Question

A dentist’s drill starts from rest. After 1.46 sof constant angular acceleration, it turns at arate of 27000 rev/min.Find the drill’s angular acceleration.Answer in units of rad/s2.

Answer 1

Answer:

616.3 rad/s²

Explanation:

Given that

t= 1.46 s

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 27000 rev/min

Angular speed in the rad/s given as

$\omega_f=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values

$\omega_f=\dfrac{2\times 27000}{60}\ rad/s$

ωf=900 rad/s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

900 = 0 + α x 1.46

$\alpha=\dfrac{900}{1.46}\ rad/s^2\\\alpha=616.43\ rad/s^2$

Therefore the acceleration will be 616.3 rad/s²