A 0.50\ kg stone is whirled on a 0.90\ m string at 4.0\ m s^-1 in a horizontal circle. Find the centripetal force. [2 marks]

SEAB Original-style practice: A car of mass 1200\ kg rounds a flat circular bend of radius 50\ m at 15\ m s^-1. (a) Calculate the centripetal force required. (b) State what provides this force and the minimum coefficient of friction needed.

(a) Centripetal force: F = mv^2r = 1200 × 15^250 = 1200 × 22550 = 5400\ N. (b) Friction between the tyres and the road provides this force. The maximum available friction is μ mg, so the minimum coefficient is μ = Fmg = 54001200 × 9.81 = 0.46. Markers reward the centripetal force from mv^2/r, identifying friction as the source, and the friction condition giving a minimum coefficient.

SingaporePhysicsSyllabus dot point

Why does a body moving in a circle at constant speed require a centripetal force, and how large must that force be?

Describe uniform circular motion using angular velocity, relate it to centripetal acceleration and force, and apply these to horizontal and vertical circular motion

A focused answer to the H2 Physics learning outcome on circular motion. Angular velocity, centripetal acceleration and force, and applications including a conical pendulum, banked tracks and vertical circles.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

SEAB wants you to describe uniform circular motion using angular velocity, to relate it to the centripetal acceleration and the centripetal force that must act toward the centre, and to apply these to situations such as a ball on a string, a banked or flat track, and a vertical circle. The central insight is that a body moving in a circle is always accelerating, even at constant speed, because its direction is changing.

The answer

Angular velocity

For an object moving in a circle of radius $r$ , the angular velocity is the rate at which the angle is swept:

\omega = \frac{\Delta\theta}{\Delta t} = 2\pi f = \frac{2\pi}{T}

where $f$ is the frequency and $T$ the period. The linear speed relates to it by:

v = r\omega

Centripetal acceleration

Even at constant speed, the velocity vector changes direction, so the body accelerates toward the centre. The centripetal acceleration is:

a = \frac{v^2}{r} = r\omega^2

directed along the radius toward the centre.

Centripetal force

By Newton's second law, this acceleration requires a resultant force toward the centre:

F = \frac{mv^2}{r} = mr\omega^2

The centripetal force is not a new kind of force. It is the name for whatever real force (tension, gravity, friction, the normal force, or a combination) points toward the centre and supplies the required value. A common error is to add a separate "centripetal force" to a free-body diagram; instead, identify which real force plays that role.

Why the centripetal force does no work

The centripetal force is always perpendicular to the velocity, so by $W = Fs\cos 90^\circ$ it does no work. This is why a body in uniform circular motion keeps constant speed and constant kinetic energy.

Vertical circles

In a vertical circle the speed changes because gravity has a component along the motion. At the top of a vertical loop, gravity and the normal (or tension) both point down toward the centre:

\frac{mv_{\text{top}}^2}{r} = mg + N

The minimum speed to maintain contact at the top is when $N = 0$ , giving $v_{\text{min}} = \sqrt{gr}$ .

Worked example

A conical pendulum has a $0.25\ \text{kg}$ bob on a $1.2\ \text{m}$ string that sweeps a horizontal circle while the string makes $30^\circ$ with the vertical. Find the speed of the bob. Take $g = 9.81\ \text{m s}^{-2}$ .

Step 1: Identify the forces and resolve

The bob has weight $mg$ down and string tension $T$ along the string. Vertically, the tension's vertical component balances the weight: $T\cos 30^\circ = mg$ .

Step 2: Write the horizontal (centripetal) equation

The tension's horizontal component provides the centripetal force: $T\sin 30^\circ = \dfrac{mv^2}{r}$ .

Step 3: Find the radius and eliminate the tension

Radius of the circle: $r = L\sin 30^\circ = 1.2 \times 0.5 = 0.60\ \text{m}$ .

Dividing the two equations: $\tan 30^\circ = \dfrac{v^2}{rg}$ , so $v^2 = rg\tan 30^\circ = 0.60 \times 9.81 \times 0.577 = 3.40$ .

Step 4: Solve for the speed

v = \sqrt{3.40} = 1.84\ \text{m s}^{-1}

The bob moves at about $1.8\ \text{m s}^{-1}$ , a result independent of the bob's mass.

Examples in context

Example 1. Banked tracks. On a frictionless banked curve, the horizontal component of the normal force supplies the centripetal force. Setting $N\sin\theta = \dfrac{mv^2}{r}$ and $N\cos\theta = mg$ gives the ideal speed $v = \sqrt{rg\tan\theta}$ for which no friction is needed. This is why race tracks and motorway bends are banked.

Example 2. A satellite in orbit. For a satellite, gravity is the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ . This single equation links orbital speed to radius and is the bridge from circular motion to the gravitational fields section.

Try this

Q1. Explain why an object moving at constant speed in a circle is accelerating. [2 marks]

Cue. Its velocity is a vector; the direction changes continuously, so the velocity changes, which is an acceleration directed toward the centre.

Q2. A $0.50\ \text{kg}$ stone is whirled on a $0.90\ \text{m}$ string at $4.0\ \text{m s}^{-1}$ in a horizontal circle. Find the centripetal force. [2 marks]

Cue. $F = \dfrac{mv^2}{r} = \dfrac{0.50 \times 4.0^2}{0.90} = 8.9\ \text{N}$ .

Q3. For a ball on a string swung in a vertical circle, find the minimum speed at the top to keep the string taut, in terms of $g$ and $r$ . [3 marks]

Cue. At minimum speed tension is zero, so gravity alone provides the centripetal force: $mg = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{gr}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA

0.30\ \text{kg}

ball on a string of length

0.80\ \text{m}

is whirled in a horizontal circle, completing

2.0

revolutions per second. (a) Find the angular velocity. (b) Find the speed of the ball. (c) Find the tension in the string (assume the string is horizontal).

Show worked answer →

(a) Angular velocity: $\omega = 2\pi f = 2\pi \times 2.0 = 12.6\ \text{rad s}^{-1}$ .

(b) Speed: $v = r\omega = 0.80 \times 12.6 = 10.1\ \text{m s}^{-1}$ .

(c) The tension provides the centripetal force: $T = m r \omega^2 = 0.30 \times 0.80 \times (12.6)^2 = 0.30 \times 0.80 \times 158 = 38\ \text{N}$ .

Markers reward $\omega = 2\pi f$ , the link $v = r\omega$ , and the tension as the centripetal force $mr\omega^2$ (or equivalently $mv^2/r$ ).

Original4 marksA car of mass

1200\ \text{kg}

rounds a flat circular bend of radius

50\ \text{m}

15\ \text{m s}^{-1}

. (a) Calculate the centripetal force required. (b) State what provides this force and the minimum coefficient of friction needed.