Chapter 1 · Mechanics
1-10 Uniform Circular Motion and Centripetal Force
❓ Lesson question
A ball is whirled in a circle on the end of a string. If the string is cut, in which direction does the ball move?
You will learn to
- 1Calculate the period (T), frequency (f), angular velocity (ω) and linear speed (v) of an object in uniform circular motion.
- 2Determine the centripetal acceleration (a_c) of the object and identify the force that provides it.
- 3Explain why no outward (centrifugal) force acts on an object moving in a circle.
Whirl a ball on a string in a horizontal circle above your head. The string stays taut, and your hand feels a constant inward pull for as long as the ball keeps moving in the circle. Now suppose the string is cut at the instant the ball is at point X. With no force from the string, the ball is free. Predict the path the ball follows after point X.
✍️ Predict before you continue
Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.
(a) When the string is cut at X, does the ball move (i) straight outward along the line of the string, (ii) along the tangent to the circle, or (iii) along a gradually curving path?
(b) While the ball moves around the circle at constant speed, is there any outward force acting on it? Explain your answer.
🔬 Explore — Testing the direction of release
What you’ll do: You will release an object moving in a circle and find the direction it then travels.
- 1Put on safety goggles and check that the space around you is completely clear. If the floor grid is not already marked, chalk a circle with its centre and the 12, 3, 6 and 9 o’clock points.
- 2Loop the free end of the string over one finger and whirl the bung gently in a horizontal circle above the grid. Feel how the string tugs your finger, and note the direction of the tug.
- 3Keeping the circle steady and slow, choose a release point on the grid (for example, the 3 o’clock mark). The instant the bung passes that point, open your finger to let the string go — releasing only toward clear space.
- 4Watch which way the bung first travels: straight outward along the line from the centre, or off along the tangent (at right angles to the string). Record the release point and the flight direction.
- 5Bring the bung to rest, then repeat for five different release points around the circle, keeping the speed roughly the same each time.
- 6Compare your five runs and decide whether the flight direction follows a consistent pattern.
Analyse what you collected. (a) Across your five releases, what is the pattern — does the bung move outward from the centre, or off along the tangent? (b) While it circled, what did your finger experience — a push outward or a pull inward — and which object pulled which?
In Explore, the string pulled the bung steadily inward, toward the centre of the circle — never outward. When the bung was released, the tension disappeared, and the bung moved in a straight line tangent to the circle at the point of release. You might have expected the bung to move outward along the radius; your observations show otherwise.
🎛️ Simulation — circular motion
Cut the string at several different release points before reading on. The green path is never the same direction twice — but it is always the direction the ball was already travelling, at right angles to the string. The grey path is the one most people expect, and nothing produces it: no force pushes the ball outward while it circles, so removing the inward tension leaves the ball with nothing but the velocity it had. There is no centrifugal force on the ball. The outward pull you feel in your hand is the force the string exerts on you, which is a different body.
When an object moves in a circle at a constant speed, the motion is called uniform circular motion. The angular velocity ω is the angular displacement per unit time, in radians per second. If the object turns through an angle θ in a time t:
The period T is the time for one revolution, and the frequency f is the number of revolutions per second (in Hz):
For a circle of radius r, the arc length l covered as the object turns through an angle θ is l=rθ, so the linear speed is:
Although the speed is constant, the velocity continuously changes direction, so the object is accelerating. In a short time interval Δt the velocity vector keeps its magnitude v but rotates through the same small angle Δθ=ωΔt. The resulting change in velocity has magnitude Δv=vΔθ and is directed toward the centre. Dividing by Δt gives the centripetal acceleration:
By Newton's second law, this acceleration requires a resultant force in the same direction — the centripetal force.
Point
- • Motion in a circle at constant speed is uniform circular motion. The angle turned per unit time is the angular velocity: ω=θ/t in rad/s.
- • The time for one revolution is the period T=2π/ω; the revolutions per unit time is the frequency f=1/T. On a circle of radius r the distance moved is l=rθ, sov=rω.
- • The direction of motion changes even when the magnitude of the velocity does not, so the object has an acceleration a=rω2=v2/r always directed toward the centre. The resultant force that produces it is the centripetal force,F=mv2/r — which is the equation F=ma written toward the centre.
