Angular Kinetics 3

Question

A wheel with a moment of inertia of 0.03 kg.m2 is accelerated from rest to 20 rad/s in 5 seconds due to application of a moment of force (torque). When the moment of force is released, the wheel stops in 1 min.

a) Calculate the moment of friction (torque due to friction).

b) Calculate the original moment of force.

Answer 1

Let

I = 0.03 kgm² be the moment of inertia,

ω = 20 s^-1 be the angular velocity,

t₁ = 5 s be the time it took to reach ω,

t₂ = 1 min = 60 s be the time it took the wheel to stop

α₁ = ω/t₁ be the acceleration from rest to ω

α₂ = ω/t₂ be the deceleration from ω to 0

The general form of Newton's second law for torque τ is

τ = Iα

a)

τ₂ = Iα₂ = 0.03×20/60 = 0.01 kgm²/s²

b)

τ₁ = Iα₁ = 0.03×20/5 = 0.12 kgm²/s²

However, we are probably supposed to assume that the same moment of friction

is active during deceleration as well as acceleration.

In that case, to overcome the moment of friction, the original moment of force is

τ₁+τ₂ = 0.13 kgm²/s²