A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

Explanation:

means initial angular velocity, which is 0 rev/min

means final angular velocity, which is

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

= (2π/60) *2.513*10^{4} rad/s

= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

so the rotational motion about a fixed axis is:

+ 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita =

be subject to a force andchange its motion

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