a dentist drill starts from rest.after 3.20s of constsant angular acceleretaion, it turns at a rate of 2.51*10 to the power of 4 rev/min. find the drills angular acceleration. determine the angle(in radians) through which the drill rotates during this period

The drill turns with a constant acceleration, so we can use kinematic equations known as SUVAT equations to solve for the angular acceleration and angular displacement (the angle of rotation). In this case, the equations we will use are:

ω = αt + ω₀

Δθ = ½ (ω + ω₀) t

where

Δθ is the angular displacement (angle of rotation)
ω₀ is the initial angular velocity
ω is the final angular velocity
α is the angular acceleration
t is the time

Given:

ω₀ = 0 rad/s

ω = 2.51×10⁴ rev/min × (2π rad/rev) × (1 min / 60 s) = 2.63×10³ rad/s

t = 3.20 s

(a) Use the first equation to find the angular acceleration α.

ω = αt + ω₀

2.63×10³ rad/s = α (3.20 s) + 0 rad/s

α = 821 rad/s²

(b) Use the second equation to find the angular displacement Δθ.

Δθ = ½ (ω + ω₀) t

Δθ = ½ (2.63×10³ rad/s + 0 rad/s) (3.20 s)

Δθ = 4.21×10³ rad