A Level Physics 12.2 Centripetal Acceleration MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice questions cover Section 12.2 of the Cambridge International AS & A Level Physics (9702) syllabus: centripetal acceleration, centripetal force and circular-motion problem solving. This is an A Level (A2) topic, so it builds on the kinematics of uniform circular motion from Section 12.1.

Formula recap: a = v2 / r = ω2r  and  F = m v2 / r = m ω2r, where the acceleration and the resultant force are always directed towards the centre of the circle. For the full A Level Physics formula sheet, see A Level Physics Formula Sheet.

Centripetal Acceleration

Q1. An object moves at constant speed in a horizontal circle. In which direction does its acceleration act?

  • (A) Along a tangent, in the direction of motion
  • (B) Directly away from the centre of the circle
  • (C) Towards the centre of the circle
  • (D) The acceleration is zero because the speed is constant
Show Answer

Answer: C – The acceleration in circular motion is centripetal: it acts along the radius towards the centre, perpendicular to the velocity, and it changes the direction of motion without changing the speed.

Q2. A car travels around a flat circular bend of radius 3.0 m at a constant speed of 6.0 m s−1. What is its centripetal acceleration?

  • (A) 2.0 m s−2
  • (B) 12 m s−2
  • (C) 18 m s−2
  • (D) 36 m s−2
Show Answer

Answer: B – a = v2 / r = (6.0)2 / 3.0 = 36 / 3.0 = 12 m s−2, directed towards the centre of the bend.

Q3. A point on a rotating wheel is 0.50 m from the axle. The wheel rotates with angular speed 4.0 rad s−1. What is the centripetal acceleration of the point?

  • (A) 8.0 m s−2
  • (B) 2.0 m s−2
  • (C) 32 m s−2
  • (D) 0.13 m s−2
Show Answer

Answer: A – a = ω2r = (4.0)2 × 0.50 = 16 × 0.50 = 8.0 m s−2.

Q4. A satellite moves in a circular orbit at constant speed, yet it is accelerating. Why?

  • (A) Its speed increases and decreases during each orbit
  • (B) A friction force acts on it at a tangent to the orbit
  • (C) The centripetal force does work on it at every instant
  • (D) Velocity is a vector: its direction changes continuously, so the velocity changes even though the speed does not
Show Answer

Answer: D – Acceleration is the rate of change of velocity. Because the direction of motion changes at every instant, the velocity changes, so there is an acceleration towards the centre. The force is perpendicular to the motion, so it does no work and the speed stays constant.

Q5. An object moves in a circle of fixed radius. Its speed is doubled. By what factor does its centripetal acceleration change?

  • (A) It doubles
  • (B) It becomes four times as large
  • (C) It halves
  • (D) It becomes sixteen times as large
Show Answer

Answer: B – a = v2 / r, so the acceleration is proportional to v2. Doubling v multiplies a by 22 = 4.

Q6. An object moves in a circle at constant speed v. The radius of the circle is doubled while v stays the same. What happens to the centripetal acceleration?

  • (A) It doubles
  • (B) It stays the same
  • (C) It halves
  • (D) It becomes one quarter as large
Show Answer

Answer: C – a = v2 / r with v constant, so a is inversely proportional to r. Doubling r halves the acceleration.

Q7. A mass rotates in a circle at constant angular speed ω. The radius of the circle is doubled while ω stays the same. What happens to the centripetal acceleration?

  • (A) It doubles
  • (B) It halves
  • (C) It becomes four times as large
  • (D) It stays the same
Show Answer

Answer: A – a = ω2r with ω constant, so a is directly proportional to r. Doubling r doubles the acceleration. (Note the contrast with Q6: the result depends on whether v or ω is held constant.)

Q8. At any instant, the velocity of an object in uniform circular motion is directed

  • (A) towards the centre of the circle
  • (B) along a tangent to the circle
  • (C) directly away from the centre of the circle
  • (D) in the same direction as the acceleration
Show Answer

Answer: B – The velocity is always tangential to the path. The acceleration (and resultant force) points along the radius towards the centre, at 90° to the velocity.

Q9. An object moving in a circle of radius 4.0 m has a centripetal acceleration of 25 m s−2. What is its speed?

  • (A) 5.0 m s−1
  • (B) 6.3 m s−1
  • (C) 10 m s−1
  • (D) 100 m s−1
Show Answer

Answer: C – a = v2 / r, so v = √(a r) = √(25 × 4.0) = √100 = 10 m s−1.

Q10. What is the angle between the velocity and the centripetal acceleration of an object in uniform circular motion?

  • (A) 90°
  • (B) 0°
  • (C) 45°
  • (D) 180°
Show Answer

Answer: A – The velocity is tangential and the acceleration is radial (towards the centre), so they are always perpendicular. This is why the force changes only the direction of the velocity, never the speed.

Centripetal Force

Q11. A ball of mass 0.50 kg is whirled in a horizontal circle of radius 2.0 m at a speed of 4.0 m s−1. What is the magnitude of the resultant force on the ball?

