A.4 · IB Physics
Rotational dynamics
Apply a force at the rim of a wheel and watch torque turn into angular acceleration. Compare a disc, a hoop and a sphere, change the lever arm and mass, and brake to slow it down.
Predict first
Start from the default settings (uniform disc, F = 5 N, r = 3.0 m, m = 4 kg). If you change the object to a hoop of the same mass and radius, what happens to the angular acceleration α?
α = τ ÷ I with the same torque. A hoop has I = mr² instead of ½mr², so I doubles and α halves (0.833 to 0.417 rad s⁻²).
Check it in the model: Change the Object select from Uniform disc to Hoop and compare the Angular acceleration α readout.
Force F acts tangentially at the rim, so the lever arm equals the radius r.
I = k m r² (k = ½ disc, 1 hoop, ⅖ sphere)
α = τ ÷ I
ω = ω₀ + α t
KE = ½ I ω² ; L = I ω
- Torque τ
- Moment of inertia I
- Angular acceleration α
- Angular velocity ω
- Kinetic energy ½Iω²
- Angular momentum L = Iω
- Time t
Try this
- Start with the uniform disc, F = 5 N, r = 3.0 m, m = 4 kg and press Reset. Double F to 10 N, then (after Reset) double m to 8 kg instead. How does each change α and the gradient of the ω–t graph?
- Keep F, r and m fixed and switch the object from disc to hoop. By what factor does I change, and what happens to ω after 5 s?
- Raise the lever arm r from 3.0 m to 5.0 m with F and m fixed. The torque grows, so why does α fall?
Exam tip: τ = Iα mirrors F = ma, but I depends on how the mass is spread, so decide I for the shape first, and remember that for a force at the rim the lever arm is the perpendicular distance r.