in the figure. Assume it to be approximately a ring with an inner radium of 0.295 m and an outer radius of 0.345m. the motorcycle is on its center stand, so that the wheel can spin freely. If the drive chain exerts a force of 2150 N at a radius of 4.95 cm, what is the angular acceleration of the wheel, in radians per square second? What is the tangential acceleration, in meters per square second, of a point on the outer edge of the tire? How long, in seconds, starting from rest, does it take to reach an angular velocity of 83 rad/s?

Explanation / Answer

PART A

I = 0.5*M*(R1^2 + R2^2)

= 0.5*10.5*(0.295^2+0.345^2)

= 1.0817 kg.m^2

T = F*r*sin(90) = 2150*0.0495*1 = 106.425 N.m

alfa = T/I = 106.425/1.0817 = 98.386 rad/s^2

PART B

atan = R2*alfa = 0.345*98.386 = 33.943 m/s^2

PART C

Apply, w = wo + alfa*t

t = (w-wo)/alfa

= (83 - 0)/98.386 = 0.8436 s

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