at t= 0 a grinding wheel has an angular velocity of 20.0 rad/s. it has a constan

Question

at t= 0 a grinding wheel has an angular velocity of 20.0 rad/s. it has a constant angular acceleration of 35.0 rad/s^2 until a circuit breaker trips at time= 1.80 s. From then on, it turns through an angle 438 rad as it coasts to a stop at a constant angular acceleration

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

correct answer: 531 rad

PART B

at what time did it stop?

- I tried 13.68 and 14.58 which were incorrect


PART C
What was its acceleration as it slowed down?

Explanation / Answer

a. use the kinematics equations for angular kinematics: From t=0 to t=2.3: total angle = angular velocity*t + 1/2 angular acceleration * t^2 =117.4 radians it then moves another 431rad, so total is 117.4+431 = 548.415 rad b. again use kinematics equation: vf = vi + 2ad 0 = 82.1 + 2(a)431 angular acceleration (c) = -.0952 rad/s^2. time can be found using another kinematics equation: 0 = 82.1 -.09521t t= 862 s

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