A machinist turns on the power to a grinding wheel at time t = 0 s.The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of $38 \mathrm { rad } / \mathrm { s }$ The wheel is run at that angular speed for 30 s and then power is shut off.The wheel slows down uniformly at 2.1 rad/s² until the wheel stops.In this situation,what is the angular acceleration of the wheel between t = 0 and t = 10s? A)3.8 rad/ $\mathrm { s }^2$ B)4.6 rad/ $\mathrm { s }^2$ C)5.3 rad/ $\mathrm { s }^2$ D)6.1 rad/ $\mathrm { s }^2$ E)6.8 rad/ $\mathrm { s }^2$

Question

Quizplus · Accepted Answer

The angular acceleration can be found using the equation:

$\theta = \frac {1}{2} \alpha t^2 + \omega _0 t$

where $\theta$ is the angular displacement, $\alpha$ is the angular acceleration, $\omega _0$ is the initial angular velocity (which is 0), and t is the time.

For the first 10 seconds, we know that $\omega = \alpha t$ since the acceleration is uniform. We are given $\omega = 38 \mathrm { rad } / \mathrm { s }$ at the end of 10 seconds, so we can solve for $\alpha$.

$\alpha = \frac {38 \mathrm { rad } / \mathrm { s }}{10 \mathrm { s }} = 3.8 \mathrm { rad } / \mathrm { s }^2$

Therefore, the answer is A, 3.8 rad/ $\mathrm { s }^2$.

A Machinist Turns on the Power to a Grinding Wheel 38rad/s38 \mathrm { rad } / \mathrm { s }38rad/s

A Machinist Turns on the Power to a Grinding Wheel $38 \mathrm { rad } / \mathrm { s }$