Question

Consider the 14.5 kg motorcycle wheel shown in the figure. Assume it to be approximately a ring with an inner radius of 0.26 m and an outer radius of 0.335 m. The motorcycle is on its center stand, so that the wheel can spin freely.

A. If the drive chain exerts a force of 2100 N at a radius of 5.1 cm, what is the angular acceleration of the wheel, in radians per second squared?

B. What is the tangential acceleration, in meters per square second, of a point on the outer edge of the tire?

C. How long, in seconds, starting from rest, does it take to reach an angular velocity of 75.5 rad/s?

Answer 1

tau =I*alpha

I = m*(r_a² +r_b² ) /2 = 14.5*(0.26^2+0.335^2)/2 =1.30373125 kg.m²

alpha = tau /I = 2100*0.051/1.30373125 =82.1488324 rad/s²

angular acceleration of the wheel = 82.1488324 rad/s²

B) tangential acceleration a = r*alpha =0.335*82.1488324 =27.5198589 m/s²

C)

ω =α*t

t=ω /α = 75.5/82.1488324 =0.91906358 sec answer