At time t =0 a grinding wheel has an angular velocity of 29.0 rad/s . It has a c

Question

At time t=0 a grinding wheel has an angular velocity of 29.0 rad/s . It has a constant angular acceleration of 29.0 rad/s2 until a circuit breaker trips at time t = 2.30 s . From then on, the wheel turns through an angle of 433 rad as it coasts to a stop at constant angular deceleration.

A) Through what total angle did the wheel turn between t=0 and the time it stopped?

Express your answer in radians.

B) At what time does the wheel stop?

Express your answer in seconds.

C) What was the wheel's angular acceleration as it slowed down?

Express your answer in rad/s2.

Explanation / Answer

A) angular dispalcement in first 2.3 s

theta = w1*t + alfa*t^2

= 29*2.3 + 0.5*29*2.3^2

= 143.4 rad

total angular dispalcement = 143.4 + 433

= 576.4 rad

B) angular speed, at t = 2.3 s

w = wo + alfa*t

= 29 + 2.3*29

= 95.7 rad

angular accelleration during slowdown, alfa = (wf^2 - wi^2)/(2*theta)

= (0 - 95.7^2)/(2*433)

= -10.58 rad/s^2

time taken to stop, t = (wf - wi)/alfa

= (0 - 95.7/(-10.58)

= 9.04 s

C) angular accelleration during slowdown, alfa = (wf^2 - wi^2)/(2*theta)

= (0 - 95.7^2)/(2*433)

= -10.58 rad/s^2

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