Question

15. A dentist's drill starts from rest. After 3.10 s of constant angular acceleration, it turns at a rate of 25.4 rad/s. Find the drills angular acceleration.

Answer 1

Solution:

Dentist's drill starts from rest, so , $\omega$ i = 0 rad/s

After 3.10s of constant accelaration , it turns at a rate of 25.4 rad/s so, $\omega$ f = 25.4 rad/s

Using Rotational Kinematics equation---->

$\omega$f = $\omega$ i + $\alpha$ *t

Or, 25.4 = 0 + $\alpha$ *(3.10)

Or, $\alpha$ = 8.1935 rad/s^2

Therefore, drill's angular accelaration is 8.1935 rad/s^2