ExaminerPrep › Question bank › IAL Physics › 4.2 Circular motion
Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
40 free, high-quality past-paper-style questions on Circular motion for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
What is an angle of 72° in radians?
| Answer | Mark | Notes |
|---|---|---|
| 360° = 2π rad, so 72° = 72 × 2π ÷ 360 = 72 × π ÷ 180 = 1.26 rad. | ✓ 1 | Answer C |
Answer: C
Exam tip: Remember one fixed pair — 180° = π rad — and scale from it. A quick check: 1 rad ≈ 57°, so 72° should be a little more than 1 rad.
The blade of a desk fan completes one revolution every 0.040 s.
What is the angular velocity of the blade?
| Answer | Mark | Notes |
|---|---|---|
| ω = 2π ÷ T = 2π ÷ 0.040 = 157 rad s−1. | ✓ 1 | Answer D |
Answer: D
Common error: 1/T is the frequency in revolutions per second. Angular velocity counts radians, so multiply by 2π.
Two children sit on a playground roundabout that rotates at a steady rate. Child P sits 1.2 m from the axis and child Q sits 2.4 m from the axis.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| The roundabout is rigid, so every point turns through the same angle in the same time: ω is the same for P and Q. Since v = ωr, Q at twice the radius has twice the speed. (Q's acceleration rω2 is also twice P's.) | ✓ 1 | Answer A |
Answer: A
Common error: Points further from the axis move faster, but they do not turn faster: angular velocity is the same everywhere on a rigid rotating body.
A car is driven round a circular roundabout at constant speed.
What is the direction of the acceleration of the car?
| Answer | Mark | Notes |
|---|---|---|
| The direction of the velocity changes continuously, so the car accelerates even at constant speed. The change in velocity, and so the acceleration, is directed towards the centre of the circle. | ✓ 1 | Answer A |
Answer: A
Exam tip: Constant speed is not constant velocity. In circular motion the velocity is along the tangent and the acceleration is along the radius, inwards.
A bicycle is turned upside down for repair and its front wheel is spun. The wheel has a radius of 0.34 m and makes 5.0 complete revolutions in 2.0 s at a steady rate.
Calculate the angle, in radians, through which the wheel turns in 2.0 s.
Calculate the angular velocity of the wheel.
Calculate the speed of a point on the rim of the wheel.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5.0 × 2π = 31 rad (31.4 rad) | ✓ 1 | |
| Part (b) | ||
| 31.4 ÷ 2.0 = 16 rad s−1 (15.7 rad s−1) | ✓ 1 | |
| Part (c) | ||
| v = ωr = 15.7 × 0.34 | ✓ 1 | |
| = 5.3 m s−1 (5.34) | ✓ 1 | |
Exam tip: One revolution is 2π rad. Counting revolutions and multiplying by 2π is the quickest route to an angle in radians.
A large observation wheel has a radius of 16 m and turns at a steady rate, making one complete revolution every 6.0 minutes. A passenger of mass 70 kg sits in one of its cabins.
Calculate the angular velocity of the wheel.
Calculate the speed of the passenger.
Calculate the resultant force on the passenger.
Explain why the passenger cannot feel any difference between the force from the seat at the top of the wheel and the force at the bottom.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| period = 6.0 × 60 = 360 s | ✓ 1 | |
| ω = 2π ÷ 360 = 0.017 rad s−1 (0.0175) | ✓ 1 | |
| Part (b) | ||
| v = ωr = 0.0175 × 16 = 0.28 m s−1 | ✓ 1 | |
| Part (c) | ||
| F = mrω2 = 70 × 16 × 0.01752 | ✓ 1 | |
| = 0.34 N towards the centre of the wheel | ✓ 1 | |
| Part (d) | ||
| the seat force is (687 − 0.34) N at the top and (687 + 0.34) N at the bottom; the difference, 0.68 N, is only about 0.1% of the weight, far too small to notice | ✓ 1 | |
Exam tip: Convert minutes to seconds before using T = 2π/ω. A quick size check — is the answer sensible for a slow wheel? — catches unit slips.
Figure 1 shows a windscreen wiper. The arm pivots at its lower end and its tip moves along the arc shown, from one end position to the other.
Through what angle θ does the arm turn?
| Answer | Mark | Notes |
|---|---|---|
| θ = s/r = 0.72 ÷ 0.45 = 1.6 rad. In degrees: 1.6 × 180° ÷ π = 91.7°. | ✓ 1 | Answer B |
Answer: B
Common error: s/r gives the angle in radians. Convert with 180°/π at the end; multiplying by 180 alone leaves out the π.
A potter's wheel turns at a constant rate. The graph shows the angle θ through which the wheel has turned against time t.
What is the period of rotation of the wheel?
| Answer | Mark | Notes |
|---|---|---|
| ω = gradient = 10 rad ÷ 4.0 s = 2.5 rad s−1. T = 2π/ω = 2π ÷ 2.5 = 2.51 s. | ✓ 1 | Answer C |
Answer: C
Exam tip: The gradient of a graph of angle (in radians) against time is the angular velocity. Use a long stretch of the line and read it where it crosses gridlines.
