A2 · Paper 4 practice · Motion in a circle

Round and round, and the force that holds it.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Motion in a circle: angular quantities and the centripetal force that keeps objects on their curved paths. Each question is linked to the lesson it draws on; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Take g = 9.81 m s⁻¹.
Keep these straight

Toward the centre, always.

01
Paper 4
[13 marks]

A horizontal turntable rotates at a constant rate of 45 revolutions per minute. A small coin rests on the turntable a distance 0.12 m from the axis, as shown in Fig. 1.1.

axis r coin ω horizontal turntable, viewed from above
Fig. 1.1

(a) State what is meant by the angular velocity of a rotating object. [1]

(b) For the turntable:

(i) show that its angular velocity is about 4.7 rad s⁻¹, [2]

(ii) calculate the linear speed of the coin, [2]

(iii) calculate the centripetal acceleration of the coin. [2]

(c) The coin has a mass of 8.0 g.

(i) State the origin of the centripetal force acting on the coin. [1]

(ii) Calculate the magnitude of this force. [2]

(iii) The coin is moved to a larger distance from the axis, with the rate of rotation unchanged. Explain why there is a maximum distance beyond which the coin cannot remain on the turntable. [3]

  • (a)Rate of change of angular displacement (angle swept out per unit time), measured in rad s⁻¹ ✓
  • (b)(i)ω = 2π × (45/60) = 2π × 0.75 = 4.71 rad s⁻¹ ✓✓
  • (b)(ii)v = rω = 0.12 × 4.71 = 0.57 m s⁻¹ ✓✓
  • (b)(iii)a = rω² = 0.12 × 4.71² = 2.66 m s⁻² (or v²/r) ✓✓
  • (c)(i)Friction between the coin and the turntable surface ✓
  • (c)(ii)F = ma = 0.0080 × 2.66 = 0.021 N ✓✓
  • (c)(iii)The required force F = mω²r grows as r increases; friction has a maximum value; beyond a certain radius the needed force exceeds the maximum friction, so the coin slides outwards (slips) ✓✓✓
02
Paper 4
[13 marks]

A ball of mass 0.15 kg is attached to a string of length 0.80 m and is swung in a vertical circle, as shown in Fig. 2.1.

fixed point L ball at top ball moving in a vertical circle
Fig. 2.1

(a) State the two forces acting on the ball when it is at the top of the circle, and the direction of each. [2]

(b) At the top of the circle the ball moves with speed v.

(i) Show that, for the string to remain taut, the speed at the top must be at least 2.8 m s⁻¹. [3]

(ii) State the name of the force that becomes zero at this minimum speed. [1]

(c) On one turn the ball passes the top of the circle with a speed of 3.5 m s⁻¹.

(i) Calculate the tension in the string at the top of the circle. [2]

(ii) Show that the speed of the ball at the bottom of the circle is 6.6 m s⁻¹. [2]

(iii) Calculate the tension at the bottom of the circle, and state why it is larger than at the top. [3]

  • (a)Weight (mg) vertically downward; tension (T) along the string, also directed downward toward the centre ✓✓
  • (b)(i)At minimum speed the tension is zero, so weight provides the centripetal force: mg = mv²/r, giving v² = gr = 9.81 × 0.80, v = 2.80 m s⁻¹ ✓✓✓
  • (b)(ii)Tension ✓
  • (c)(i)T = m(v²/r − g) = 0.15(3.5²/0.80 − 9.81) = 0.83 N ✓✓
  • (c)(ii)Energy: ½mvb² = ½mvt² + mg(2r), so vb² = vt² + 4gr = 3.5² + 4 × 9.81 × 0.80 = 43.6; vb = 6.6 m s⁻¹ ✓✓
  • (c)(iii)T = m(vb²/r + g) = 0.15(43.6/0.80 + 9.81) = 9.7 N ✓✓; larger because at the bottom the tension must both provide the centripetal force and support the weight, and the ball moves faster there ✓
03
Paper 4
[12 marks]

A small bob of mass 0.20 kg is attached to a string of length 1.2 m. The bob moves in a horizontal circle so that the string makes a constant angle of 35° with the vertical, as shown in Fig. 3.1.

