Questions on torque and angular acceleration

Question 1

State what is meant by the torque of a force about an axis, and give its unit.

Accepted Answer

Torque is the turning effect of a force about the axis, the product of the force and its perpendicular distance from the axis, T = Fr for a force applied tangentially at radius r (1). Its unit is N m (1).

Question 2

Write the equation linking the net torque on a rotating body to its angular acceleration, and define each symbol.

Accepted Answer

T = Iα (1), where T is the net torque in N m, I the moment of inertia about that axis in kg m² and α the angular acceleration in rad s⁻² (1). It is the rotational form of F = ma, and it is the net torque, not the drive torque, that belongs in it.

Question 3

State the principle of conservation of angular momentum.

Accepted Answer

If no resultant external torque acts on a system, its total angular momentum Iω stays constant (1). Internal forces, muscles, clutches or a collapsing structure, cannot change it, however much they rearrange the mass (1).

Question 4

Explain why an ice skater spinning with her arms outstretched speeds up when she pulls her arms in.

Accepted Answer

No external torque acts, so her angular momentum Iω is conserved (1). Pulling her arms in moves mass closer to the axis, reducing I, so ω must rise to keep the product constant (1).

Question 5

Explain how a heavy flywheel mounted on the crankshaft of a piston engine smooths out the fluctuations in torque from the power strokes.

Accepted Answer

Its large moment of inertia means a given torque fluctuation produces only a tiny change in ω, since α = T/I (1). It absorbs angular momentum during each power stroke and returns it during the gaps between them, so the shaft turns almost uniformly (1).

Question 6

Give the unit of angular momentum.

Accepted Answer

kg m² s⁻¹, equivalently N m s (1), following straight from Iω with I in kg m² and ω in rad s⁻¹.

Question 7

Calculate the torque produced by a force of 250 N applied tangentially at the rim of a wheel of diameter 0.60 m.

Accepted Answer

The radius is half the diameter: r = 0.60/2 = 0.30 m (1). T = Fr = 250 × 0.30 = 75 N m (1). The moment arm is the radius: using the diameter gives 150 N m, double the true torque.

Question 8

A flywheel of moment of inertia 8.0 kg m² is acted on by a constant net torque of 20 N m, starting from rest. Calculate its angular acceleration and its angular velocity after 15 s.

Accepted Answer

α = T/I (1) = 20/8.0 = 2.5 rad s⁻² (1). ω₂ = ω₁ + αt = 0 + 2.5 × 15 = 37.5 rad s⁻¹ (1).

Question 9

An engine develops an output power of 45 kW at 2400 revolutions per minute. Calculate the torque at the output shaft.

Accepted Answer

ω = 2400 × 2π/60 = 251 rad s⁻¹ (1). P = Tω, so T = P/ω (1) = 45 000/251 = 179 N m (1). P = Tω requires ω in rad s⁻¹: dividing by 2400 instead gives an answer about 9.5 times too small.

Question 10

A disc of moment of inertia 0.75 kg m² spins at 24 rad s⁻¹. A brake applies a steady friction torque of 3.0 N m. Calculate the time taken for the disc to stop.

Accepted Answer

Angular momentum to remove: Δ(Iω) = 0.75 × 24 = 18 kg m² s⁻¹ (1). Angular impulse TΔt = Δ(Iω) (1), so Δt = 18/3.0 = 6.0 s (1). Half the torque would take twice as long, delivering the same angular impulse.

Question 11

A skater spinning at 1.5 revolutions per second has a moment of inertia of 5.6 kg m². She pulls her arms in, reducing it to 2.0 kg m². Calculate her new rate of spin and the factor by which her kinetic energy changes.

Accepted Answer

Iω is conserved: 5.6 × 1.5 = 2.0 × f (1), so f = 4.2 rev s⁻¹ (1). Energy ratio = (2.0 × 4.2²)/(5.6 × 1.5²) = 2.8 (1). Revolutions per second are safe here because only the ratio matters, but the energy rises even though the momentum did not: her muscles did the work of hauling her arms inward.

Question 12

A motor applies a constant torque of 12 N m to a shaft, turning it through 500 rad in 25 s. Calculate the work done and the average power delivered.

