Senior 6 practice book · Unit 2 of 10

Rotational Motion

50 questions that complete the Senior 6 quiz for this unit: 27 core and 23 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A wheel turning at a steady rate has no acceleration.Every point except the axle moves in a circle, so it has a centripetal acceleration ω²r towards the axis even when ω is constant.
The moment of inertia of a body is a fixed number, like its mass.Moment of inertia depends on the axis chosen: a rod is four times harder to turn about one end than about its centre.
A bigger force always gives a bigger torque.Torque also depends on the perpendicular distance from the axis: a small force far from the hinge can beat a large force applied close to it.
A moving bicycle stays up because the rider is clever, not because of physics.Spinning wheels have angular momentum along the axle; a torque is needed to change its direction, which helps keep the bicycle steady while the rider steers.
A ball rolling down a ramp reaches the same speed as a box sliding down without friction.Part of the rolling ball's energy goes into spinning, so less is left for forward motion; the frictionless box arrives faster.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Angular displacement, angular velocity and angular acceleration; rad, rev/min and v = ωr
Equations of uniformly accelerated rotational motion
Tangential and centripetal acceleration of a point on a rotating body (a = rα, a = ω²r)
Torque, couples and rotational equilibrium
Moment of inertia of point masses and systems of masses
Moment of inertia of standard bodies, radius of gyration and the parallel-axis theorem
Newton's second law for rotation, τ = Iα (including friction torque)
Rotational kinetic energy, work done by a torque and rotational power
Rolling without slipping: speeds and energy shares
Angular momentum and its conservation
Angular momentum as a vector: stability of spinning bodies (bicycle, top, gyroscope)
Angular impulse: torque × time = change in angular momentum

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 23 harder ones are marked ★ Challenge.

1True or false
A spinning top stays upright longer when it is spun faster.
Show answer
Answer: True
A faster spin gives more angular momentum, so gravity's torque changes its direction more slowly and the top wobbles less.
Common mistakeThinking speed makes no difference ignores angular momentum: the larger Iω, the harder it is for a torque to tip the axis.
2True or false · ★ Challenge
Moving a small mass twice as far from the axis doubles its contribution to the moment of inertia.
Show answer
Answer: False
The contribution is mr², so doubling r multiplies it by 2² = 4.
Common mistakeThinking it doubles forgets that the distance is squared in I = mr².
3True or false
A point on a wheel turning at a constant angular velocity has zero acceleration.
Show answer
Answer: False
Its direction of motion keeps changing, so it has a centripetal acceleration ω²r towards the axis.
Common mistakeThinking a steady spin means no acceleration forgets that acceleration includes any change in direction, not only in speed.
4Multiple choice · ★ Challenge
Seen from above, a horizontal wheel turns anticlockwise. Using the right-hand grip rule, in which direction does its angular momentum vector point?
1. AVertically upwards, along the axle
2. BVertically downwards, along the axle
3. CHorizontally, along the direction a point on the rim moves
4. DTowards the centre of the wheel
Show answer
Answer: A. Vertically upwards, along the axle
Curl the fingers of the right hand in the direction of rotation (anticlockwise from above); the thumb points up, along the axis.
Common mistakeChoosing the direction of the rim's motion confuses the linear velocity of a point with the angular momentum, which lies along the axis.
5True or false
If a wheel's angular acceleration is constant, its angular velocity is also constant.
Show answer
Answer: False
A constant angular acceleration means the angular velocity changes by the same amount every second.
Common mistakeConfusing 'constant acceleration' with 'constant velocity' is like saying a car that speeds up steadily has a steady speed.
6Fill in the blank · ★ Challenge
A flywheel of mass 20 kg has a moment of inertia of 0.80 kg m². Its radius of gyration is ______ m.
Show answer
Answer: 0.20
I = Mk², so k = √(I/M) = √(0.80 ÷ 20) = √0.040 = 0.20 m.
Common mistakeWriting 0.040 m forgets the square root; the radius of gyration is a distance, so k² = I/M.
7True or false
For a wheel rolling without slipping, the point of the tyre touching the ground is momentarily at rest.