L θ bob r bob moving in a horizontal circle
Fig. 3.1

(a) State the two forces acting on the bob, and the direction of each. [2]

(b) The tension in the string is T. By resolving the forces on the bob, show that tan θ = v² / (rg), where v is the speed of the bob and r is the radius of its circular path. [3]

(c) For this motion:

(i) show that the radius r of the circular path is 0.69 m, [1]

(ii) calculate the speed v of the bob, [2]

(iii) calculate the time taken for one complete revolution. [2]

(d) The bob is then made to move faster. State and explain what happens to the angle θ. [2]

  • (a)Weight mg vertically downward; tension T along the string, directed up toward the support ✓✓
  • (b)Vertical: T cosθ = mg. Horizontal (centripetal): T sinθ = mv²/r. Dividing the two: tanθ = (v²/r)/g = v²/(rg) ✓✓✓
  • (c)(i)r = L sinθ = 1.2 × sin 35° = 0.69 m ✓
  • (c)(ii)v² = rg tanθ = 0.688 × 9.81 × tan 35° = 4.73; v = 2.2 m s⁻¹ ✓✓
  • (c)(iii)t = 2πr/v = 2π × 0.688 / 2.17 = 2.0 s ✓✓
  • (d)θ increases ✓; a greater speed needs a greater centripetal force, so the horizontal component T sinθ must increase, which requires a larger angle (the string rises toward the horizontal) ✓
04
Paper 4
[10 marks]

A flywheel rotates at 300 revolutions per minute; its rim is at a radius of 0.40 m.

(a) Define the radian. [1]

(b) Show that 300 revolutions per minute is an angular speed of about 31 rad s⁻¹. [2]

(c) Calculate the frequency and period of the rotation. [2]

(d) Calculate the linear speed of a point on the rim. [2]

(e) Calculate the linear speed of a point at radius 0.20 m. [1]

(f) State and explain how the angular speed of the inner point compares with that of the rim. [2]

  • (a)The angle subtended at the centre of a circle by an arc equal in length to the radius ✓
  • (b)ω = 300 × 2π / 60 ✓ = 31 rad s⁻¹ ✓
  • (c)f = 300/60 = 5.0 Hz ✓; T = 1/f = 0.20 s ✓
  • (d)v = rω = 0.40 × 31.4 ✓ = 13 m s⁻¹ ✓
  • (e)v = 0.20 × 31.4 = 6.3 m s⁻¹ ✓
  • (f)The angular speed is the same (it is one rigid body) ✓; the linear speed is proportional to radius, so the inner point is slower ✓
05
Paper 4
[10 marks]

A turntable completes one revolution every 1.8 s. Point P lies 12 cm from the centre and point Q lies 24 cm from the centre.

(a) Calculate the angular speed of the turntable. [2]

(b) Calculate the linear speed of P and of Q. [2]

(c) State and explain how the angular speeds of P and Q compare. [2]

(d) Calculate the angle, in radians, turned through in 0.50 s. [2]

(e) Calculate the centripetal acceleration of Q. [2]

  • (a)ω = 2π/T = 2π/1.8 ✓ = 3.5 rad s⁻¹ ✓
  • (b)vₚ = rω = 0.12 × 3.49 = 0.42 m s⁻¹ ✓; vₜ = 0.24 × 3.49 = 0.84 m s⁻¹ ✓
  • (c)Equal ✓: both complete a revolution in the same time, so they share the same angular speed ✓
  • (d)θ = ωt = 3.49 × 0.50 ✓ = 1.7 rad ✓
  • (e)a = rω² = 0.24 × 3.49² ✓ = 2.9 m s⁻² ✓
06
Paper 4
[11 marks]

A 0.50 kg mass on a string moves in a horizontal circle of radius 0.80 m on a frictionless table, at a constant speed of 3.0 m s⁻¹.

(a) State what provides the centripetal force. [1]

(b) State the direction of the centripetal force. [1]

(c) Calculate the centripetal acceleration. [2]

(d) Calculate the tension in the string. [2]

(e) The string will break when the tension reaches 20 N. Calculate the maximum speed. [3]

(f) State and explain the direction in which the mass moves immediately after the string breaks. [2]

  • (a)The tension in the string ✓
  • (b)Toward the centre of the circle ✓
  • (c)a = v²/r = 3.0²/0.80 ✓ = 11 m s⁻² ✓
  • (d)F = mv²/r = 0.50 × 11.25 ✓ = 5.6 N ✓
  • (e)v = √(Fr/m) = √(20 × 0.80 / 0.50) ✓✓ = 5.7 m s⁻¹ ✓
  • (f)Along a straight line, tangent to the circle ✓: with no resultant force it continues at constant velocity (Newton’s first law) ✓
07
Paper 4
[10 marks]

A car of mass 1200 kg rounds a flat (unbanked) bend of radius 50 m. The maximum frictional force available is 0.60 of the car’s weight.

