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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.

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Answer:

616.3 rad/s²

Step-by-step 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=(2\pi N)/(60)\ rad/s

Now by putting the values


\omega_f=(2* 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=(900)/(1.46)\ rad/s^2\\\alpha=616.43\ rad/s^2

Therefore the acceleration will be 616.3 rad/s²

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