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 6.9 rad/sec. The angular acceleration continues until t= 16 sec, after which the angular velocity remains constant. The tire has a radius of 22cm.

1. What is the angular acceleration at t= 16 sec? =

2. What is angular velocity at t= 16 sec? =

3. What is the total angular displacement of the tire from t=0sec to t=16 sec? =

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=16 sec?s =

5. What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=16 sec? x =

Explanation / Answer

given data

1. What is the angular acceleration at t= 16 sec? =

By omega(final) = omega(initial) + alfa x time
=>6.9 = 0 + alfa x 5
=>alfa =1.38 rad/sec^2

2. What is angular velocity at t= 16 sec? =

By omega(final) = omega(initial) + alfa x time
=>omega(final) = 0 + 1.38x 16 = 22.08 rad/sec

3. What is the total angular displacement of the tire from t=0sec to t=16 sec? =

theta = omega(initial) x t + 1/2 x alfa x t^2
=>theta = 0 + 1/2 x 1.38 x (16)^2 = 176.64 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=16 sec?s =

By theta = s/r
=>s = r x theta = 22 x 10^-2 x 176.64 = 38.86m

5. What would be the linear displacement of the tire's axle as it moves forward from t=0sec to t=16 sec? x =

delta x = s/(2 x pi x r) = 38.86/(2 x 3.14 x 22 x 10^-2) = 28.12m

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