Accepted Answer

W = Tθ (1) = 12 × 500 = 6000 J (1). P = W/t = 6000/25 = 240 W (1). The same answer comes from P = Tω with ω = 500/25 = 20 rad s⁻¹: 12 × 20 = 240 W.

Question 13

A drum of moment of inertia 4.0 kg m² is driven by a torque of 30 N m and opposed by a friction torque of 6.0 N m. Starting from rest, calculate its angular acceleration and the number of revolutions it makes in the first 10 s.

Accepted Answer

Net torque = 30 − 6.0 = 24 N m (1). α = T/I = 24/4.0 = 6.0 rad s⁻² (1). θ = αt²/2 = 6.0 × 10²/2 = 300 rad (1). Revolutions = 300/2π = 48 (1). T = Iα takes the net torque: using 30 N m gives 7.5 rad s⁻².

Question 14

A disc of moment of inertia 1.8 kg m² spinning at 120 rad s⁻¹ is lowered onto a stationary disc of moment of inertia 1.2 kg m² mounted on the same axis. Friction between them brings the pair to a common angular velocity, the friction torque between the faces being 9.0 N m. Calculate the common angular velocity, the kinetic energy before and after, the energy lost, and the time the coupling takes.

Accepted Answer

No external torque acts on the pair, so Iω is conserved: 1.8 × 120 = (1.8 + 1.2)ω (1), giving ω = 72 rad s⁻¹ (1). E_k before = ½ × 1.8 × 120² = 12960 J (1). E_k after = ½ × 3.0 × 72² = 7776 J (1). Energy lost = 12960 − 7776 = 5184 J, heating the slipping faces (1). For the time, apply the angular impulse to the disc that was at rest: TΔt = Δ(Iω) = 1.2 × 72 = 86.4 kg m² s⁻¹, so Δt = 86.4/9.0 = 9.6 s (1). Angular momentum survives a coupling; kinetic energy does not survive the slip, so an answer that conserves energy here has used the wrong law.

Question 15

A turbine rotor of moment of inertia 220 kg m² is run up from rest to 3000 revolutions per minute by a constant torque of 1100 N m. Calculate the time taken and the angle turned through, and show that the work done by the torque equals the rotor's final kinetic energy.

Accepted Answer

ω = 3000 × 2π/60 = 314.2 rad s⁻¹ (1). Angular impulse: TΔt = Δ(Iω) = 220 × 314.2 = 6.91 × 10⁴ kg m² s⁻¹, so t = 6.91 × 10⁴/1100 = 62.8 s (1). θ = (ω₁ + ω₂)t/2 = 314.2 × 62.8/2 = 9.87 × 10³ rad (1). W = Tθ = 1100 × 9.87 × 10³ = 1.09 × 10⁷ J (1). E_k = ½Iω² = ½ × 220 × 314.2² = 1.09 × 10⁷ J, the same, as energy conservation requires (1).

Question 16

A spinning skater pulls her arms inward and her rate of spin rises. Explain why her angular momentum is unchanged while her kinetic energy increases, and state where the extra energy comes from.

Accepted Answer

Her arms are pulled in by internal forces, and no resultant external torque acts on her, so Iω is conserved (1). Her kinetic energy can be written as ½(Iω)ω; since Iω is fixed and ω has risen, the kinetic energy must rise too (1). The extra energy is work done by her muscles (1), hauling her arms inward against the outward pull needed to keep them circling (1).

Question 17

A wheel of moment of inertia 0.40 kg m² spinning at 30 rad s⁻¹ is brought to rest in 2.5 s by a brake pad pressed against its rim at a radius of 0.25 m. Calculate the friction force at the rim and the heat generated at the pad.

Accepted Answer

Δ(Iω) = 0.40 × 30 = 12 kg m² s⁻¹ (1). TΔt = Δ(Iω), so T = 12/2.5 = 4.8 N m (1). T = Fr, so F = 4.8/0.25 = 19.2 N (1). All the rotational kinetic energy becomes heat: E_k = ½ × 0.40 × 30² = 180 J (1). The third step divides the torque by the rim radius: dividing by the diameter, or quoting the torque itself as the force, gives the wrong force.