At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 25.0 rad/s^2 until a circuit breaker trips at time t = 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.

Through what total angle did the wheel turn between t=0 and the time it stopped? At what time does the wheel stop? What was the wheel's angular acceleration as it slowed down?

Question

Rene Dunn

Physics

22 February, 22:08

At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 25.0 rad/s^2 until a circuit breaker trips at time t = 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.

Through what total angle did the wheel turn between t=0 and the time it stopped? At what time does the wheel stop? What was the wheel's angular acceleration as it slowed down?

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Answers (1)

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Adyson Owen · Answer 1

The equations of motions will be applied in this question; except that in this case it will be angular motion instead of linear motion.

We use the formula

v = u + at; to determine the final velocity of before the circuit breaker trips.

v = 26 + 2.5 x 25

= 88.5 rad/s

Total angle covered before circuit breaker trips:

2as = v² - u²

s = (88.5² - 26²) / 2 (25)

s = 143.125 rad

Angle covered before stopping after trip = 440 rad

Total angle covered from start to finish:

143.125 + 440

= 583.125 rad

Acceleration as wheel stops:

2as = v² - u²; v = 0

a = - (88.5²) / 2 (440)

a = 0.1 rad/s²

Time to stop:

v = u + at

0 = 88.5 - 0.1t

t = 885 seconds

Total time: 2.5 + 885

= 887.5 seconds