Questions on rotational motion and the angular equations

Question 1

State what is meant by the angular velocity of a rotating body, and give its SI unit.

Accepted Answer

Angular velocity is the angle swept out per unit time about the axis, ω = Δθ/Δt (1). Its unit is rad s⁻¹ (1). Every point of a rigid body shares the same ω, however far from the axis it sits.

Question 2

Calculate the angular velocity, in rad s⁻¹, of a shaft turning at 2400 revolutions per minute.

Accepted Answer

One revolution is 2π rad and one minute is 60 s, so ω = 2400 × 2π/60 (1) = 251 rad s⁻¹ (1). Leaving the speed in rev min⁻¹ carries the error through every line that follows.

Question 3

State what the moment of inertia of a body measures, and the two properties of the body that fix its value.

Accepted Answer

It measures the body's resistance to angular acceleration about that axis, the rotational counterpart of mass (1). It depends on how much mass the body has and on how that mass is distributed about the axis, since I = Σmr² weights each particle by the square of its distance (1).

Question 4

Explain why a hoop has a larger moment of inertia than a uniform disc of the same mass and the same radius, about the same central axis.

Accepted Answer

I = Σmr², so each particle counts by the square of its distance from the axis (1). The hoop carries all its mass at the full radius, while the disc spreads its mass inward where r² is small, so the hoop's sum is larger, in fact twice as large (1).

Question 5

Name the rotational quantity that corresponds to each of the following linear quantities: displacement, velocity, acceleration, mass.

Accepted Answer

Displacement s becomes angular displacement θ; velocity v becomes angular velocity ω; acceleration a becomes angular acceleration α (1); mass m becomes moment of inertia I (1).

Question 6

Write the equation for the kinetic energy of a body of moment of inertia I rotating with angular velocity ω.

Accepted Answer

E_k = ½Iω² (1), the exact twin of ½mv² with I standing in for m and ω for v.

Question 7

A grinding wheel accelerates uniformly from rest to 900 revolutions per minute in 6.0 s. Calculate its angular acceleration.

Accepted Answer

Convert first: ω₂ = 900 × 2π/60 = 94.2 rad s⁻¹ (1). α = (ω₂ − ω₁)/t (1) = 94.2/6.0 = 15.7 rad s⁻² (1). Dividing 900 by 6.0 without converting gives 150, a number in the wrong units entirely.

Question 8

A turntable has a moment of inertia of 0.045 kg m² and rotates at 45 revolutions per minute. Calculate its rotational kinetic energy.

Accepted Answer

ω = 45 × 2π/60 = 4.71 rad s⁻¹ (1). E_k = ½Iω² (1) = ½ × 0.045 × 4.71² = 0.50 J (1). The conversion is again the step the rest depends on: a raw 45 in the formula inflates the energy about ninety-fold.

Question 9

Two 0.60 kg masses are fixed to the ends of a light rod of length 0.80 m, which rotates about an axis through the rod's centre, perpendicular to it. Calculate the moment of inertia of the arrangement, and its value when each mass is moved in to 0.10 m from the axis.

Accepted Answer

Each mass sits at r = 0.80/2 = 0.40 m, not 0.80 m (1). I = Σmr² = 2 × 0.60 × 0.40² = 0.192 kg m² (1). At 0.10 m, I = 2 × 0.60 × 0.10² = 0.012 kg m² (1), smaller by a factor of 16, because r² rewards distance so heavily.

Question 10

A flywheel slows uniformly from 120 rad s⁻¹ to rest in 250 revolutions. Calculate the magnitude of its angular deceleration.

Accepted Answer

θ = 250 × 2π = 1571 rad (1). Use ω₂² = ω₁² + 2αθ (1): 0 = 120² + 2α × 1571, so α = −4.58 rad s⁻², a deceleration of 4.58 rad s⁻² (1).

Question 11

A wheel of diameter 0.90 m turns at 12 rad s⁻¹ with an angular acceleration of 3.0 rad s⁻². Calculate the speed of a point on the rim and its acceleration along its circular path.

Accepted Answer

Halve the diameter: r = 0.90/2 = 0.45 m (1). v = ωr = 12 × 0.45 = 5.4 m s⁻¹ (1). a = αr = 3.0 × 0.45 = 1.35 m s⁻² (1). Both formulae take the radius: feeding the diameter straight in doubles both answers.

Question 12

A flywheel is a uniform disc of mass 25 kg and radius 0.30 m, for which I = ½mr². Calculate the energy it stores when spun at 2000 revolutions per minute.

Accepted Answer

I = ½ × 25 × 0.30² = 1.125 kg m² (1). ω = 2000 × 2π/60 = 209 rad s⁻¹ (1). E_k = ½Iω² = ½ × 1.125 × 209² = 2.5 × 10⁴ J (1).

