Question

A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +52 N·m is applied to the wheel for 23 s, giving the wheel an angular velocity of +640 rev/min. The external torque is then removed, and the wheel comes to rest 120 s later. (Include the sign in your answers.)

(a) Find the moment of inertia of the wheel.   kg·m²
(b) Find the frictional torque, which is assumed to be constant.   N·m

Answer 1

Torque (τ) = I α

Here there are two torques, the external torque and the frictional torque.

For the accelerative phase:

τ.e + τ.f = I α.1 ... (1)
α.1 = Δ ω.1 / Δ t.1
Δ ω.1 = 640 rev/min * 2&pi rad/rev * 1 min/60sec
Δ t.1 = 23 s
α.1 = 2.914 rads/s²


For the frictional phase
τ.f = I α.2 ... (2)
α.2 = Δ ω.2 / Δ t.2
Δ ω.2 = - 640 rev/min * 2&pi rad/rev * 1 min/60sec
Δ t.2 = 120 s

τ.f = I Δ ω.2 / Δ t.2
τ.f = - I * 0.5585 rads/s² ... (3)

We can now substitute &tau.f into equation (1), above:

τ.e + τ.f = I α.1
τ.e - I * 0.5585 rads/s² = I α.1
τ.e = I (0.5585 rads/s² + 2.914 rad/s²)
I = +52 N·m / (0.5585 rads/s² + 2.914 rad/s²)
I = 14.97 kg·m²

τf = 14.97 kg.m² x .5585 rad/m² = - 8.36 N⋅m (the negative sign is directional, only)