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 33.0 rad/s2 until a circuit breaker trips at time t = 1.80 s . From then on, the wheel turns through an angle of 432 rad as it coasts to a stop at constant angular deceleration.

Part A

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

Express your answer in radians.

Part B

At what time does the wheel stop?

Express your answer in seconds.

Part C

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

Express your answer in radians per second per second.

Explanation / Answer

a. accel is +ve from t=0 to t=1.8s
so angle turned from t=0 to 1.8s is s = u*t + 0.5*a*t^2
s = (29.0*1.8) + 0.5*33.0*(1.8)^2 = 105.66 rad
Total angle from t=0 to rest = 105.66 rad + 432 rad = 537.66 rad
c. velocity at t=1.8s is v = u + a*t = (29.0 + 33.0*1.8) rad/s
v = 88.4 rad/s
accel from t=1.8s is negative (slowing down)
so a = -v^2 / 2s = 88.4^2 / (2*432) rad/s^2 = 9.044 rad/s^2 (negative)
b. time slowing down is v /a = 88.4 / 9.044 = 9.77 s
so time of stopping is t = 9.77 s + 1.8 s = 11.57 sec

Comment below if you have any doubt.

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