The figure shows a belt drive. The motor pulley turns at an angular velocity of 30 rad s−1 and the belt does not slip on either pulley.
What is the angular velocity of the large pulley?
| Answer | Mark | Notes |
|---|---|---|
| The belt moves at one speed, so the rims of both pulleys have the same speed: v = ωr = 30 × 0.040 = 1.2 m s−1. For the large pulley ω = v/r = 1.2 ÷ 0.12 = 10 rad s−1. | ✓ 1 | Answer B |
Answer: B
Common error: Points joined by a belt share the same speed, not the same angular velocity. The larger pulley turns more slowly.
The figure shows a car, seen from above, moving at constant speed round a circular test track with centre O. Four arrows are drawn at the car's position P.
Which arrow shows the direction of the resultant force on the car at P?
| Answer | Mark | Notes |
|---|---|---|
| At constant speed the velocity changes only in direction, and that change is towards the centre. So the resultant force (sideways friction from the road) acts towards O: arrow 1. Arrow 2 shows the direction of the velocity. | ✓ 1 | Answer A |
Answer: A
Common error: Drawing the resultant force along the direction of motion. In uniform circular motion the resultant force is at 90° to the velocity, so it does no work and the speed stays the same.
The figure shows a revolving door seen from above. One wing of the door, 1.1 m from the central axis to its outer edge E, turns at a steady rate from position 1 to position 2 in 3.5 s.
Convert the angle shown in the figure to radians.
Calculate the angular velocity of the wing.
Calculate the speed of edge E.
Calculate the time the door takes to make one complete revolution.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 105 × π ÷ 180 = 1.83 rad | ✓ 1 | |
| Part (b) | ||
| ω = 1.83 ÷ 3.5 = 0.52 rad s−1 (0.524) | ✓ 1 | |
| Part (c) | ||
| v = ωr = 0.524 × 1.1 | ✓ 1 | |
| = 0.58 m s−1 (0.576) | ✓ 1 | |
| Part (d) | ||
| T = 2π ÷ 0.524 = 12 s | ✓ 1 | |
Exam tip: Convert the angle to radians before dividing by the time, and keep the unrounded ω (0.5236 rad s−1) for the later parts.
The figure shows a hammer throw seen from above. The athlete turns and the hammer head, of mass 7.3 kg, moves in a horizontal circle. Assume that the wire is horizontal, and ignore the weight of the hammer head and the mass of the wire.
The athlete lets go of the hammer when the head is at P. State which arrow, A, B, C or D, shows the direction in which the head moves just after release.
Just before release the hammer head moves at 26 m s−1. Calculate its angular velocity.
Calculate the tension in the wire just before release.
State the direction of the force that the wire exerts on the athlete.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| C, along the tangent in the direction of motion | ✓ 1 | |
| Part (b) | ||
| ω = v/r = 26 ÷ 1.9 | ✓ 1 | |
| = 14 rad s−1 (13.7) | ✓ 1 | |
| Part (c) | ||
| T = mv2/r = 7.3 × 262 ÷ 1.9 | ✓ 1 | |
| = 2.6 kN (2600 N) | ✓ 1 | |
| Part (d) | ||
| along the wire, away from the athlete towards the hammer head | ✓ 1 | |
Common error: Thinking the hammer flies off along the wire, directly away from the athlete. Once it is released there is no horizontal force on it, so it carries on along the tangent.
A track cyclist rides round a curve of radius 25 m at a constant speed of 12 m s−1.
What is the magnitude of the cyclist's centripetal acceleration?
| Answer | Mark | Notes |
|---|---|---|
| a = v2/r = 122 ÷ 25 = 5.8 m s−2 (5.76). | ✓ 1 | Answer C |
Answer: C
Exam tip: Check the units: (m s−1)2 ÷ m = m s−2. Only v2/r gives the units of an acceleration.
A puck of mass 0.30 kg on a smooth horizontal table is attached to a string fixed at the centre of the table. It moves in a circle of radius 0.80 m with an angular velocity of 5.0 rad s−1.
What is the tension in the string?
| Answer | Mark | Notes |
|---|---|---|
| The tension is the only horizontal force, so it provides the centripetal force: T = mrω2 = 0.30 × 0.80 × 5.02 = 6.0 N. | ✓ 1 | Answer C |
Answer: C
Exam tip: Choose the form of the equation that matches the data: with ω given, use mrω2 rather than converting to a speed first.
A car drives at constant speed over the top of a hump-backed bridge whose surface is an arc of a vertical circle. The car stays in contact with the road.
At the top of the bridge, how does the normal contact force on the car compare with its weight?
| Answer | Mark | Notes |
|---|---|---|
| At the top the centre of the circle is below the car, so the resultant force must be downwards: mg − N = mv2/r. So N = mg − mv2/r, which is less than the weight. It would only be zero at the speed √(gr), when the car would just lose contact. | ✓ 1 | Answer B |
Answer: B
Common error: Assuming the forces balance because the speed is constant. The car is accelerating towards the centre of the arc, so the forces cannot balance.
