At time t =0 a grinding wheel has an angular velocity of 26.0 rad/s . It has a c

Question

At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until a circuit breaker trips at time t = 2.40 s . From then on, the wheel turns through an angle of 431 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

wi = 26 rad/s ; alpha = 35 rad/s^2 ; t = 2.4 s ; theta(i) = 431 rad

A)from eqn in circular motion,

theta = wit + 1/2 alpha t^2

theta = 26 x 2.4 + 0.5 x 35 x 2.4^2 = 163.2 rad

So the angle through which the wheel rotates:

theta(f) = theta(i) + theta

theta(f) = 431 + 163.2 = 594.2 rad

Hence, theta(f) = 594.2 rad

B)When the circuit breaker trips, the angular velocity becomes,

from eqn in circular motion

w = wi + alpha t

w = 26 + 35 x 2.4 = 110 rad/s

also, w^2 = wi^2 + 2 alpha' theta(i)

alpha' = 0 - 110^2/2 x 431 = 14.04 ras/s^2

theta(i) = 1/2 alpha' t^2

t = sqrt (2 theta(i))/alpha') = sqrt (2 x 431/14.04) = 7.84 s

T = 2.4 + t = 2.4 + 7.84 = 10.24 s

Hence, t = 10.24 s

c)alpha' = 14.04 rad/s^2 (calculated in B)

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