Question

A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 96 rad/s. The wheel is run at that angular velocity for 40 s and then power is shut off. The wheel slows down uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the angular displacement in Revolutions during the slowdown is:

Answer 1

Answer:489 Revolutions

Explanation:

Given

Angular deceleration$(\alpha ) =1.5rad/s^2$

Given wheel angular velocity =96 rad/s when machine is turned off

time taken by machine to reach zero angular velocity

$0=\omega _0+(\alpha)t$

0=96+(-1.5)t

t=64 sec

angular displacement is given by

$\theta =\omega_0t+\frac{1}{2}\alpha t^2$

$\theta =96(64)-\frac{1}{2}(-1.5)(64^2)=3072 degree$

For revolutions =$\frac{3072}{2\cdot \pi}=488.86 \approx 489 revolution during Slowdown$