Answer:
The magnitude of the drills angular acceleration is
.
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](https://tex.z-dn.net/?f=%5Cdfrac%7B2400%5C%3Arev%7D%7B60s%7D%20%3D40%5C%3Arev%2Fs)
And the angular velocity is
(<em>one revolution is </em>
<em> radians)</em>
![\boxed{\omega =80\pi\:s^{-1}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Comega%20%3D80%5Cpi%5C%3As%5E%7B-1%7D%7D)
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}](https://tex.z-dn.net/?f=a%3D%20%5Cdfrac%7B%5CDelta%20%5Comega%7D%7B%5CDelta%20s%7D)
![a=\dfrac{80\pi-0 }{0-2.5s}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B80%5Cpi-0%20%7D%7B0-2.5s%7D)
![\boxed{a=-32\pi \:s^{-2}}](https://tex.z-dn.net/?f=%5Cboxed%7Ba%3D-32%5Cpi%20%5C%3As%5E%7B-2%7D%7D)
In other words, the magnitude of the angular acceleration is ![-32\pi\: s^{-2}.](https://tex.z-dn.net/?f=-32%5Cpi%5C%3A%20s%5E%7B-2%7D.)
Now, we find the angular displacement of the drill which is given by the equation
![\theta = \omega t +\dfrac{1}{2} \alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega%20t%20%2B%5Cdfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2)
putting in
,
, and
, we get:
![\theta = (80\pi s^{-1} *2.5s)+\dfrac{1}{2} (-32\pi s^{-2})(2.5s)^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%2880%5Cpi%20s%5E%7B-1%7D%20%2A2.5s%29%2B%5Cdfrac%7B1%7D%7B2%7D%20%28-32%5Cpi%20s%5E%7B-2%7D%29%282.5s%29%5E2)
![\theta = 100\pi](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20100%5Cpi)
which is
<em>(one revolution is </em>
<em> radians)</em>
50 revolutions.
In other words ,the drill makes 50 revolutions before it stops.