  • (A) 16 N
  • (B) 8.0 N
  • (C) 2.0 N
  • (D) 4.0 N
Show Answer

Answer: D – F = m v2 / r = 0.50 × (4.0)2 / 2.0 = 0.50 × 16 / 2.0 = 4.0 N, directed towards the centre.

Q12. Which statement about centripetal force is correct?

  • (A) It is an extra force that must be added to the other forces acting on the object
  • (B) It is the resultant of all the real forces acting on the object, directed towards the centre
  • (C) It is a new type of force produced by rotation itself
  • (D) It acts outwards, balancing the inward forces
Show Answer

Answer: B – “Centripetal force” is not a separate force. It is the name given to the resultant (net) force towards the centre, which is always provided by real forces such as tension, friction, gravity or a normal contact force.

Q13. A stone is whirled in a horizontal circle on the end of a string. Which real force provides the centripetal force?

  • (A) The tension in the string
  • (B) The weight of the stone
  • (C) The friction between the stone and the air
  • (D) The normal reaction of the ground
Show Answer

Answer: A – The string pulls the stone towards the centre of the circle. This tension is the resultant horizontal force, so it provides m v2 / r.

Q14. A car drives at constant speed around a flat, unbanked circular bend. Which force provides the centripetal force?

  • (A) The weight of the car
  • (B) The normal reaction of the road
  • (C) The friction between the tyres and the road
  • (D) The forward force from the engine
Show Answer

Answer: C – Weight and the normal reaction are vertical and cancel. The only horizontal force is friction between the tyres and the road, acting towards the centre of the bend.

Q15. A mass of 0.40 kg is whirled in a horizontal circle of radius 0.80 m on a string that breaks when the tension reaches 20 N. What is the maximum speed of the mass before the string breaks?

  • (A) 6.3 m s−1
  • (B) 4.0 m s−1
  • (C) 8.0 m s−1
  • (D) 40 m s−1
Show Answer

Answer: A – The tension provides the centripetal force: T = m v2 / r, so v = √(T r / m) = √(20 × 0.80 / 0.40) = √40 = 6.3 m s−1.

Q16. A planet moves in a circular orbit around the Sun. Which force provides the centripetal force?

  • (A) Air resistance
  • (B) The thrust of the planet
  • (C) Tension in the orbit
  • (D) The gravitational attraction of the Sun
Show Answer

Answer: D – Gravity is the only force acting on the planet. It points towards the Sun (the centre of the orbit) and provides the centripetal force m v2 / r.

Q17. A mass of 2.0 kg moves in a circle of radius 1.5 m with angular speed 3.0 rad s−1. What is the centripetal force on the mass?

  • (A) 9.0 N
  • (B) 18 N
  • (C) 27 N
  • (D) 54 N
Show Answer

Answer: C – F = m ω2r = 2.0 × (3.0)2 × 1.5 = 2.0 × 9.0 × 1.5 = 27 N.

Q18. A car of mass 1000 kg goes around a flat bend of radius 40 m. The maximum friction force between the tyres and the road is 3000 N. What is the maximum speed at which the car can take the bend without skidding?

  • (A) 5.5 m s−1
  • (B) 11 m s−1
  • (C) 22 m s−1
  • (D) 120 m s−1
Show Answer

Answer: B – Friction provides the centripetal force: F = m v2 / r, so v = √(F r / m) = √(3000 × 40 / 1000) = √120 = 11 m s−1. Above this speed the available friction cannot supply the required centripetal force.

Q19. A ball on a string is moving in a horizontal circle when the string suddenly breaks. What does the ball do immediately after the break?

  • (A) It moves off along a tangent to the circle
  • (B) It moves directly outwards along the radius
  • (C) It continues around the circle for a short time
  • (D) It stops where it is
Show Answer

Answer: A – With the tension gone there is no resultant horizontal force, so by Newton’s first law the ball continues with its velocity at that instant, which is tangential. It does not fly outwards: there is no outward force.

Q20. An object moves in a circle of radius r at speed v. A second, identical object of twice the mass moves in the same circle at the same speed. How does the centripetal force on the second object compare with the first?

  • (A) It is the same
  • (B) It is twice as large
  • (C) It is half as large
  • (D) It is four times as large
Show Answer

Answer: B – F = m v2 / r is directly proportional to m when v and r are unchanged. Doubling the mass doubles the required centripetal force.


Circular Motion Problems

Q21. In a conical pendulum, a mass moves in a horizontal circle of radius 0.60 m with the string at 37° to the vertical (tan 37° ≈ 0.75). What is the speed of the mass? (g = 9.81 m s−2)

  • (A) 1.2 m s−1
  • (B) 4.4 m s−1
  • (C) 6.6 m s−1
  • (D) 2.1 m s−1
Show Answer

Answer: D – Resolving the tension: the vertical component balances the weight (T cos θ = m g) and the horizontal component provides the centripetal force (T sin θ = m v2 / r). Dividing gives tan θ = v2 / (r g), so v2 = r g tan θ = 0.60 × 9.81 × 0.75 = 4.4, and v = 2.1 m s−1.

Q22. A disc rotates at 300 revolutions per minute. A small object is fixed to the disc 0.10 m from the centre. What is the centripetal acceleration of the object?