Show answer
Answer: True
At the contact point the backward speed due to rotation (ωR) exactly cancels the forward speed v of the axle.
Common mistakeThinking the bottom moves fastest mixes up the bottom and the top: the top moves at 2v, the bottom at zero.
8Fill in the blank · ★ Challenge
The Earth has a moment of inertia of 8.0 × 10³⁷ kg m² and an angular velocity of 7.3 × 10⁻⁵ rad/s. Its angular momentum is about ______ × 10³³ kg m²/s.
Show answer
Answer: 5.8
L = Iω = 8.0 × 10³⁷ × 7.3 × 10⁻⁵ ≈ 5.8 × 10³³ kg m²/s.
Common mistakeUsing ½Iω² gives an energy in joules, not angular momentum; L = Iω has no ½ and no square.
9True or false
A body has the same moment of inertia about every axis.
Show answer
Answer: False
I = Σmr² depends on how far the mass is from the chosen axis, so it changes with the axis.
Common mistakeTreating I like mass is the error: mass is fixed, but the moment of inertia depends on the axis of rotation.
10True or false
Doubling the angular velocity of a flywheel doubles the energy stored in it.
Show answer
Answer: False
KE = ½Iω², so doubling ω multiplies the energy by 4.
Common mistakeThinking the energy doubles forgets that ω is squared in the kinetic-energy formula.
11Multiple choice · ★ Challenge
The grinding disc of a maize mill turns at 300 revolutions per minute. What is the centripetal acceleration of a point on the disc 0.20 m from the axis?
1. A6.3 m/s²
2. B20 m/s²
3. C197 m/s²
4. D1.8 × 10⁴ m/s²
Show answer
Answer: C. 197 m/s²
ω = 2π × 300 ÷ 60 = 31.4 rad/s; a = ω²r = 31.4² × 0.20 ≈ 197 m/s².
Common mistake6.3 m/s² is ωr, which is the speed in m/s, not an acceleration; 1.8 × 10⁴ m/s² uses 300 rev/min directly as if it were rad/s.
12True or false
To give two wheels the same angular acceleration, the wheel with the larger moment of inertia needs the larger torque.
Show answer
Answer: True
τ = Iα, so for the same α the torque must be proportional to I.
Common mistakeThinking a heavier-rimmed wheel needs less torque mixes up I with speed: a larger I resists changes in rotation more.
13Short answer · ★ Challenge
A learner throws a frisbee with a strong spin and it flies steadily; thrown without spin, it wobbles and tumbles. Explain the difference.
Show answer
Model answer: A spinning frisbee has a large angular momentum along its axis. Air forces give small torques, but these only slowly change the direction of a large angular momentum, so the frisbee keeps its tilt and flies steadily. Without spin, the same torques simply turn the frisbee over.
Common mistakeSaying the spin makes the frisbee lighter or gives it lift misses the point: the spin's angular momentum resists changes in the direction of the axis.
14Fill in the blank
Two equal, opposite, parallel forces whose lines of action do not coincide form a ______, which turns a body without moving it along.
Show answer
Answer: couple
A couple has zero resultant force but a non-zero torque.
Common mistakeAnswering 'equilibrium' is wrong: a couple gives zero resultant force but a resultant torque, so the body is not in rotational equilibrium.
15Multiple choice · ★ Challenge
On a see-saw, a 30 kg child sits 2.0 m from the pivot. Where must a 40 kg child sit on the other side to balance it?
1. A2.7 m from the pivot
2. B2.0 m from the pivot
3. C1.5 m from the pivot
4. D1.3 m from the pivot
Show answer
Answer: C. 1.5 m from the pivot
Clockwise moment = anticlockwise moment: 40g × d = 30g × 2.0, so d = 60 ÷ 40 = 1.5 m.
Common mistake2.7 m puts the heavier child further out; to balance, the heavier child must sit closer to the pivot.
16Fill in the blank
Torque is measured in newton ______, which must not be called joules even though the units look the same.
Show answer
Answer: metres
Torque = force × perpendicular distance, unit N m; the joule is used for energy, not torque.
