At time t=0 a grinding wheel has an angular velocity of 20.0 rad/s . It has a co

Question

At time t=0 a grinding wheel has an angular velocity of 20.0 rad/s . It has a constant angular acceleration of 25.0 rad/s2 until a circuit breaker trips at time t = 2.20 s . From then on, the wheel turns through an angle of 440 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 radians per second per second.

thank you in advance.

Explanation / Answer

  a.

accel is +ve from t=0 to t=2.2s
so angle turned from t=0 to 2.2s is s = u*t + 0.5*a*t^2
s = (20.0*2.2) + 0.5*25.0*(2.2)^2 = 44 rad + 60.5 rad =104.5 rad
Total angle from t=0 to rest = 104.5 rad + 440 rad = 588.5 rad (3 s.f.)

c.

velocity at t=2.2s is v = u + a*t = 20.0 * 25.0*2.2 m/s
v = 75 m/s

accel from t=2.2s is negative (slowing down)
so a = -v^2 / 2s = 75^2 / (2*440) rad/s^2 = 6.39 rad/s^2 (negative)

b.

time slowing down is v /a = 75 / 6.39 = 11.73 s
so time of stopping is t = 11.73 s + 2.2s = 13.93 s

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