A bucket of water is swung in a vertical circle of radius 0.95 m. g = 9.81 m s−2.
What is the minimum speed of the bucket at the top of the circle if the water is to stay in the bucket?
| Answer | Mark | Notes |
|---|---|---|
| At the minimum speed the bucket exerts no force on the water, so the weight alone provides the centripetal force: mg = mv2/r, giving v = √(gr) = √(9.81 × 0.95) = 3.1 m s−1 (3.05). | ✓ 1 | Answer B |
Answer: B
Exam tip: At the minimum speed at the top of a vertical circle the contact force or tension is zero. Write mg = mv2/r; the mass cancels.
The drum of a washing machine has a radius of 0.25 m. During the spin cycle it rotates at 1200 revolutions per minute. The curved wall of the drum has many small holes in it.
Show that the angular velocity of the drum is about 130 rad s−1.
Calculate the centripetal acceleration of a piece of clothing pressed against the wall of the drum.
Explain why water leaves the clothes through the holes in the drum during the spin cycle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| frequency = 1200 ÷ 60 = 20 Hz | ✓ 1 | |
| ω = 2π × 20 = 126 rad s−1 (125.7) | ✓ 1 | |
| Part (b) | ||
| a = rω2 = 0.25 × 125.72 | ✓ 1 | |
| = 3.9 × 103 m s−2 (3950) | ✓ 1 | |
| Part (c) | ||
| the water needs a (large) resultant force towards the centre to keep moving in a circle | ✓ 1 | |
| at a hole there is no wall to push on the water / adhesion to the cloth is too small to provide it, so the water moves off along a tangent, through the hole | ✓ 1 | |
Common error: Saying that the water is “thrown outwards” by a force. Nothing pushes the water out; it leaves because nothing pushes it in.
A car of mass 980 kg travels round a level roundabout on a circular path of radius 18 m at a constant speed of 8.5 m s−1.
Calculate the resultant force on the car.
State what provides this force.
On a dry day the largest sideways friction the road can provide is 6.5 kN. Calculate the greatest speed at which the car can travel round this path.
On an icy day the largest sideways friction is only 3.0 kN. Explain what happens if the car tries to follow the same path at 8.5 m s−1.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = mv2/r = 980 × 8.52 ÷ 18 | ✓ 1 | |
| = 3.9 kN (3930 N) towards the centre | ✓ 1 | |
| Part (b) | ||
| (sideways) friction between the tyres and the road | ✓ 1 | |
| Part (c) | ||
| 6500 = 980v2 ÷ 18 | ✓ 1 | |
| v = 10.93 m s−1, so 10.9 m s−1 (rounded down: at 11 m s−1 the car would skid) | ✓ 1 | |
| Part (d) | ||
| the friction available (3.0 kN) is less than the 3.9 kN needed | ✓ 1 | |
| so the car cannot keep to the 18 m circle; it follows a path of larger radius / slides towards the outside of the roundabout | ✓ 1 | |
Common error: Writing that the car slides “outwards because of centrifugal force”. It slides out because the inward force is too small to turn it as sharply as the path requires.
A light aircraft of mass 1100 kg flies in a horizontal circle at a constant speed of 62 m s−1. To do this the pilot banks the aircraft so that its wings make an angle of 28° with the horizontal. The lift force acts at right angles to the wings. Air resistance is balanced by the thrust and can be ignored. g = 9.81 m s−2.
Explain why the lift force must be greater than the weight of the aircraft during the turn.
Show that the lift force is about 1.2 × 104 N.
Calculate the radius of the circle.
Calculate the time taken for the aircraft to fly once round the circle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the lift is tilted at 28° to the vertical, and only its vertical component balances the weight | ✓ 1 | |
| L cos 28° = mg, and cos 28° < 1, so L > mg | ✓ 1 | |
| Part (b) | ||
| L = 1100 × 9.81 ÷ cos 28° | ✓ 1 | |
| = 1.22 × 104 N (12 200 N) | ✓ 1 | |
| Part (c) | ||
| L sin 28° = mv2/r, so r = 1100 × 622 ÷ (12 220 × sin 28°) | ✓ 1 | |
| = 740 m (737 m) | ✓ 1 | |
| Part (d) | ||
| 2π × 737 ÷ 62 = 75 s (74.7 s) | ✓ 1 | |
Exam tip: Resolve the lift vertically (it balances the weight) and horizontally (it provides mv2/r). Dividing the two gives tan 28° = v2/(rg) — a useful check.
A student whirls a rubber bung of mass 0.025 kg in a horizontal circle. The string passes through a short glass tube held in her hand and a hanging mass of 0.100 kg is attached to its lower end, so the tension in the string is equal to the weight of the hanging mass. A small clip fixed to the string keeps the radius at 0.50 ± 0.01 m. She times 20 revolutions as 14.2 ± 0.2 s. Assume the string between the tube and the bung is horizontal and that friction at the tube is negligible. g = 9.81 m s−2.
