Specification: AQA A-level Physics (7408) 3.6.1.1 | Paper: 1 | Data and formulae booklet: given in the exam
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Centripetal acceleration: an object moving in a circle at constant speed is accelerating, because its direction (and so its velocity) is continually changing. The acceleration is directed towards the centre of the circle, a = v² ÷ r = ω²r.
Centripetal force: the resultant force on an object moving in a circle, directed towards the centre, F = mv² ÷ r. It is not an extra force: it is provided by a real force such as tension, friction, gravity or the normal reaction.
No work done: the centripetal force is always perpendicular to the velocity, so it does no work and the speed (and kinetic energy) stays constant.
The equation
centripetal acceleration = (speed)² ÷ radius = (angular speed)² × radius a = v² ÷ r = ω²r and centripetal force = mass × (speed)² ÷ radius F = mv² ÷ r = mω²r
| Quantity | Symbol | Unit |
|---|---|---|
| centripetal acceleration | a | metres per second squared (m s⁻²) |
| centripetal (resultant) force | F | newtons (N) |
| mass | m | kilograms (kg) |
| linear speed | v | metres per second (m s⁻¹) |
| angular speed | ω | radians per second (rad s⁻¹) |
| radius of the circle | r | metres (m) |
Rearranged: v = √(Fr ÷ m) | r = mv² ÷ F | ω = √(F ÷ mr) | m = Fr ÷ v² | Always ask “which real force points towards the centre?” and set the resultant of the forces towards the centre equal to mv² ÷ r.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A car travels at a constant speed of 20 m s⁻¹ round a bend of radius 50 m. Calculate the centripetal acceleration of the car.
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F: a = v² ÷ r
I: a = 20² ÷ 50 = 400 ÷ 50
F: a is already the subject
A: 8.0 m s⁻², directed towards the centre of the bend
Q2 (Core) A 0.20 kg puck on a smooth horizontal table is attached to a string and moves in a horizontal circle of radius 0.60 m at 5.0 m s⁻¹. Calculate the tension in the string, to 2 significant figures.
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F: F = mv² ÷ r (the tension provides the centripetal force)
I: F = 0.20 × 5.0² ÷ 0.60
F: F is already the subject
A: 8.333… N = 8.3 N (2 s.f.)
Q3 (Core) A 1200 kg car drives round a level bend at 54 km h⁻¹. The friction between the tyres and the road provides a centripetal force of 4.5 kN. Calculate the radius of the bend.
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Convert first: 54 km h⁻¹ = 54 ÷ 3.6 = 15 m s⁻¹; 4.5 kN = 4500 N
F: F = mv² ÷ r
I: 4500 = 1200 × 15² ÷ r
F: r = mv² ÷ F = 1200 × 225 ÷ 4500
A: 60 m
Watch out: forgetting to square the speed, or squaring 54 instead of 15. Convert km h⁻¹ to m s⁻¹ before you square.
Q4 (Core) The drum of a washing machine has a radius of 25 cm and spins at 1200 rpm. A wet towel of mass 500 g is pressed against the drum wall. Calculate, to 2 significant figures, (a) the centripetal acceleration of the towel and (b) the centripetal force on it. (c) Name the force that provides it.
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Convert first: 1200 rpm = 20 rev s⁻¹, so ω = 2π × 20 = 125.66… rad s⁻¹; 25 cm = 0.25 m; 500 g = 0.500 kg
Step 1 – (a) acceleration
F: a = ω²r
I: a = 125.66…² × 0.25
A: 3947.8… m s⁻² = 3.9 × 103 m s⁻² (2 s.f.), about 400 times g
Step 2 – (b) force
F: F = mω²r = ma
I: F = 0.500 × 3947.8…
A: 1973.9… N = 2.0 × 103 N (2 s.f.)
(c) the normal reaction force of the drum wall pushing inwards on the towel
Q5 (Exam) An athlete swings a 7.26 kg hammer in a horizontal circle of radius 1.9 m at 2.0 revolutions per second. Treat the hammer head as a point mass and ignore its weight. Calculate (a) the tension in the wire and (b) the speed of the hammer when it is released, both to 2 significant figures.
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Step 1 – angular speed
ω = 2πf = 2π × 2.0 = 12.56… rad s⁻¹
Step 2 – (a) tension
F: F = mω²r
I: F = 7.26 × 12.56…² × 1.9
F: F is already the subject
A: 2178.2… N = 2.2 kN (2 s.f.)
Step 3 – (b) release speed
v = ωr = 12.56… × 1.9 = 23.87… m s⁻¹ = 24 m s⁻¹ (2 s.f.), at a tangent to the circle
Watch out: on release the hammer flies off along the tangent, not outwards along the radius. There is no “outward force”.
Q6 (Exam) The Moon orbits the Earth in a circle of radius 3.84 × 108 m with a period of 27.3 days. The mass of the Moon is 7.35 × 1022 kg. Calculate, to 3 significant figures, (a) the centripetal acceleration of the Moon and (b) the gravitational force the Earth exerts on it.
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Convert first: T = 27.3 × 86 400 = 2 358 720 s
Step 1 – angular speed
ω = 2π ÷ T = 2π ÷ 2 358 720 = 2.663… × 10−6 rad s⁻¹
Step 2 – (a) acceleration
F: a = ω²r
I: a = (2.663… × 10−6)² × 3.84 × 108
A: 2.724… × 10−3 m s⁻² = 2.72 × 10−3 m s⁻² (3 s.f.)
Step 3 – (b) force
gravity provides the centripetal force, so F = ma = 7.35 × 1022 × 2.724… × 10−3
A: 2.002… × 1020 N = 2.00 × 1020 N (3 s.f.)
