Physics
posted by Sam .
A dentist's drill starts from rest. After 3.20 s of constant angular acceleration it turns at a rate of 2.7 104 rev/min.
(a) Find the drill's angular acceleration.
(b) Determine the angle (in radians) through which the drill rotates during this period.

V = 2.7104 rev / min.
V = 2.7104 rev/min * 1 / 60 min/s = 0.04517 rev/s.
a. a = 0.04517 rev/s / 3.2 s =
0.01412 rev/s^2.
b. A = 0.04517 rev/s * 3.2 s * 6.28 rad/rev = 0.91 Radians. 
i've a question, where did u get 6.28 rad/rev?

2pie equals 6.28

I have a question. 2.7104 was that suppose to actually be 2.7*10^4?

A dentist's drill starts from rest it truns at a rate of 2.22*10^5 rpm at 3.5s.
1)find the angular acceleration of the drill.
2)through how many revolution the drill rotates during this period.
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