Calculate the period and the angular velocity of the bung.
Calculate mrω2 for the bung and compare it with the tension in the string.
Determine the percentage uncertainty in her value of mrω2. Assume the uncertainty in the mass of the bung is negligible.
Explain why she timed 20 revolutions rather than one.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| T = 14.2 ÷ 20 = 0.710 s | ✓ 1 | |
| ω = 2π ÷ 0.710 = 8.85 rad s−1 | ✓ 1 | |
| Part (b) | ||
| mrω2 = 0.025 × 0.50 × 8.852 = 0.979 N | ✓ 1 | |
| tension = 0.100 × 9.81 = 0.981 N, so the two agree (to within 0.2%), as F = mrω2 predicts | ✓ 1 | |
| Part (c) | ||
| % uncertainty in r = 0.01 ÷ 0.50 × 100 = 2.0% | ✓ 1 | |
| % uncertainty in time (and so in ω) = 0.2 ÷ 14.2 × 100 = 1.4%, doubled for ω2 = 2.8% | ✓ 1 | |
| total = 2.0 + 2.8 = 4.8% | ✓ 1 | |
| Part (d) | ||
| the same timing (reaction-time) uncertainty is spread over 20 periods, so the percentage uncertainty in the period is much smaller | ✓ 1 | |
Exam tip: When a quantity is squared, its percentage uncertainty is doubled. Add the percentage uncertainties of the quantities that are multiplied or divided.
A student investigates when a coin slides off a rotating turntable. For each angular velocity ω she finds the greatest distance r from the axis at which the coin stays in place. Her results: ω = 3.0 rad s−1, r = 0.300 m; ω = 4.0 rad s−1, r = 0.169 m; ω = 5.0 rad s−1, r = 0.108 m; ω = 6.0 rad s−1, r = 0.075 m. The maximum friction force Fmax between the coin and the turntable is the same at every radius.
Explain why, at a given angular velocity, the coin slides if it is placed further than a certain distance from the axis.
Show that r = (Fmax/m) × 1/ω2 at the limit, and state what the gradient of a graph of r against 1/ω2 represents.
Use the results for ω = 3.0 rad s−1 and ω = 6.0 rad s−1 to show that the gradient is about 2.7 m s−2.
The coin has a mass of 7.5 g. Calculate Fmax.
Calculate the greatest distance from the axis at which the coin stays in place when the turntable rotates at 45 revolutions per minute.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the centripetal force needed, mrω2, increases with r | ✓ 1 | |
| friction has a maximum value, so beyond a certain radius it cannot provide the force needed and the coin slides | ✓ 1 | |
| Part (b) | ||
| at the limit Fmax = mrω2, which rearranges to r = (Fmax/m)(1/ω2) | ✓ 1 | |
| gradient = Fmax/m (the greatest centripetal acceleration the friction can give the coin) | ✓ 1 | |
| Part (c) | ||
| 1/ω2 = 0.111 s2 and 0.0278 s2 | ✓ 1 | |
| gradient = (0.300 − 0.075) ÷ (0.111 − 0.0278) = 2.70 m s−2 | ✓ 1 | |
| Part (d) | ||
| 0.0075 × 2.70 = 0.020 N | ✓ 1 | |
| Part (e) | ||
| ω = 2π × 45 ÷ 60 = 4.71 rad s−1 | ✓ 1 | |
| r = 2.70 ÷ 4.712 = 0.12 m (0.122 m) | ✓ 1 | |
Model sketch
Exam tip: To find what a gradient means, rearrange the physics equation into the form y = mx with your plotted quantities as y and x; everything multiplying x is the gradient.
An object moves at a constant speed of 5.0 m s−1 round a circle. The figure shows its velocity at P and, a quarter of a revolution later, at Q.
What is the magnitude of the change in velocity between P and Q?
| Answer | Mark | Notes |
|---|---|---|
| Δv = vQ − vP. The two velocities are at right angles and each is 5.0 m s−1, so |Δv| = √(5.02 + 5.02) = 7.1 m s−1 (7.07). | ✓ 1 | Answer D |
Answer: D
Exam tip: To subtract one vector from another, add its reverse: vQ + (−vP). Drawn nose to tail, these two form a right-angled triangle.
A toy car runs at different steady speeds round a circular track of radius 0.40 m. A sensor measures the resultant force F on the car towards the centre. The graph shows F against v2.
What is the mass of the car?
| Answer | Mark | Notes |
|---|---|---|
| F = mv2/r, so the gradient of F against v2 is m/r. Gradient = 1.5 N ÷ 5.0 m2 s−2 = 0.30 kg m−1, so m = 0.30 × 0.40 = 0.12 kg. | ✓ 1 | Answer A |
Answer: A
Exam tip: Match the equation to y = mx: with F on the y-axis and v2 on the x-axis, the gradient is everything multiplying v2, which is m/r.
The figure shows a half-pipe whose curved part is an arc of a circle. A skateboarder of total mass 55 kg passes the lowest point at 6.0 m s−1. g = 9.81 m s−2.