Common mistakeWriting joules is the trap: N m for torque is a turning effect, while the joule measures energy transferred.
17Multiple choice · ★ Challenge
Friction reduces the angular momentum of a spinning wheel from 30 kg m²/s to 6.0 kg m²/s in 8.0 s. What is the average friction torque?
1. A4.5 N m
2. B3.0 N m
3. C0.33 N m
4. D24 N m
Show answer
Answer: B. 3.0 N m
τ = ΔL/Δt = (30 − 6.0) ÷ 8.0 = 24 ÷ 8.0 = 3.0 N m.
Common mistake4.5 N m adds the two values (36 ÷ 8); the torque depends on the CHANGE in angular momentum.
18Fill in the blank
The product of a torque and the time for which it acts is equal to the change in angular ______ of the body.
Show answer
Answer: momentum
τΔt = ΔL, the rotational form of impulse = change in momentum.
Common mistakeWriting 'velocity' leaves out the moment of inertia: torque × time changes Iω, not ω alone.
19Short answer · ★ Challenge
A tightrope walker carries a long, heavy pole held across her body. Use the idea of moment of inertia to explain how this helps her balance.
Show answer
Model answer: The mass of the pole is far from her axis of rotation, so her total moment of inertia becomes much larger. For the same disturbing torque, α = τ/I is then much smaller, so she tips slowly and has time to correct her balance.
Common mistakeSaying the pole only makes her heavier misses the point: it is the mass placed far from the axis (large r²) that greatly increases I.
20Fill in the blank
A bicycle wheel turns through 3 complete revolutions. Its angular displacement is ______ rad (give your answer in terms of π).
Show answer
Answer: 6π
One revolution is 2π rad, so 3 revolutions = 3 × 2π = 6π rad (about 18.8 rad).
Common mistakeWriting 3π uses π rad per turn; a full turn is 2π rad, half a turn is π rad.
21Multiple choice
Why is a moving bicycle much easier to keep upright than a stationary one?
1. AThe moving bicycle is lighter
2. BAir flowing past pushes it upright
3. CThe road pushes up harder on the tyres of a moving bicycle, so friction holds it upright
4. DIts spinning wheels have angular momentum that resists a change in direction
Show answer
Answer: D. Its spinning wheels have angular momentum that resists a change in direction
Angular momentum is a vector along the axle; tilting the wheel changes its direction, which needs a torque, so small disturbances have less effect.
Common mistakeChoosing 'air pushes it upright' has no physical basis at cycling speeds; the stabilising effect comes from the spinning wheels (and the rider's steering).
22Multiple choice · ★ Challenge
A star of radius 7.0 × 10⁸ m turns once every 25 days. It collapses to a neutron star of radius 10 km. Treating both as uniform spheres of the same mass, what is the new period of rotation?
1. A4.4 × 10⁻⁴ s
2. B31 s
3. C3.1 × 10⁻² s
4. D6.1 × 10⁻⁹ s
Show answer
Answer: A. 4.4 × 10⁻⁴ s
L = Iω with I ∝ R², so T ∝ R²: T₂ = 25 × 86 400 s × (1.0 × 10⁴ ÷ 7.0 × 10⁸)² = 2.16 × 10⁶ × 2.0 × 10⁻¹⁰ ≈ 4.4 × 10⁻⁴ s.
Common mistake31 s uses T ∝ R without squaring; the moment of inertia, and hence the period, depends on R².
23Multiple choice
The Earth turns once on its axis every 24 hours. What is its angular velocity?
1. A4.4 × 10⁻³ rad/s
2. B0.26 rad/s
3. C7.3 × 10⁻⁵ rad/s
4. D1.2 × 10⁻⁵ rad/s
Show answer
Answer: C. 7.3 × 10⁻⁵ rad/s
ω = 2π/T = 2π ÷ (24 × 3600 s) = 2π ÷ 86 400 ≈ 7.3 × 10⁻⁵ rad/s.
Common mistake4.4 × 10⁻³ rad/s divides 2π by the number of minutes in a day; T must be in seconds for rad/s.
