Question

A dentist’s drill starts from rest. After 1.28 sof constant angular acceleration it turns at a rate of 24040 rev/min

Answer 1

This question is an incomplete question, here is a complete question.

A dentist’s drill starts from rest. After 1.28 s of constant angular acceleration it turns at a rate of 24040 rev/min. Find the drill's angular acceleration in rad/s²

Answer : The angular acceleration of the drill is,
`1965.77rad/s^2`

Explanation :

Formula used for angular acceleration of the drill is:

`a=(\omega_2-\omega_1)/(t)`

where,

a = angular acceleration

`\omega_1` = initial angular velocity = 0

`\omega_2` = final angular velocity =
`24040rev/min* (2\pi rad)/(rev)* (1min)/(60s)=2516.19rad/s`

t = time = 1.28 s

Now put all the given values in the above formula, we get:

`a=((2516.19-0)rad/s)/(1.28s)=1965.77rad/s^2`

Therefore, the angular acceleration of the drill is,
`1965.77rad/s^2`