What is the normal contact force on the skateboarder at the lowest point?
| Answer | Mark | Notes |
|---|---|---|
| At the lowest point the centre of the arc is directly above, so the resultant force is upwards: N − mg = mv2/r. N = 55 × 9.81 + 55 × 6.02 ÷ 3.2 = 540 + 619 = 1200 N (1160 N). | ✓ 1 | Answer D |
Answer: D
Common error: Writing N = mv2/r. At the bottom the normal force has two jobs: it supports the weight and provides the resultant force towards the centre.
A car travels round a bend on a banked track at the speed at which no sideways friction is needed. Each diagram shows the car from behind, with the centre of the bend to the left.
Which diagram correctly shows all the forces acting on the car?
| Answer | Mark | Notes |
|---|---|---|
| Only two forces act: the weight W and the normal contact force N, which is at right angles to the track. Because the track is tilted, N has a horizontal component towards the centre of the bend; this provides the centripetal force, while its vertical component balances W. Diagram 3. | ✓ 1 | Answer C |
Answer: C
Common error: Adding “mv2/r” or “centripetal force” as an extra arrow. The centripetal force is the resultant of the real forces, not an additional force.
A laboratory centrifuge spins sample tubes about a vertical axis. The graph shows how the angular velocity ω of the rotor varies with time t after it is switched on. The bottom of each sample tube is 0.080 m from the axis.
Show that the rotor turns through about 160 revolutions in the first 5.0 s.
Calculate the period of rotation once the rotor has reached its full speed.
Calculate the centripetal acceleration at the bottom of a tube at full speed.
A sample of mass 2.5 g sits at the bottom of a tube. Calculate the resultant force on the sample at full speed and state what provides it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| angle turned = area under the graph = ½ × 5.0 × 400 = 1000 rad | ✓ 1 | |
| number of revolutions = 1000 ÷ 2π = 159 | ✓ 1 | |
| Part (b) | ||
| T = 2π ÷ 400 = 0.016 s (0.0157 s) | ✓ 1 | |
| Part (c) | ||
| a = rω2 = 0.080 × 4002 | ✓ 1 | |
| = 1.3 × 104 m s−2 (12 800) | ✓ 1 | |
| Part (d) | ||
| F = ma = 0.0025 × 12 800 | ✓ 1 | |
| = 32 N | ✓ 1 | |
| the normal contact force (push) of the bottom of the tube on the sample, directed towards the axis | ✓ 1 | |
Exam tip: The area under an ω–t graph is the angle turned in radians, just as the area under a v–t graph is a displacement. Divide by 2π to count revolutions.
The figure shows a plan view of a level kart circuit, drawn to scale. A kart and its driver have a total mass of 160 kg. On this surface the largest sideways friction force the tyres can provide is 0.80 times the weight of the kart and driver. g = 9.81 m s−2.
State what provides the centripetal force on the kart as it goes round a bend.
Calculate the greatest speed at which the kart can go round bend X without sliding.
The kart goes round bend Y at 11 m s−1. Determine whether it can do so without sliding.
Explain why the greatest speed at which a bend can be taken does not depend on the mass of the kart.
State which bend can be taken at the highest speed. Give a reason.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (sideways) friction between the tyres and the track surface | ✓ 1 | |
| Part (b) | ||
| 0.80mg = mv2/r, so v = √(0.80 × 9.81 × 12) | ✓ 1 | |
| = 9.7 m s−1 (9.70) | ✓ 1 | |
| Part (c) | ||
| friction needed = 160 × 112 ÷ 20 = 970 N (968 N) | ✓ 1 | |
| greatest friction available = 0.80 × 160 × 9.81 = 1260 N, which is more than needed, so the kart does not slide | ✓ 1 | |
| Part (d) | ||
| the force needed (mv2/r) and the greatest friction (0.80mg) are both proportional to m, so m cancels: v2 = 0.80gr | ✓ 1 | |
| Part (e) | ||
| bend Z, because it has the largest radius and the greatest speed is proportional to √r | ✓ 1 | |
Common error: Saying the kart is “thrown outwards” on a tight bend. It slides outwards because friction cannot supply the larger inward force that a smaller radius needs.
The figure shows a conical pendulum. A small mass of 0.15 kg on a light string moves at constant speed in a horizontal circle. g = 9.81 m s−2.
Show that the radius of the circle is about 0.66 m.
Calculate the tension in the string.
Show that ω2 = g/h, where h is the vertical distance from the support to the plane of the circle.