24Multiple choice · ★ Challenge
A potter's wheel starts from rest with a constant angular acceleration of 2.0 rad/s². Through what angle does it turn during the third second (from t = 2 s to t = 3 s)?
1. A9.0 rad
2. B6.0 rad
3. C4.0 rad
4. D5.0 rad
Show answer
Answer: D. 5.0 rad
θ = ½αt²: after 3 s, θ = ½ × 2 × 9 = 9 rad; after 2 s, θ = 4 rad; during the third second 9 − 4 = 5 rad.
Common mistake9.0 rad is the total angle in the first 3 s; the third second is only the angle between t = 2 s and t = 3 s.
25Short answer
State the parallel-axis theorem and give one situation where it is needed.
Show answer
Model answer: I = I_cm + Md², where I_cm is the moment of inertia about a parallel axis through the centre of mass, M is the mass and d is the distance between the two axes. It is needed when a body turns about an axis that does not pass through its centre, such as a door on its hinges or a leg swinging from the hip.
Common mistakeWriting I = I_cm + Md (without squaring d) gives the wrong units; Md² has units kg m² like I.
26Multiple choice · ★ Challenge
A solid ball, a solid disc, a thin ring and a small block (sliding with no friction) are released together from the same height on the same slope. Which reaches the bottom first?
1. AThe ring
2. BThe solid ball
3. CThe solid disc
4. DThe block
Show answer
Answer: D. The block
The block has no rotational energy, so all its lost PE becomes forward KE; of the rolling bodies the ball (smallest I/MR²) is fastest.
Common mistakeChoosing the ball forgets the block: any rolling body puts some energy into spinning, while the frictionless block puts all of it into speed.
27Multiple choice
A bicycle wheel with almost all its mass in the rim and a solid disc have the same mass and radius. What is I(wheel) ÷ I(disc) about their central axes?
1. A2
2. B½
3. C1
4. D4
Show answer
Answer: A. 2
Ring: I = MR²; uniform disc: I = ½MR²; ratio = MR² ÷ ½MR² = 2.
Common mistakeChoosing 1 assumes the same mass and radius give the same I; the disc has mass close to the axis, so its I is only half.
28Multiple choice · ★ Challenge
A wheel (I = 2.0 kg m²) spinning freely slows from 20 rad/s to rest in 40 s because of friction. What torque must a motor supply to give it an angular acceleration of 2.0 rad/s² against the same friction?
1. A4.0 N m
2. B5.0 N m
3. C3.0 N m
4. D1.0 N m
Show answer
Answer: B. 5.0 N m
Friction torque = Iα = 2.0 × (20 ÷ 40) = 1.0 N m. The motor must provide Iα + friction = 2.0 × 2.0 + 1.0 = 5.0 N m.
Common mistake4.0 N m ignores friction; the resultant torque Iα is the motor torque MINUS the friction torque, so the motor must supply more.
29Multiple choice
The motor of an electric moto taxi gives 3.0 kW while turning at 1500 revolutions per minute. What torque does it give?
1. A2 N m
2. B120 N m
3. C19 N m
4. D4.7 × 10⁵ N m
Show answer
Answer: C. 19 N m
ω = 2π × 1500 ÷ 60 ≈ 157 rad/s; τ = P/ω = 3000 ÷ 157 ≈ 19 N m.
Common mistake120 N m divides by 25 rev/s without the 2π; P = τω needs ω in rad/s.
30Multiple choice · ★ Challenge
Kigali is almost on the equator, 6.37 × 10⁶ m from the Earth's axis. How fast is a building in Kigali carried round by the Earth's daily spin?
1. A74 m/s
2. B463 m/s
3. C1.1 × 10⁴ m/s
4. D7.3 × 10⁻⁵ m/s
Show answer
Answer: B. 463 m/s
v = ωr = (2π ÷ 86 400) × 6.37 × 10⁶ ≈ 463 m/s.
Common mistake74 m/s forgets the factor 2π (one turn is 2π rad, not 1 rad); 7.3 × 10⁻⁵ is ω itself, not a speed.
31Multiple choice
A wheel of radius 0.40 m has an angular acceleration of 5.0 rad/s². What is the tangential acceleration of a point on its rim?
