Answer:
The magnitude of the drills angular acceleration is [tex]-32\pi s^{-2}[/tex].
The drill makes 50 revolutions before it stops.
Explanation:
The revolutions that the drill makes in 1 second is
[tex]\dfrac{2400\:rev}{60s} =40\:rev/s[/tex]
And the angular velocity is
[tex]\omega= \dfrac{40 (2\pi )}{1s}[/tex] (one revolution is [tex]2\pi[/tex] radians)
[tex]\boxed{\omega =80\pi\:s^{-1}}[/tex]
Now, the drill comes to a halt in 2.5 s, which means the magnitude of it's angular acceleration is
[tex]a= \dfrac{\Delta \omega}{\Delta s}[/tex]
[tex]a=\dfrac{80\pi-0 }{0-2.5s}[/tex]
[tex]\boxed{a=-32\pi \:s^{-2}}[/tex]
In other words, the magnitude of the angular acceleration is [tex]-32\pi\: s^{-2}.[/tex]
Now, we find the angular displacement of the drill which is given by the equation
[tex]\theta = \omega t +\dfrac{1}{2} \alpha t^2[/tex]
putting in [tex]\omega = 80\pi s^{-1}[/tex], [tex]\alpha = -32\pi s^{-2}[/tex], and [tex]t=2.5s[/tex], we get:
[tex]\theta = (80\pi s^{-1} *2.5s)+\dfrac{1}{2} (-32\pi s^{-2})(2.5s)^2[/tex]
[tex]\theta = 100\pi[/tex]
which is
[tex]\dfrac{100\pi }{2\pi } =50\:rev[/tex] (one revolution is [tex]2\pi[/tex] radians)
50 revolutions.
In other words ,the drill makes 50 revolutions before it stops.
