Question

A high-speed drill rotating ccw at 2400 rpm comes to a halt in 2.5 s. What is the magnitude of the drill’s angular acceleration? How many revolutions does it make as it stops?

Answer 1

Answer:

The magnitude of the drills angular acceleration is $-32\pi s^{-2}$.

The drill makes 50 revolutions before it stops.

Explanation:

The revolutions that the drill makes in 1 second is

$\dfrac{2400\:rev}{60s} =40\:rev/s$

And the angular velocity is

$\omega= \dfrac{40 (2\pi )}{1s}$          (one revolution is $2\pi$ radians)

$\boxed{\omega =80\pi\:s^{-1}}$

Now, the drill comes to a halt in 2.5 s, which means the magnitude of it's angular acceleration is

$a= \dfrac{\Delta \omega}{\Delta s}$

$a=\dfrac{80\pi-0 }{0-2.5s}$

$\boxed{a=-32\pi \:s^{-2}}$

In other words, the magnitude of the angular acceleration is $-32\pi\: s^{-2}.$

Now, we find the angular displacement of the drill which is given by the equation

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

putting in $\omega = 80\pi s^{-1}$, $\alpha = -32\pi s^{-2}$, and $t=2.5s$, we get:

$\theta = (80\pi s^{-1} *2.5s)+\dfrac{1}{2} (-32\pi s^{-2})(2.5s)^2$

$\theta = 100\pi$

which is

$\dfrac{100\pi }{2\pi } =50\:rev$                (one revolution is $2\pi$ radians)

50 revolutions.

In other words ,the drill makes 50 revolutions before it stops.