1. A12.5 m/s²
2. B0.080 m/s²
3. C5.0 m/s²
4. D2.0 m/s²
Show answer
Answer: D. 2.0 m/s²
a = rα = 0.40 × 5.0 = 2.0 m/s².
Common mistake12.5 m/s² divides α by r; tangential acceleration grows with distance from the axis, so multiply.
32Multiple choice · ★ Challenge
A flywheel (I = 2.0 kg m²) spinning at 30 rad/s is stopped by a brake that gives a constant torque of 15 N m. How many revolutions does it make while stopping?
1. A9.5 revolutions
2. B60 revolutions
3. C4.8 revolutions
4. D30 revolutions
Show answer
Answer: A. 9.5 revolutions
KE = ½Iω² = ½ × 2.0 × 30² = 900 J; θ = W/τ = 900 ÷ 15 = 60 rad = 60 ÷ 2π ≈ 9.5 revolutions.
Common mistake60 revolutions forgets that W = τθ gives θ in radians; divide by 2π to get turns.
33Short answer
Explain why a helicopter with one main rotor needs a small tail rotor.
Show answer
Model answer: When the engine turns the main rotor one way, the rotor pushes back on the helicopter body with an equal and opposite torque. With no external torque, total angular momentum stays zero, so the body would spin the opposite way. The tail rotor gives a sideways thrust whose torque cancels this.
Common mistakeSaying the tail rotor helps the helicopter fly forwards misses its job: it balances the reaction torque so the body does not spin.
34Multiple choice · ★ Challenge
A uniform rod 1.2 m long with mass 0.50 kg has I = ML²/12 about its centre. Use the parallel-axis theorem to find I about an axis through one end, perpendicular to the rod.
1. A0.060 kg m²
2. B0.72 kg m²
3. C0.24 kg m²
4. D0.12 kg m²
Show answer
Answer: C. 0.24 kg m²
I_cm = 0.50 × 1.2² ÷ 12 = 0.060 kg m²; I_end = I_cm + Md² = 0.060 + 0.50 × 0.60² = 0.060 + 0.18 = 0.24 kg m² (= ML²/3).
Common mistake0.72 kg m² is M × 1.2², using d = 1.2 m and leaving out I_cm; the shift is from the centre to the end, which is half the length, 0.60 m.
35Multiple choice
A mechanic in a Gikondo garage must tighten a wheel nut to a torque of 60 N m with a spanner 0.25 m long. What is the smallest force she can use?
1. A240 N
2. B15 N
3. C60 N
4. D480 N
Show answer
Answer: A. 240 N
The smallest force acts at right angles at the end of the spanner: F = τ/r = 60 ÷ 0.25 = 240 N.
Common mistake15 N multiplies 60 by 0.25 instead of dividing; a short spanner needs a LARGER force for the same torque.
36Short answer · ★ Challenge
Show that a thin hoop (I = MR²) rolling without slipping always has equal translational and rotational kinetic energy. Hence find how fast a hoop is moving after rolling from rest through a vertical drop of 0.80 m (g = 9.8 m/s²).
Show answer
Model answer: KE_rot = ½(MR²)(v/R)² = ½Mv², the same as KE_trans = ½Mv². Then Mgh = ½Mv² + ½Mv² = Mv², so v = √(gh) = √(9.8 × 0.80) ≈ 2.8 m/s.
Common mistakeUsing v = √(2gh) ≈ 4.0 m/s ignores rotation; for a ring half of the energy goes into spinning.
37Multiple choice
A football, treated as a uniform solid sphere (I = ⅖MR²), is kicked so that it rolls along the pitch without slipping. What share of its kinetic energy is stored in its spinning?
1. A2/5
2. B2/7
3. C5/7
4. D1/3
Show answer
Answer: B. 2/7
KE_rot = ½ × ⅖MR² × (v/R)² = ⅕Mv²; total = ½Mv² + ⅕Mv² = 7/10 Mv²; fraction = (1/5) ÷ (7/10) = 2/7.
