Question

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?

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

Answer 2

Angle turned by the wheel is given by the kinematics as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

now here we know that

$\omega = 26 rad/s$

$\alpha = 25 rad/s^2$

t = 2.50 seconds

$\theta = 26(2.50) + \frac{1}{2}(25)(2.50)^2$

$\theta = 143.1 rad$

So here total angle that the wheel turn is given as

$\theta = 143.1 + 440 = 583.1 rad$

Speed of the wheel after 2.5 s

$\omega = \omega_0 + \alpha t$

$\omega = 26 + (25)(2.5)$

$\omega = 88.5 rad/s$

now angular deceleration is given as

$\alpha_1 = \frac{\omega^2 - \omega_0^2}{2\theta}$

$\alpha_1 = \frac{0 - 88.5^2}{2(440)}$

$\alpha_1 = -8.9 rad/s^2$

time taken to stop

$t = \frac{\omega - \omega_0}{\alpha}$

$t = \frac{0 - 88.5}{-8.9} = 9.94 s$

now total time is given as

$T = 2.50 + 9.94 = 12.44 s$