At time t0 a grinding wheel has an angular velocity of 290 r

At time t=0 a grinding wheel has an angular velocity of 29.0 rad/s . It has a constant angular acceleration of 30.0 rad/s2 until a circuit breaker trips at time t = 2.40 s . From then on, the wheel turns through an angle of 435 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?

B) At what time does the wheel stop?

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

Solution

v = v° + 1/2 a t^2

v° = 29 rad /s

a = 30 rad /s2

t = 2.4 s

e = v°t + 1/2 a t^2 = 29 *2.4 + 0.5 * 30 * 2.4^2 = 156 rad

v = v° + a t = 29 + 30 *2.4 = 101 rad /s

total angle did the wheel turn between t=0 and the time it stopped 156 + 435 = 591 rad

t = 435 / vmoy = 435 /0.5*101 = 8.61 +2.4 = 11.01 s

wheel\'s angular acceleration as it slowed down

a = d v / d t = 101 / 8.61 = - 11.73 rad /s2