Common mistake2/5 is just the factor in I; the fraction must compare the rotational KE with the TOTAL KE.
38Multiple choice
A steady torque of 4.0 N m acts for 3.0 s on a wheel that starts at rest. What angular momentum does the wheel gain?
1. A1.3 kg m²/s
2. B0.75 kg m²/s
3. C12 kg m²/s
4. D6 kg m²/s
Show answer
Answer: C. 12 kg m²/s
Angular impulse = τΔt = 4.0 × 3.0 = 12 kg m²/s = ΔL.
Common mistake1.3 kg m²/s divides torque by time; like linear impulse FΔt, angular impulse multiplies torque and time.
39Short answer · ★ Challenge
A driver turns a steering wheel of radius 0.18 m by pushing up with 25 N on one side and pulling down with 25 N on the other side. Calculate the torque, and explain why the steering wheel turns but is not pushed sideways.
Show answer
Model answer: The two forces form a couple: τ = F × d = 25 × (2 × 0.18) = 25 × 0.36 = 9.0 N m. The forces are equal and opposite, so the resultant force is zero (no sideways push), but their lines of action are apart, so there is a resultant torque that turns the wheel.
Common mistakeUsing 0.18 m instead of 0.36 m gives only half the torque; the torque of a couple is one force times the distance between the two forces.
40Multiple choice
A diver leaves the board spinning at 2.0 rad/s with I = 12 kg m², then tucks so that I = 4.0 kg m². What is her new angular velocity?
1. A6.0 rad/s
2. B0.67 rad/s
3. C4.0 rad/s
4. D24 rad/s
Show answer
Answer: A. 6.0 rad/s
I₁ω₁ = I₂ω₂: ω₂ = 12 × 2.0 ÷ 4.0 = 6.0 rad/s.
Common mistake0.67 rad/s inverts the ratio; making I smaller must make ω larger to keep Iω constant.
41Short answer · ★ Challenge
At one instant, a point on the rim of a wheel of radius 0.50 m has ω = 4.0 rad/s and the wheel's angular acceleration is 6.0 rad/s². Find the tangential and centripetal accelerations of the point and the size of its total acceleration.
Show answer
Model answer: Tangential a_t = rα = 0.50 × 6.0 = 3.0 m/s². Centripetal a_c = ω²r = 4.0² × 0.50 = 8.0 m/s². They are at right angles, so a = √(3.0² + 8.0²) = √73 ≈ 8.5 m/s².
Common mistakeAdding 3.0 + 8.0 = 11 m/s² ignores direction: the tangential and centripetal parts are perpendicular, so combine them with Pythagoras.
42Multiple choice
A ceiling fan turning at 30 rad/s is switched off and turns through 45 rad while slowing uniformly to rest. What is the size of its angular deceleration?
1. A20 rad/s²
2. B10 rad/s²
3. C0.67 rad/s²
4. D5.0 rad/s²
Show answer
Answer: B. 10 rad/s²
ω² = ω₀² + 2αθ: 0 = 30² + 2α × 45, so α = −900 ÷ 90 = −10 rad/s².
Common mistake20 rad/s² forgets the 2 in 2αθ; 0.67 rad/s² divides ω by θ, which is not a valid equation.
43Short answer · ★ Challenge
A flywheel energy store (I = 50 kg m²) is used to keep a clinic's fridge running during a power cut. It slows from 300 rad/s to 100 rad/s in 20 s. How much energy does it give out, and what is the average power?
Show answer
Model answer: ΔKE = ½I(ω₁² − ω₂²) = ½ × 50 × (300² − 100²) = 25 × 80 000 = 2.0 × 10⁶ J. Average power = 2.0 × 10⁶ ÷ 20 = 1.0 × 10⁵ W (100 kW).
Common mistakeUsing ½I(ω₁ − ω₂)² = ½ × 50 × 200² gives 1.0 × 10⁶ J, which is wrong: subtract the squares, not square the difference.
44Short answer
On a turning merry-go-round, a child sits near the centre and another near the edge. Explain why both have the same angular velocity but different linear speeds.
