A sanding disk with rotational inertia 1.3 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of 9 N·m about the central axis of the disk. What are the following values about the central axis at the instant the torque has been applied for 120 ms? (a) the angular momentum of the disk kg·m2/s (b) the angular speed of the disk rad/s

a) Angular momentum, ΔL = 1.08  kg·m²/s

b) Angular speed, ω = 830.77 rad/s

Explanation:

a) Torque = Change in angular momentum/Change in time

Let Torque = τ

L = Angular momentum

t = time

τ = ΔL/Δt

ΔL = τ * Δt

τ = 9 N.m

Δt = 120 ms = 0.12 s

ΔL = 9 * 0.12

ΔL = 1.08  kg·m²/s

b) Angular momentum = Inertia * Angular speed

Let I = Inertia

I = 1.3 * 10-3 kg·m²

ω = Angular speed

L = I * ω

ω = L/I

ω = 1.08/(1.3 * 10-3)

ω = 830.77 rad/s