Q7 (Exam) A 1200 kg car approaches a level, unbanked bend of radius 45 m. The maximum frictional force the road can exert sideways on the tyres is 9.0 kN. Calculate the maximum speed at which the car can take the bend without skidding, in m s⁻¹ and km h⁻¹, to 2 significant figures.
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Convert first: 9.0 kN = 9000 N
F: F = mv² ÷ r (friction provides the centripetal force)
I: 9000 = 1200 × v² ÷ 45
F: v² = Fr ÷ m = 9000 × 45 ÷ 1200 = 337.5, so v = √337.5
A: 18.37… m s⁻¹ = 18 m s⁻¹ (2 s.f.), and 18.37… × 3.6 = 66 km h⁻¹
Watch out: any faster and the friction needed (mv² ÷ r) exceeds 9.0 kN, so the car skids outwards along a tangent. On a wet road the maximum friction falls and so does the safe speed.
Q8 (Exam) A 900 kg car drives at 12 m s⁻¹ over the top of a hump-backed bridge. The top of the bridge is a circular arc of radius 30 m. (a) Calculate the normal reaction force of the road on the car at the top, to 2 significant figures. (b) Calculate the maximum speed at which the car can go over the top without losing contact with the road, to 2 significant figures. (g = 9.81 m s⁻²)
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Step 1 – forces at the top
weight mg = 900 × 9.81 = 8829 N acts down (towards the centre); normal reaction N acts up
Step 2 – (a) resultant towards the centre
F: mg − N = mv² ÷ r
I: 8829 − N = 900 × 12² ÷ 30 = 4320
F: N = 8829 − 4320
A: 4509 N = 4.5 kN (2 s.f.)
Step 3 – (b) leaving the road
contact is lost when N = 0, so mg = mv² ÷ r and v = √(gr) = √(9.81 × 30) = √294.3
A: 17.15… m s⁻¹ = 17 m s⁻¹ (2 s.f.)
Watch out: adding the centripetal force to the weight. At the top the resultant downward force is mg − N, so the road pushes up with less than the weight.
Q9 (Stretch) A 0.20 kg mass on a light string is whirled in a vertical circle of radius 0.50 m. At the top of the circle its speed is 4.0 m s⁻¹. Ignore air resistance. (g = 9.81 m s⁻²) (a) Calculate the tension in the string at the top. (b) Use conservation of energy to find the speed at the bottom. (c) Calculate the tension at the bottom. (d) Calculate the minimum speed at the top for the string to stay taut. Give answers to 2 significant figures.
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Step 1 – (a) tension at the top
at the top, tension T and weight mg both act down, towards the centre
F: T + mg = mv² ÷ r
I: T + 0.20 × 9.81 = 0.20 × 4.0² ÷ 0.50, so T + 1.962 = 6.4
F: T = 6.4 − 1.962
A: 4.438 N = 4.4 N (2 s.f.)
Step 2 – (b) speed at the bottom
the mass falls 2r = 1.0 m, so ½mv² = ½mu² + mg(2r) and v² = 4.0² + 2 × 9.81 × 1.0 = 35.62
A: v = √35.62 = 5.968… m s⁻¹ = 6.0 m s⁻¹ (2 s.f.)
Step 3 – (c) tension at the bottom
at the bottom, T acts up (towards the centre) and mg acts down
F: T − mg = mv² ÷ r
I: T − 1.962 = 0.20 × 35.62 ÷ 0.50 = 14.248
F: T = 14.248 + 1.962
A: 16.21 N = 16 N (2 s.f.)
Step 4 – (d) minimum speed at the top
the string just goes slack when T = 0, so mg = mv² ÷ r and v = √(gr) = √(9.81 × 0.50) = 2.214… m s⁻¹ = 2.2 m s⁻¹ (2 s.f.)
Check: Tbottom − Ttop = 16.21 − 4.438 = 11.772 N = 6mg
Watch out: writing T = mv² ÷ r. The tension is not the centripetal force; the resultant of tension and weight is. The string is most likely to snap at the bottom, where the tension is greatest.
Q10 (Stretch) A conical pendulum consists of a 0.30 kg bob on a light string of length 0.80 m. The bob moves in a horizontal circle with the string at 30° to the vertical. (g = 9.81 m s⁻²) Calculate, to 3 significant figures, (a) the tension in the string, (b) the angular speed of the bob and (c) the period of its motion.
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Step 1 – (a) vertical components balance
T cos 30° = mg, so T = 0.30 × 9.81 ÷ cos 30° = 2.943 ÷ 0.8660…
A: 3.398… N = 3.40 N (3 s.f.)
Step 2 – radius of the circle
r = L sin 30° = 0.80 × 0.5 = 0.40 m (not the string length)
Step 3 – (b) horizontal component provides the centripetal force
F: T sin 30° = mω²r
I: 3.398… × 0.5 = 1.699… N = 0.30 × ω² × 0.40
F: ω² = 1.699… ÷ (0.30 × 0.40) = 14.15… (equivalently ω² = g tan 30° ÷ r)
A: ω = √14.15… = 3.762… rad s⁻¹ = 3.76 rad s⁻¹ (3 s.f.)
Step 4 – (c) period
T = 2π ÷ ω = 2π ÷ 3.762… = 1.669… s = 1.67 s (3 s.f.)
Watch out: using r = 0.80 m. The radius is the horizontal distance from the bob to the axis, L sin θ. Also, there is no separate “centripetal force” on the diagram: it is the horizontal component of the tension.
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