Calculate the time for one revolution.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| r = √(1.202 − 1.002) = 0.663 m | ✓ 1 | |
| Part (b) | ||
| the vertical component of the tension balances the weight: T × (1.00 ÷ 1.20) = 0.15 × 9.81 | ✓ 1 | |
| T = 1.8 N (1.77 N) | ✓ 1 | |
| Part (c) | ||
| horizontally T sin θ = mrω2 and vertically T cos θ = mg, so tan θ = rω2/g | ✓ 1 | |
| from the figure tan θ = r/h, so r/h = rω2/g, giving ω2 = g/h | ✓ 1 | |
| Part (d) | ||
| ω = √(9.81 ÷ 1.00) = 3.13 rad s−1 | ✓ 1 | |
| time = 2π ÷ 3.13 = 2.0 s (2.01 s) | ✓ 1 | |
Exam tip: No angle is given, but the figure fixes it: cos θ = 1.00/1.20 and tan θ = r/h. Take the geometry from the diagram before resolving the tension.
A student fixes a smartphone flat on a horizontal rotating platform. The phone's accelerometer measures the centripetal acceleration a of its sensor, and she finds the angular velocity ω of the platform by timing 10 revolutions. The graph shows her results, with a line of best fit.
State the direction of the acceleration measured by the phone.
Explain how the graph supports the relationship a = rω2.
Use the graph to determine the distance of the sensor from the axis.
Calculate the speed of the sensor when the platform turns at 8.0 rad s−1.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| horizontal, towards the axis of rotation | ✓ 1 | |
| Part (b) | ||
| the points lie close to a straight line through the origin, so a ∝ ω2 | ✓ 1 | |
| the constant gradient is consistent with a fixed radius r, which equals the gradient | ✓ 1 | |
| Part (c) | ||
| gradient from a large triangle on the line, e.g. 9.0 ÷ 60 | ✓ 1 | |
| gradient = 0.150 (allow 0.145–0.155) | ✓ 1 | |
| r = gradient = 0.15 m | ✓ 1 | |
| Part (d) | ||
| v = ωr = 8.0 × 0.150 | ✓ 1 | |
| = 1.2 m s−1 (allow 1.16–1.24) | ✓ 1 | |
Exam tip: Take the gradient from points on the best-fit line, not from the data points, using a triangle that covers most of the line.
The figure shows part of a roller-coaster track in side view. The lowest point P of the dip lies on an arc of radius 12 m, and the highest point Q of the crest lies on an arc of radius 18 m. A rider of mass 60 kg travels in a car along the track. g = 9.81 m s−2.
Explain how the contact force from the seat on the rider compares with the rider's weight at P and at Q.
The car passes P at 15 m s−1. Calculate the contact force from the seat on the rider at P.
Calculate the greatest speed at which the car can pass Q without the rider losing contact with the seat.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Indicative content: | — | |
| at P the centre of the circle is above the rider, so the resultant force on the rider must be upwards | ✓ 1 | |
| so at P the contact force is greater than the weight: N − mg = mv2/r | ✓ 1 | |
| at Q the centre of the circle is below the rider, so the resultant force must be downwards | ✓ 1 | |
| so at Q the contact force is less than the weight: mg − N = mv2/r | ✓ 1 | |
| Part (b) | ||
| N = 60 × 9.81 + 60 × 152 ÷ 12 | ✓ 1 | |
| = 589 + 1125 = 1.7 kN (1710 N) | ✓ 1 | |
| Part (c) | ||
| at the greatest speed the contact force is zero: mg = mv2/r | ✓ 1 | |
| v = √(9.81 × 18) = 13 m s−1 (13.28) | ✓ 1 | |
Common error: Assuming the forces balance because the car simply follows the track. At P and Q the car moves on a circle, so there is a resultant force towards the centre of that circle.
An object moves in a horizontal circle. Its speed is doubled and the radius of its circle is halved.
By what factor does the resultant force on the object change?
| Answer | Mark | Notes |
|---|---|---|
| F = mv2/r. Doubling v multiplies F by 4; halving r multiplies it by 2 more. Overall factor = 4 × 2 = 8. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a ratio question, write the new force as m(2v)2/(r/2) and simplify; do not try to do it in your head.
A small mass on a string moves as a conical pendulum in a horizontal circle of radius 0.40 m. The string makes an angle of 30° with the vertical. g = 9.81 m s−2.
What is the speed of the mass?
| Answer | Mark | Notes |
|---|---|---|
| Vertically T cos 30° = mg; horizontally T sin 30° = mv2/r. Dividing: tan 30° = v2/(rg), so v = √(0.40 × 9.81 × tan 30°) = 1.5 m s−1 (1.51). | ✓ 1 | Answer B |
Answer: B
Exam tip: Resolve vertically and horizontally, then divide one equation by the other: the tension and the mass both cancel.
An object moves at constant speed v round a circle of radius r. In a short time Δt it moves from point A to point B, and the radius turns through a small angle Δθ.
Explain why the object is accelerating even though its speed is constant.
Using a vector diagram, show that the magnitude of the acceleration is v2/r.
