Question

A dentists drill starts from rest after 1.28 s of constant angular acceleration, it turns at a rate of 22,540 rev/min.

a. Find the drill's angular acceleration.
b. Throughout what angle does the drill rotate during this period?

Answer 1

The drill's angular acceleration is approximately 1840.16 rad/s² and it rotates through an angle of approximately 1481.08 radians.

Step-by-step explanation:

The angular acceleration of the drill can be found using the kinematic formula:

ω = ω0 + αt

Where:

ω = final angular velocity = 22,540 rev/min = (22540 rev/min) * (2π rad/rev) * (1 min/60 s) = 2357.82 rad/s

ω0 = initial angular velocity = 0 rad/s

t = time = 1.28 s

Solving for α:

α = (ω - ω0) / t

α = (2357.82 rad/s - 0 rad/s) / 1.28 s

α ≈ 1840.16 rad/s²

The angle of rotation can be calculated using the formula:

θ = ω0t + 0.5αt²

θ = (0 rad/s)(1.28 s) + 0.5(1840.16 rad/s²)(1.28 s)²

θ ≈ 1481.08 rad