A dentist using a dental drill brings it from rest to maximum operating speed of 382,000 rpm in 3.0 s. Assume that the drill accelerates at a constant rate during this time. A) What is the angular acceleration of the drill in rev/s2?
B) Find the number of revolutions the drill bit makes during the 2.6 s time interval. rev

a.

b.

Explanation:

The computation is shown below:

data provided in the question

The initial angular velocity   = 0 rev/s,

f = Final angular velocity = = 382000 rpm i.e = = 6,366.67

And time = t = 3.0s

Based on the above information

a. For angular acceleration of drill

b. For the number of revolutions

We simply applied the above formulas for determining each parts

be subject to a force andchange its motion

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