(a) State what provides the centripetal force on the car. [1]

(b) Write the condition for the car not to skid. [2]

(c) Derive an expression for the maximum cornering speed. [2]

(d) Calculate the maximum speed. [2]

(e) State and explain what happens if the car exceeds this speed. [2]

(f) State one change that would raise the maximum cornering speed. [1]

  • (a)Friction between the tyres and the road ✓
  • (b)The friction must supply the centripetal force: μmg ≥ mv²/r ✓✓
  • (c)At the limit μmg = mv²/r, so v_max = √(μrg) ✓✓
  • (d)v = √(0.60 × 50 × 9.81) ✓ = 17 m s⁻¹ ✓
  • (e)The required centripetal force exceeds the maximum friction ✓, so the car skids outward and leaves the bend ✓
  • (f)A larger radius, a higher coefficient of friction, or banking the road ✓
08
Paper 4
[11 marks]

A bend is banked at an angle θ so that a car can round it at a design speed without relying on friction, as shown in Fig. 8.1.

θ N mg banked track, angle θ
Fig. 8.1

(a) Explain why banking allows a higher cornering speed than a flat road. [2]

(b) For a frictionless banked track, show that tanθ = v²/(rg). [3]

(c) A bend of radius 80 m is to be taken at 25 m s⁻¹. Calculate the banking angle. [3]

(d) State and explain what tends to happen if a car travels below the design speed on a frictionless bank. [2]

(e) State the direction of the centripetal force on the car. [1]

  • (a)The horizontal component of the normal contact force can supply the centripetal force ✓, so the car does not depend on friction, which can be exceeded ✓
  • (b)Horizontally N sinθ = mv²/r ✓; vertically N cosθ = mg ✓; dividing gives tanθ = v²/(rg) ✓
  • (c)tanθ = 25²/(80 × 9.81) = 625/784.8 = 0.796 ✓✓; θ = 39° ✓
  • (d)It tends to slide down the bank, toward the centre ✓, because the horizontal force component now exceeds the centripetal force needed ✓
  • (e)Horizontal, directed toward the centre of the circular path ✓
09
Paper 4
[10 marks]

A centrifuge spins a 5.0 g sample at a radius of 8.0 cm, rotating at 4000 revolutions per minute. This question links the angular quantities of Q4 and Q5 with the centripetal force ideas of Q6 to Q8.

(a) Calculate the angular speed in rad s⁻¹. [2]

(b) Calculate the linear speed of the sample. [1]

(c) Calculate the centripetal acceleration of the sample. [2]

(d) Express this acceleration as a multiple of g. [2]

(e) Calculate the centripetal force on the sample. [2]

(f) State the direction of this force. [1]

  • (a)ω = 4000 × 2π/60 ✓ = 4.2 × 10² rad s⁻¹ ✓
  • (b)v = rω = 0.080 × 418.9 = 34 m s⁻¹ ✓
  • (c)a = rω² = 0.080 × 418.9² ✓ = 1.4 × 10⁴ m s⁻² ✓
  • (d)a/g = 1.4 × 10⁴ / 9.81 ≈ 1.4 × 10³ g ✓✓
  • (e)F = ma = 0.0050 × 1.40 × 10⁴ ✓ = 70 N ✓
  • (f)Toward the axis of rotation (the centre) ✓
10
Paper 4
[11 marks]

In a fairground "rotor" ride, riders stand against the inside wall of a vertical cylinder of radius 3.0 m. The cylinder spins, the floor drops away, and friction holds the riders up, as shown in Fig. 10.1. The coefficient of friction between clothing and wall is 0.40.

wall wall N (centripetal) friction mg rotating cylinder (top spins)
Fig. 10.1

(a) State what provides the centripetal force on a rider. [1]

(b) State the direction of the centripetal force. [1]

(c) Write the condition for a rider not to slip down the wall. [2]

(d) Show that the minimum angular speed is ω = √(g/(μr)). [3]

(e) Calculate the minimum angular speed and the linear speed of a rider at the wall. [3]

(f) State why the rider’s mass does not affect the minimum speed. [1]

  • (a)The normal (contact) force from the wall ✓
  • (b)Horizontally, toward the axis of the cylinder ✓
  • (c)The friction must at least support the weight: μN ≥ mg, with N = mrω² ✓✓
  • (d)μ(mrω²) ≥ mg ✓; ω² ≥ g/(μr) ✓; so ω_min = √(g/(μr)) ✓
  • (e)ω = √(9.81/(0.40 × 3.0)) = √8.18 = 2.9 rad s⁻¹ ✓✓; v = rω = 3.0 × 2.86 = 8.6 m s⁻¹ ✓
  • (f)The mass cancels: both the friction needed and the weight are proportional to m ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.