College Physics

Angular displacement, θ = 0.75 X 2π = 1.5π
rad

Initial angular speed, ω = 1.42 rad/s

Final angular speed, ω’ = 0

By ω’^2 = ω^2 + 2αθ

0 = (1.42)^2 + 2α(1.5π)

Average angular acceleration, α = -0.214
rads^-2

As the wheel is a disk, the moment of
inertia, I is given by

I = 1/2 mr^2 = 1/2 (5.4)(0.56)^2 = 0.84672
kgm^2

By Γ = Iα

Torque required, Γ = (0.84672)(-0.214) =
-0.181 Nm

So, the magnitude of average torque
required = 0.181 Nm


Alternative method:

Moment of inertia of the wheel, I = 0.84672
kgm^2

By conservation of energy,

Work done by the torque = rotational K.E.
lost

Γθ = 1/2 Iω'^2 – 1/2 Iω^2

Γ(1.5π) = 1/2 (0.84672)(1.42)^2
– 0

Torque required, Γ
= -0.181 Nm