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A2 · Practice questions · Centripetal acceleration and force

The resultant force always points to the centre.

Six original Cambridge-style questions on centripetal acceleration, the force that supplies it, banked turns and motion in a vertical circle.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Centripetal force is a resultant, not a new force.

01
Calculation
[3 marks]

A car of mass 1200 kg travels round a flat, horizontal bend of radius 45 m at a steady 15 m s⁻¹. Calculate the centripetal acceleration of the car and the frictional force the road must provide, and state the direction of this force.

  • a = v² / r = 15² / 45 = 5.0 m s⁻². ✓
  • F = ma = 1200 × 5.0 = 6.0 × 10³ N. ✓
  • The friction acts horizontally toward the centre of the bend, since it is the only horizontal force on the car and so must be the centripetal force. ✓
02
Analysis
[2 marks]

A satellite moves in a circular orbit at constant speed. Explain why the satellite is nevertheless accelerating, and state the direction of its acceleration.

  • Velocity is a vector; its direction changes continuously as the satellite moves round the circle, so the velocity changes and there is an acceleration even though the speed (magnitude) is constant. ✓
  • The acceleration is directed toward the centre of the circle, perpendicular to the velocity. ✓
03
Calculation
[3 marks]

The drum of a washing machine has a radius of 0.25 m and spins at 1200 revolutions per minute. Calculate the angular velocity of the drum and the centripetal acceleration of a sock pressed against the drum wall, and express this acceleration as a multiple of g.

  • ω = 2πf = 2π × (1200 / 60) = 2π × 20 = 126 rad s⁻¹. ✓
  • a = rω² = 0.25 × 126² = 3.9 × 10³ m s⁻². ✓
  • a / g = 3.95 × 10³ / 9.81 = 400, so the wall pushes on the sock with about 400 times its weight. ✓
04
Analysis
[3 marks]

A passenger on a bus states that, as the bus turns a corner, "a centrifugal force pushes me outward against the window". Explain what actually happens to the passenger, naming the force that acts on the passenger and stating what would happen if that force were too small.

  • There is no outward force on the passenger: by inertia (Newton's first law) the passenger tends to continue in a straight line while the bus turns beneath them, which feels like being pushed outward. ✓
  • The real force on the passenger is the inward push from the window (a normal contact force), together with friction from the seat; this resultant inward force is the centripetal force mv² / r. ✓
  • If the available inward force were less than mv² / r the passenger would not follow the circle but would slide outward relative to the bus, continuing along a straighter path. ✓
05
Calculation
[3 marks]

An aircraft of mass 8.0 × 10³ kg flies in a horizontal circle of radius 1.2 km at a constant speed of 85 m s⁻¹. The only forces on it are its weight and the lift L, which is perpendicular to the wings. Calculate the angle θ at which the wings must be banked from the horizontal, and the magnitude of the lift.

  • Vertically L cos θ = mg and horizontally L sin θ = mv² / r, so tan θ = v² / (rg) = 85² / (1200 × 9.81) = 0.614. ✓
  • θ = tan⁻¹(0.614) = 32° (31.5°). ✓
  • L = mg / cos θ = (8.0 × 10³ × 9.81) / cos 31.5° = 9.2 × 10⁴ N, larger than the weight of 7.8 × 10⁴ N because the lift must also supply the inward force. ✓
06
Calculation
[4 marks]

A roller-coaster car of mass 450 kg travels round the inside of a vertical circular loop of radius 8.0 m. At the top of the loop its speed is 12 m s⁻¹.

(a) Calculate the centripetal acceleration of the car at the top of the loop. (b) Calculate the normal contact force exerted by the track on the car at the top. (c) Calculate the minimum speed at the top for which the car just stays in contact with the track. (d) Explain why passengers feel pressed hard into their seats at the bottom of the loop.

  • (a) a = v² / r = 12² / 8.0 = 18 m s⁻², directed downward toward the centre. ✓
  • (b) At the top both the weight and the contact force act toward the centre: N + mg = ma, so N = m(a − g) = 450 × (18 − 9.81) = 3.7 × 10³ N. ✓
  • (c) Contact is just lost when N = 0, so mg = mv² / r and v = √(gr) = √(9.81 × 8.0) = 8.9 m s⁻¹. ✓
  • (d) At the bottom the centripetal force is upward, so the seat must push up with N = mg + mv² / r, more than the weight, and the car is also moving fastest there (it has lost height), making v² / r larger still. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.