### Question Description

A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of ^{4}

rad/s

^{2}

(b) Determine the angle (in radians) through which the drill rotates during this period.

rad

## Final Answer

α = (ω - ωo)/t

α = (2 e4 rev/min)(2π rad/rev)(min / 60 s) / (3s)

α = 698.13 rad/s^2

(b)

θ = αt^2 / 2

θ = (698.13rad/s^2)(3 s)^2 / 2

θ = 3141.59 rad = 3.142 * 10^3 rad

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