Use your answer to (b) to show that the acceleration can also be written as rω2.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| velocity is a vector / has direction, and the direction of motion is changing | ✓ 1 | |
| a change in velocity (per unit time) is an acceleration | ✓ 1 | |
| Part (b) | ||
| diagram: velocity vectors at A and B, each of magnitude v, drawn from one point with angle Δθ between them; Δv joins their tips | ✓ 1 | |
| for small Δθ, |Δv| = vΔθ (Δθ in radians) | ✓ 1 | |
| time taken Δt = arc length ÷ speed = rΔθ/v | ✓ 1 | |
| a = Δv/Δt = vΔθ × v/(rΔθ) = v2/r | ✓ 1 | |
| Part (c) | ||
| substitute v = ωr: a = (ωr)2/r = rω2 | ✓ 1 | |
Model sketch
Exam tip: In a derivation, label your diagram and give a reason for every line. State that Δθ is small and measured in radians — that is what makes the short side of the triangle equal to vΔθ.
The bends of a cycle-racing track are steeply banked, so the outer edge of the track is much higher than the inner edge.
Explain how banking the track allows cyclists to travel round the bends at high speed, and why there is one speed at which no sideways friction is needed. Your answer should refer to the forces acting on the cyclist and bicycle.
| Answer | Mark | Notes |
|---|---|---|
| Indicative content: | — | |
| a resultant force of mv2/r towards the centre of the bend is needed to move in a circle | ✓ 1 | |
| on a flat track this force can only come from sideways friction, which has a maximum value, limiting the speed | ✓ 1 | |
| on a banked track the normal reaction acts at right angles to the surface, so it is tilted towards the centre of the bend | ✓ 1 | |
| the vertical component of the normal reaction balances the weight | ✓ 1 | |
| the horizontal component of the normal reaction provides some or all of the centripetal force, so less friction is needed | ✓ 1 | |
| at one speed the horizontal component of the normal reaction equals mv2/r exactly, so no friction is needed; faster or slower, friction acts down or up the slope | ✓ 1 |
Model sketch
Exam tip: For a long explanation, plan the chain before writing: what is needed (mv2/r inwards) → what supplies it on a flat track → what changes when the track is tilted → the special speed. One idea per sentence.
A ball of mass 0.20 kg on the end of a light string is swung in a vertical circle of radius 0.60 m. Air resistance is negligible. g = 9.81 m s−2.
Calculate the minimum speed the ball must have at the top of the circle for the string to stay taut.
The ball passes the top of the circle at 3.0 m s−1. Calculate the tension in the string at this point.
Calculate the speed of the ball at the bottom of the circle.
Calculate the tension in the string at the bottom of the circle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| tension zero, so mg = mv2/r | ✓ 1 | |
| v = √(9.81 × 0.60) = 2.426 m s−1, so 2.43 m s−1 (rounded up: at 2.4 m s−1 the string would go slack) | ✓ 1 | |
| Part (b) | ||
| T + mg = mv2/r: T = 0.20 × 3.02 ÷ 0.60 − 0.20 × 9.81 | ✓ 1 | |
| = 3.00 − 1.96 = 1.0 N (1.04 N) | ✓ 1 | |
| Part (c) | ||
| gain in kinetic energy = loss in potential energy over 2r = 1.20 m: v2 = 3.02 + 2 × 9.81 × 1.20 | ✓ 1 | |
| v = 5.7 m s−1 (5.70) | ✓ 1 | |
| Part (d) | ||
| T − mg = mv2/r: T = 0.20 × 9.81 + 0.20 × 32.5 ÷ 0.60 | ✓ 1 | |
| = 1.96 + 10.8 = 13 N (12.8 N) | ✓ 1 | |
Exam tip: At each point take towards the centre as positive. The weight helps at the top and opposes at the bottom, which is why the tension is greatest at the bottom of the circle.
The figure shows a fairground wheel. It turns at a steady rate, making one revolution every 8.0 s, and the riders stay upright in their seats. g = 9.81 m s−2.
What is the ratio of the contact force from the seat on a rider at the bottom to the contact force from the seat on the same rider at the top?
| Answer | Mark | Notes |
|---|---|---|
| ω = 2π ÷ 8.0 = 0.785 rad s−1, so rω2 = 9.0 × 0.7852 = 5.55 m s−2. Top (centre below): mg − N = mrω2, so N = m(9.81 − 5.55). Bottom (centre above): N − mg = mrω2, so N = m(9.81 + 5.55). Ratio = 15.36 ÷ 4.26 = 3.6. | ✓ 1 | Answer D |
Answer: D
Exam tip: At each point take the direction towards the centre as positive. In a ratio the mass cancels, so it is never needed.
A student swings a small mass on a string in a vertical circle. A force sensor records the tension T in the string as the mass passes the lowest point, and a light gate records its speed v there. The graph shows T against v2 for several runs. g = 9.81 m s−2.
What is the radius of the circle?
| Answer | Mark | Notes |
|---|---|---|
| At the lowest point T − mg = mv2/r, so T = mg + (m/r)v2. Intercept: mg = 2.0 N, so m = 2.0 ÷ 9.81 = 0.204 kg. Gradient: (18 − 2.0) ÷ 40 = 0.40 kg m−1 = m/r, so r = 0.204 ÷ 0.40 = 0.51 m. | ✓ 1 | Answer B |
Answer: B
Exam tip: Compare T = mg + (m/r)v2 with y = c + mx: the intercept gives the weight and the gradient gives m/r. You need both.