Show answer
Model answer: The merry-go-round is rigid, so every point turns through the same angle in the same time: ω is the same. In one turn, the child at the edge goes round a larger circle (2πr is bigger), so v = ωr is greater for the larger radius.
Common mistakeThinking the outer child turns faster mixes angular and linear speed: both make one turn in the same time; only the distance travelled differs.
45Short answer · ★ Challenge
A learner pulls a string wound round the rim of a bicycle wheel (radius 0.30 m, I = 0.12 kg m²) with a steady force of 12 N. Ignoring friction, find the angular acceleration of the wheel and the length of string pulled off in the first 2.0 s from rest.
Show answer
Model answer: τ = Fr = 12 × 0.30 = 3.6 N m; α = τ/I = 3.6 ÷ 0.12 = 30 rad/s². θ = ½αt² = ½ × 30 × 2.0² = 60 rad. Length = rθ = 0.30 × 60 = 18 m.
Common mistakeGiving 60 m as the length forgets to convert the angle into an arc length with s = rθ.
46Multiple choice
A grinding wheel with I = 0.20 kg m² is brought from rest to 150 rad/s in 5.0 s by a motor. What constant torque does the motor supply (ignore friction)?
1. A30 N m
2. B750 N m
3. C3.0 N m
4. D6.0 N m
Show answer
Answer: D. 6.0 N m
α = Δω/Δt = 150 ÷ 5.0 = 30 rad/s²; τ = Iα = 0.20 × 30 = 6.0 N m.
Common mistake30 N m multiplies I by ω (0.20 × 150), which gives angular momentum, not torque; first find α.
47Short answer · ★ Challenge
Two learners spin identical roundabouts of radius 2.0 m from rest. Ama pushes tangentially at the edge with 50 N for 4.0 s; Kalisa pushes with 100 N for 2.0 s. Compare the angular momentum each roundabout gains.
Show answer
Model answer: Ama: τΔt = (50 × 2.0) × 4.0 = 400 kg m²/s. Kalisa: τΔt = (100 × 2.0) × 2.0 = 400 kg m²/s. Both gain the same angular momentum, so the identical roundabouts end with the same angular velocity.
Common mistakeAssuming the bigger force always wins forgets time: angular impulse is torque × time, so half the force for twice as long gives the same result.
48Multiple choice
Three 0.50 kg masses are fixed to a light rod at 0.10 m, 0.20 m and 0.30 m from one end. What is the moment of inertia of the system about an axis through that end, perpendicular to the rod?
1. A0.30 kg m²
2. B0.070 kg m²
3. C0.060 kg m²
4. D0.14 kg m²
Show answer
Answer: B. 0.070 kg m²
I = Σmr² = 0.50 × (0.10² + 0.20² + 0.30²) = 0.50 × 0.14 = 0.070 kg m².
Common mistake0.30 kg m² uses Σmr without squaring r; 0.060 kg m² puts all 1.5 kg at the middle mass, but each distance must be squared separately.
49Multiple choice
A moto taxi moves at 12 m/s on wheels of radius 0.30 m that roll without slipping. How fast is the top of a tyre moving relative to the road?
1. A12 m/s
2. B0 m/s
3. C40 m/s
4. D24 m/s
Show answer
Answer: D. 24 m/s
The top moves at v (forward motion of the axle) + ωR (rotation) = 12 + 12 = 24 m/s; the bottom is momentarily at rest.
Common mistake12 m/s forgets that rotation adds to the forward motion at the top; 40 m/s is v/R, which is ω, not a speed.
50Short answer · ★ Challenge
A graph of angular velocity against time for a grinding stone is a straight line falling from 24 rad/s at t = 0 to zero at t = 8.0 s. Use the graph to find the angular deceleration and the number of revolutions made while stopping.
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Model answer: Gradient = (0 − 24) ÷ 8.0 = −3.0 rad/s², so the deceleration is 3.0 rad/s². Angle = area under the graph = ½ × 24 × 8.0 = 96 rad = 96 ÷ 2π ≈ 15 revolutions.
Common mistakeReading 96 as the number of revolutions forgets that the area under an ω–t graph gives radians; divide by 2π to get turns.