Starting from rest at t=0sec, a tire on a mountain bike has a constant angular a

Question

Starting from rest at t=0sec, a tire on a mountain bike has a constant angular acceleration. When t= 4.0 sec, the angular velocity of the wheel is 7.4 rad/sec. The angular acceleration continues until t= 17.0 sec, after which the angular velocity remains constant. The tire has a radius of 22cm. 1) What is the angular acceleration at t= 17.0 sec? ? = rad/sec2 2) What is angular velocity at t= 17.0 sec? ? = rad/sec 3) What is the total angular displacement of the tire from t=0sec to t=17.0 sec? ? = rad 4) What would be the total displacement of a small stone stuck in the tread of the tire as it goes around from t=0sec to t=17.0 sec? ?s = m 5) What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=17.0 sec? ?x = m

Explanation / Answer

1)Constant angular acceleration=/t=1.85rad/s2

2)=t=31.45 rad/s

3)=0.5t2=267.33 rad

4)d=r=58.8m

5)Linear displacement=angular displacement/2=9.36m

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