Question

A sanding disk with rotational inertia 1.2 × 10−3 kg ⋅ m2 is attached to an electric drill whose motor delivers a torque of magnitude 16 N ⋅ m about the central axis of the disk. About that axis and with the torque applied for 33 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

Answer 1

Answer:

L = 0.528kgm²/s

ω = 440 s^-1

Explanation:

given

Inertia, I = 1.2 * 10^-3kgm²

Torque, τ = 16Nm

Time, t = 33ms

Angular momentum is gotten using the relation

τ = ΔL/ΔT

Where τ = Torque

ΔL = change in angular momentum

ΔT = change in time taken

ΔL = τΔT

ΔL = 16Nm * 33*10^-3s

ΔL = 0.528kgm²/s

Angular velocity of the disk will be gotten by using the formula

L = Iω where

L = angular momentum

I = rotational inertia

ω = angular velocity

ω = L/I

ω = 0.538/1.2*10^-3

ω = 440 s^-1