An object moves anticlockwise at a constant speed of 4.0 m s−1 round a circle of radius 2.0 m. The left of the figure shows the object at P and, a short time later, at Q; the radii to P and Q are 40° apart. The vector diagram on the right shows the velocities at P and Q drawn from one point.
Calculate the time taken to move from P to Q.
Use the vector diagram to calculate the magnitude of the change in velocity between P and Q.
Calculate the average acceleration of the object between P and Q.
Calculate v2/r. Explain why it is slightly larger than your answer to (c), and what happens to the difference as Q is taken closer to P.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| arc PQ = rθ = 2.0 × 0.698 = 1.40 m | ✓ 1 | |
| t = 1.40 ÷ 4.0 = 0.35 s (0.349 s) | ✓ 1 | |
| Part (b) | ||
| the triangle is isosceles: two sides of 4.0 m s−1 with 40° between them | ✓ 1 | |
| |Δv| = 2 × 4.0 × sin 20° = 2.7 m s−1 (2.74) | ✓ 1 | |
| Part (c) | ||
| 2.74 ÷ 0.349 = 7.8 m s−2 (7.84) | ✓ 1 | |
| Part (d) | ||
| v2/r = 4.02 ÷ 2.0 = 8.0 m s−2 | ✓ 1 | |
| (c) is an average over a 40° turn: |Δv| = 2v sin 20° is slightly less than vΔθ (the chord is shorter than the arc); as Q moves closer to P the average acceleration approaches v2/r, the instantaneous value | ✓ 1 | |
Exam tip: Δv runs from the tip of the first velocity to the tip of the second. It points half-way between the two radii, inwards — towards the centre of the circle.
In a rotor ride, riders stand against the inside wall of a large vertical cylinder that spins about its central axis. Once the cylinder is spinning fast enough, the floor is lowered. The figure shows a cross-section, with a rider of mass 65 kg and the three forces X, Y and Z acting on the rider. The cylinder makes one revolution every 1.9 s. g = 9.81 m s−2.
Name the forces X, Y and Z.
State which of the forces provides the centripetal force on the rider.
Calculate the magnitude of force Z.
The friction force between the rider and the wall can be at most 0.50 times force Z. Deduce whether the rider slides down the wall when the floor is lowered.
Explain why the floor must not be lowered until the cylinder is spinning quickly.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| X: weight; Y: friction from the wall | ✓ 1 | |
| Z: normal contact force (reaction) from the wall | ✓ 1 | |
| Part (b) | ||
| Z, the normal contact force from the wall | ✓ 1 | |
| Part (c) | ||
| ω = 2π ÷ 1.9 = 3.31 rad s−1, so Z = mrω2 = 65 × 2.2 × 3.312 | ✓ 1 | |
| = 1.6 kN (1560 N) | ✓ 1 | |
| Part (d) | ||
| greatest friction = 0.50 × 1560 = 780 N; weight = 65 × 9.81 = 640 N (638 N) | ✓ 1 | |
| the friction needed to balance the weight is less than the greatest friction available, so the rider does not slide | ✓ 1 | |
| Part (e) | ||
| Z = mrω2, and so the greatest friction, is small when ω is small; friction could not balance the weight and the rider would slide down | ✓ 1 | |
Exam tip: Treat the two directions separately: vertically the rider is in equilibrium (friction balances weight); horizontally there is a resultant force, Z, towards the axis.
A student whirls a small mass on a string in a vertical circle. A force sensor at the centre of the circle records the tension in the string, and the graph shows the tension against time. Assume that the speed of the mass is constant. g = 9.81 m s−2.
Explain why the tension is greatest when the mass is at the lowest point of the circle.
Use the graph to show that the mass is about 0.20 kg.
Use the graph to determine the time for one revolution.
Determine the radius of the circle.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at the lowest point the weight acts away from the centre: T − mg = mv2/r, so T = mv2/r + mg | ✓ 1 | |
| at the highest point the weight acts towards the centre: T = mv2/r − mg; mv2/r is the same everywhere, so the tension is greatest at the bottom | ✓ 1 | |
| Part (b) | ||
| greatest − least tension = (mv2/r + mg) − (mv2/r − mg) = 2mg = 6.0 − 2.0 = 4.0 N | ✓ 1 | |
| m = 4.0 ÷ (2 × 9.81) = 0.204 kg | ✓ 1 | |
| Part (c) | ||
| time between successive maxima = 0.80 s | ✓ 1 | |
| Part (d) | ||
| mrω2 = mean of greatest and least tension = (6.0 + 2.0) ÷ 2 = 4.0 N | ✓ 1 | |
| ω = 2π ÷ 0.80 = 7.85 rad s−1 | ✓ 1 | |
| r = 4.0 ÷ (0.204 × 7.852) = 0.32 m (0.318 m) | ✓ 1 | |
Exam tip: Write one equation for the top and one for the bottom, taking towards the centre as positive. Subtracting them removes mv2/r; adding them removes mg.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy