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

Question

Starting from rest at t=0sec, a tire on a mountain bike has a constant angular acceleration. When t= 5 sec, the angular velocity of the wheel is 7.3 rad/sec. The angular acceleration continues until t= 19 sec, after which the angular velocity remains constant. The tire has a radius of 22cm.

1) What is the angular acceleration at t= 19 sec?

? = rad/sec2

2) What is angular velocity at t= 19 sec?

? = rad/sec

3) What is the total angular displacement of the tire from t=0sec to t=19 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 tot=19 sec?

?s = m

5) What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=19 sec?

?x =

Explanation / Answer

1)alpha = w/t

alpha = 7.3/5 = 1.46 rad/s^2

alpha = 1.46 rad/s^2

2)w = w0 + alpha t

w = 1.46 x 19 = 27.74 rad/s

w = 27.74 rad/s

3)theta = wt + 1/2 alpha t^2

theta = 0 + 0.5 x 1.46 x 19^2 = 263.53 rad4

theta = 263.53 rad

4)s = 263.53/3.14 = 83.93 m

s = 83.93 m

5)x = s/2 pi r = 83.93/2 x 3.14 x 0.22 = 60.75 m

x = 60.75

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