Spinning Wheel on a Rotating Stool
In a laboratory, a student sits on a low-friction rotating stool and holds a bicycle wheel by its axle. The axle is vertical and aligned with the stool's rotation axis, so the wheel's spin angular momentum is vertical. The wheel is spinning at 18 rad/s; its moment of inertia about the axle is 0.35 kg m². The student plus stool have moment of inertia 2.8 kg m² about the vertical stool axis. Initially the stool is at rest. The student then rotates the wheel through 180° about a horizontal axis through its centre, reversing the wheel's spin direction, and holds it fixed relative to her body. The stool starts turning. After 4.0 s, a lab partner applies a gentle brake to the stool axle and brings the stool to rest while the wheel continues spinning. Bearing friction is negligible unless stated.
1. Determine the angular speed of the stool immediately after the wheel is flipped, and state its direction relative to the wheel's original spin.
Read the worked answer
Take the vertical axis as positive in the direction of the wheel's original spin. Before the flip, the total angular momentum about that axis is L_i = I_w ω_w because the stool is at rest. After the flip, the wheel's spin angular momentum is -I_w ω_w and the stool has angular momentum I_s Ω. No external torque acts about the vertical axis, so
I_w ω_w = -I_w ω_w + I_s Ω.
Thus Ω = 2 I_w ω_w / I_s = 2(0.35)(18)/(2.8) = 4.5 rad/s.
The positive sign means the stool turns in the same sense as the wheel's original spin. This assumes negligible bearing friction and that the student's flip is internal to the student-stool-wheel system.
2. The student claims the flip requires no work because the wheel's spin speed is unchanged. Evaluate this claim and estimate the minimum work she must do.
Read the worked answer
The claim is false. The wheel's rotational kinetic energy is unchanged, but the stool gains kinetic energy. Initial energy: K_i = 1/2 I_w ω_w² = 0.5(0.35)(18)² = 56.7 J. Final energy: K_f = 56.7 + 1/2 I_s Ω² = 56.7 + 0.5(2.8)(4.5)² = 56.7 + 28.35 = 85.1 J. The increase is about 28 J. The student must therefore do at least 28 J of work; in practice more, because muscles also do internal work and some energy is dissipated. Reversing the spin changes the angular momentum vector, not the wheel's spin speed, so the wheel's kinetic energy alone is not the whole energy budget.
3. The brake is applied to the stool axle for 4.0 s. Calculate the average braking torque. Explain whether the wheel's spin affects this braking time if the wheel bearing is frictionless.
Read the worked answer
The stool's angular momentum to be removed is L_s = I_s Ω = 2.8(4.5) = 12.6 kg m²/s. Average torque magnitude = ΔL/Δt = 12.6/4.0 = 3.15 N m. If the wheel bearing is frictionless, the wheel exerts no torque on the stool, so the braking time depends only on the stool's angular momentum and the brake torque; the wheel's spin does not affect it. If the bearing had friction, the spinning wheel would couple to the stool and the brake would also have to remove wheel angular momentum, lengthening the stopping time.