Answer:
a. [tex]\alpha =2122.22\: rev/s^{2}[/tex]
b. [tex]\Delta \theta =9,550.02\: rev[/tex]
Explanation:
The computation is shown below:
data provided in the question
The initial angular velocity [tex]\small \omega o[/tex] = 0 rev/s,
f = Final angular velocity = [tex]\small \omega[/tex] = 382000 rpm i.e = [tex]\frac{382,000}{60}[/tex] = 6,366.67
And time = t = 3.0s
Based on the above information
a. For angular acceleration of drill
[tex]\small \omega =\omega _{o}+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^{2}[/tex]
b. For the number of revolutions
[tex]\small \omega ^{2}-\omega _{o}^{2}=2\alpha \Delta \theta \\\\(6,366.67) ^{2}-(0)^{2}=2(2122.22) \Delta \theta \\\\ \Delta \theta =9,550.02\: rev[/tex]
We simply applied the above formulas for determining each parts
