Question

A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 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?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev

Answer 1

a

`\alpha = 2327.7 \ rev/s^2`

b

`\theta = 9124.5 \ rev`

Step-by-step explanation:

From the question we are told that

The maximum angular speed is
`w_(max) = 391000 \ rpm = (2 \pi * 391000)/(60) = 40950.73 \ rad/s`

The time taken is
`t = 2.8 \ s`

The minimum angular speed is
`w_(min)= 0 \ rad/s` this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

`w_(max) = w_(mini) + \alpha * t`

=>
`\alpha = ( w_(max))/(t)`

substituting values

`\alpha = (40950.73)/(2.8)`

`\alpha = 14625 .3 \ rad/s^2`

converting to
`rev/s^2`

We have

`\alpha = 14625 .3 * 0.159155 \ rev/s^2`

`\alpha = 2327.7 \ rev/s^2`

According to the first equation of motion the angular displacement is mathematically represented as

`\theta = w_(min) * t + (1)/(2) * \alpha * t^2`

substituting values

`\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2`

`\theta = 57331.2 \ radian`

converting to revolutions

`revolution = 57331.2 * 0.159155`

`\theta = 9124.5 \ rev`