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:

$a=\frac{\omega_2-\omega_1}{t}$

where,

a = angular acceleration

$\omega_1$ = initial angular velocity = 0

$\omega_2$ = final angular velocity = $24040rev/min\times \frac{2\pi rad}{rev}\times \frac{1min}{60s}=2516.19rad/s$

t = time = 1.28 s

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

$a=\frac{(2516.19-0)rad/s}{1.28s}=1965.77rad/s^2$

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