  • (A) 3.1 m s−2
  • (B) 9.9 m s−2
  • (C) 31 m s−2
  • (D) 99 m s−2
Show Answer

Answer: D – 300 rev min−1 = 5.0 rev s−1, so ω = 2π × 5.0 = 31.4 rad s−1. Then a = ω2r = (31.4)2 × 0.10 = 987 × 0.10 ≈ 99 m s−2.

Q23. A mass on a string moves at constant speed around a vertical circle. The tension in the string is greatest at the bottom of the circle because

  • (A) at the bottom the tension must both support the weight and provide the centripetal resultant: T = m g + m v2 / r
  • (B) at the bottom the speed of the mass is smallest
  • (C) at the bottom the string is stretched the longest
  • (D) the centripetal force is zero at the bottom
Show Answer

Answer: A – At the bottom, the resultant force towards the centre (upwards) is T − m g = m v2 / r, so T = m g + m v2 / r. At the top, T + m g = m v2 / r, so the tension there is smaller by 2 m g.

Q24. A mass of 0.30 kg on a string of length 1.0 m passes the bottom of its vertical circle at 5.0 m s−1. What is the tension in the string at the bottom? (g = 9.81 m s−2)

  • (A) 7.5 N
  • (B) 10.4 N
  • (C) 2.9 N
  • (D) 4.6 N
Show Answer

Answer: B – At the bottom, T − m g = m v2 / r, so T = m v2 / r + m g = (0.30 × 25 / 1.0) + (0.30 × 9.81) = 7.5 + 2.9 = 10.4 N. (7.5 N forgets the weight; 4.6 N wrongly subtracts it, which is the top-of-circle arrangement.)

Q25. For a satellite in a circular orbit, gravity provides the centripetal force, giving v = √(G M / r). How does the orbital speed change as the orbital radius r increases?

  • (A) It increases in proportion to r
  • (B) It stays the same for all orbits
  • (C) It increases as √r
  • (D) It decreases as r increases
Show Answer

Answer: D – Setting G M m / r2 = m v2 / r gives v = √(G M / r). The speed falls as r grows: satellites in higher orbits move more slowly (and take longer to complete each orbit).

Q26. Why are bends on high-speed roads and velodrome tracks banked (tilted)?

  • (A) Banking increases the force of friction between the tyres and the road
  • (B) Banking reduces the value of g on the bend
  • (C) A horizontal component of the normal reaction can then provide some or all of the centripetal force
  • (D) Banking increases the radius of the bend
Show Answer

Answer: C – On a banked surface the normal reaction is tilted, so it has a horizontal component pointing towards the centre of the bend. At the ideal speed this component alone provides the centripetal force, with no friction needed.

Q27. An aeroplane turns in a horizontal circle by banking its wings. What provides the centripetal force for the turn?

  • (A) A horizontal component of the lift force, because the lift acts perpendicular to the wings
  • (B) A sideways push from the rudder
  • (C) The weight of the aeroplane
  • (D) The thrust of the engines, which points towards the centre
Show Answer

Answer: A – Lift acts at right angles to the wings. When the plane banks, the lift tilts with it: its vertical component balances the weight and its horizontal component points towards the centre of the turn, providing the centripetal force.

Q28. A motorcycle loops a vertical circle of radius 2.0 m. What is the minimum speed at the top of the loop for the motorcycle to stay in contact with the track? (g = 9.81 m s−2)

  • (A) 19.6 m s−1
  • (B) 14 m s−1
  • (C) 4.4 m s−1
  • (D) 2.2 m s−1
Show Answer

Answer: C – At the minimum speed the normal reaction at the top is just zero, so the weight alone provides the centripetal force: m g = m v2 / r, giving v = √(g r) = √(9.81 × 2.0) = √19.6 = 4.4 m s−1.

Q29. A coin of mass 0.050 kg rests on a turntable 0.15 m from the centre. The maximum friction force between the coin and the turntable is 0.45 N. At what angular speed does the coin start to slip?

  • (A) 60 rad s−1
  • (B) 3.9 rad s−1
  • (C) 0.13 rad s−1
  • (D) 7.7 rad s−1
Show Answer

Answer: D – Friction provides the centripetal force: F = m ω2r, so ω2 = F / (m r) = 0.45 / (0.050 × 0.15) = 0.45 / 0.0075 = 60, and ω = √60 = 7.7 rad s−1. Above this, friction cannot hold the coin in its circle.

Q30. A satellite orbits the Earth in a circle of radius 7.0 × 106 m at a speed of 7.5 × 103 m s−1. What is its orbital period?

  • (A) 9.3 × 102 s
  • (B) 5.9 × 103 s
  • (C) 3.7 × 104 s
  • (D) 5.3 × 107 s
Show Answer

Answer: B – One orbit covers 2πr, so T = 2πr / v = (2π × 7.0 × 106) / (7.5 × 103) = 5.9 × 103 s, about 98 minutes – typical of a low Earth orbit.

Keep practising: Work through more A Level Physics topic questions on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open while you revise circular motion.

These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.