Question

A wheel free to rotate about its axis that is not frictionless is initially at rest....

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

Homework Answers

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)

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