Question 13

A disc starts from rest and accelerates uniformly at 4.0 rad s⁻² for 8.0 s. Calculate the angle it turns through and the number of revolutions it makes.

Accepted Answer

θ = ω₁t + αt²/2 (1) = 0 + 4.0 × 8.0²/2 = 128 rad (1). Revolutions = 128/2π = 20.4 (1). Dividing by 2π is the return trip of the usual conversion, and forgetting it leaves an answer 6.3 times too large.

Question 14

Explain why an engineer designing a flywheel energy store gains far more by doubling its angular velocity than by doubling its mass, and explain why the mass that is used is placed at the rim rather than near the axis.

Accepted Answer

The store is E_k = ½Iω². Doubling ω quadruples the energy, because ω is squared (1), while doubling the mass only doubles I and so only doubles the energy (1). Placing mass at the rim raises I for the same mass, because I = Σmr² counts each particle by the square of its distance from the axis (1), so a rim-loaded wheel stores more energy at the same spin rate than a solid one of equal mass (1). Real flywheel batteries are therefore modest rim-heavy discs spun very fast in a vacuum, not heavy ones spun slowly.

Question 15

A tram recovers braking energy in a flywheel, a uniform disc of mass 180 kg and radius 0.50 m for which I = ½mr². The flywheel is charged to 3600 revolutions per minute and is regarded as discharged when it has slowed to 1800 revolutions per minute. Calculate the moment of inertia, the energy stored at each speed, the usable energy, and the time for which that usable energy could supply 40 kW.

Accepted Answer

I = ½ × 180 × 0.50² = 22.5 kg m² (1). ω₁ = 3600 × 2π/60 = 377 rad s⁻¹ and ω₂ = 1800 × 2π/60 = 188 rad s⁻¹ (1). E₁ = ½ × 22.5 × 377² = 1.60 × 10⁶ J (1). E₂ = ½ × 22.5 × 188² = 4.00 × 10⁵ J (1). Usable energy = 1.60 × 10⁶ − 4.00 × 10⁵ = 1.20 × 10⁶ J (1). Time = 1.20 × 10⁶/40 000 = 30 s (1). Note that halving the spin rate empties three quarters of the store, not half, because E_k ∝ ω²: an answer of half the energy has forgotten the square.

Question 16

Two flywheels each have a mass of 60 kg and a radius of 0.40 m and are spun at 150 rad s⁻¹. One is a uniform disc, for which I = ½mr²; the other is a rim-loaded ring, for which I = mr². Calculate the energy stored by each and state which is the better design for an energy store.

Accepted Answer

Disc: I = ½ × 60 × 0.40² = 4.8 kg m², so E_k = ½ × 4.8 × 150² = 5.4 × 10⁴ J (1). Ring: I = 60 × 0.40² = 9.6 kg m², so E_k = ½ × 9.6 × 150² = 1.1 × 10⁵ J (1). The ring stores twice as much for the same mass and the same spin rate (1), so the ring is the better store: mass held at full radius counts fully in Σmr², whereas the disc's inner mass contributes little (1).

Question 17

A wheel accelerates uniformly from 8.0 rad s⁻¹ to 20 rad s⁻¹ while turning through 42 rad, and its kinetic energy rises by 1680 J during this change. Determine the angular acceleration, the time taken and the moment of inertia of the wheel.

Accepted Answer

ω₂² = ω₁² + 2αθ gives α = (20² − 8.0²)/(2 × 42) (1) = 4.0 rad s⁻² (1). Then t = (ω₂ − ω₁)/α = 12/4.0 = 3.0 s (1). Energy: ΔE_k = ½I(ω₂² − ω₁²) (1), so 1680 = ½ × I × (400 − 64) and I = 10.0 kg m² (1). The energy step must use the difference of the squares: ½I(Δω)² with Δω = 12 gives an I more than twice too large.

Question 18

A flywheel is mounted on a horizontal axle of radius 0.050 m. A string wound round the axle carries a 4.0 kg mass, which is released from rest and falls 2.0 m, by which time the flywheel is turning at 24 rad s⁻¹. Friction at the bearings is negligible and the axle itself has a negligible moment of inertia. Taking g = 9.81 m s⁻², determine the moment of inertia of the flywheel.

Accepted Answer

The potential energy lost is shared between the moving mass and the spinning wheel: mgh = ½mv² + ½Iω² (1). The mass descends at the speed of the string on the axle, v = ωr = 24 × 0.050 = 1.2 m s⁻¹ (1). mgh = 4.0 × 9.81 × 2.0 = 78.48 J (1). ½mv² = ½ × 4.0 × 1.2² = 2.88 J, so the wheel holds 78.48 − 2.88 = 75.60 J (1). ½Iω² = 75.60, so I = 2 × 75.60/24² = 0.26 kg m² (1). The radius in v = ωr is the axle's, not the flywheel's, and giving the wheel all of the lost potential energy overstates I by about four per cent.