Question

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

Answer 1

a.
`\alpha =2122.22\: rev/s^(2)`

b.
`\Delta \theta =9,550.02\: rev`

Step-by-step explanation:

The computation is shown below:

data provided in the question

The initial angular velocity
`\small \omega o` = 0 rev/s,

f = Final angular velocity =
`\small \omega` = 382000 rpm i.e =
`(382,000)/(60)` = 6,366.67

And time = t = 3.0s

Based on the above information

a. For angular acceleration of drill

`\small \omega =\omega _(o)+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^(2)`

b. For the number of revolutions

`\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`

We simply applied the above formulas for determining each parts