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At time t=0 a grinding wheel has an angular velocity of 25
At time t=0 a grinding wheel has an angular velocity of 25.0rad/s . It has a constant angular acceleration of 35.0rad/s^2 until a circuit breaker trips at time t = 2.00s . From then on, the wheel turns through an angle of 440rad 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.
Expert Solution
a.θ = (25 x 2) + 0.5 x 35.0 x (2)^2 = 120 rad
so total angle from t=0 to rest = 120 rad + 440 rad = 560 rad
c. velocity at t=2 is v =25 + 35.0x 2 m/s
v = 95 m/s
now a2 = -v^2 / 2s = 90^2 / (2x 440) rad/s^2 = - 9.2 rad/s^2 (
b. time of slowing down =90 / 9.2= 9.78 s
so time of stopping t = 9.78s + 2s = 11.78s
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