📐 Equation box
Angular velocity ω
Centripetal acceleration a
Centripetal force F
Period and frequency
🛠️ Worked example (book p.59)
Situation
A 0.50 kg ball is attached to a spring and moves in uniform circular motion on a smooth horizontal surface. The spring is stretched from its natural length, and its elastic force provides the centripetal force required for the circular motion.
Answer. (a) T=5.0/20=0.25 s. (b) f=1/T=4.0 Hz. (c) ω=2π/T=2π/0.25=25.13 rad/s. (d) v=ωr=25.13×0.20=5.03 m/s.
Note the order: everything follows from the period, so find it first and the rest are one step each.
In a New Context
The washing-machine spin cycle. During the spin cycle, the wet clothes move with the rotating drum in a circular path. The normal force exerted by the inner surface of the drum provides most of the inward resultant force required for this motion. At a perforation in the drum, there is no solid wall to exert the required inward contact force on the water. The water therefore continues with its instantaneous tangential velocity. Because the drum curves away from this straight-line path, the water passes through the perforation and is collected by the outer tub.
Your turn: Why does water pass through the holes in the spinning drum while the clothes remain pressed against the inner surface?
Answer. Because the clothes have a wall in front of them and the water, at a hole, does not. Circular motion needs a continuous inward force; the drum supplies it to the clothes through the normal contact force, so they keep turning. Where a perforation interrupts the wall, nothing can push inward on the water there, so it simply keeps going in the direction it was already moving — along the tangent — and the drum curves away from under it. Note the phrasing that matters: the water is not flung out. Nothing pushes it outward. It goes straight while the drum turns, which is the same thing that happens to the bung when the string is cut.
Because work needs a component of force along the displacement, and there is none. The centripetal force points at the centre; the displacement at every instant is along the tangent, at right angles to it. A force perpendicular to the motion changes the direction of the velocity and never its magnitude — which is exactly why the speed can be constant while the object is accelerating.
🧪 Nature of Science — Theories — a relationship before a broader framework
In 1659, Christiaan Huygens derived a relationship for uniform circular motion. Expressed in modern notation, his result is equivalent in magnitude toFc=mv2/r. Huygens obtained this relationship geometrically, before Newton's laws provided a broader framework linking resultant force to acceleration. Newton's theory of universal gravitation later showed that gravity provides the inward force that keeps planets in orbit.
An answer to "your turn". It could calculate the size of the inward force needed to hold a given mass on a given circle at a given speed — a complete, usable, correct number. What it could not answer is what supplies that force in any particular case, or why the same expression should apply to a stone on a string and to the Moon. That took Newton: once resultant force is tied to acceleration in general, and gravity is identified as a force that acts between any two masses, Huygens's formula stops being a rule about whirling objects and becomes one instance of a law about everything. A relationship tells you how much; a framework tells you why, and where else to look.
Your turn: Give one quantity that Huygens's relationship could calculate, and one question it could not answer.
✅ Lesson question, answered
A ball is whirled in a circle on the end of a string. If the string is cut, in which direction does the ball move?
It moves along the tangent to the circle at the release point — the direction it was already moving — not outward, away from the centre. While it moved in the circle, the string's inward tension provided the centripetal force that kept the object on its circular path.
In an inertial frame, no outward (centrifugal) force acts on the object; the outward pull felt by your hand is the force exerted by the string on your hand. When the string is cut, the inward tension disappears, so no inward force remains to maintain the circular motion, and the object continues in a straight line.
- uniform circular motion
- — Motion in a circle at constant speed.
- angular velocity ω
- — The angle turned per unit time, in rad/s.
- period T
- — The time for one revolution.
- frequency f
- — The number of revolutions per second, f = 1/T.
- centripetal acceleration
- — The acceleration directed toward the centre, a = v²/r = ω²r.
- centripetal force
- — The resultant force toward the centre, F = mv²/r — not an extra force.
How the real paper is built
The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.
One step. You are given the formula; put the numbers in.
Two steps, or you must choose the right component before you can start.
Multi-step, or a case the textbook does not set and examiners do.
📝 Exam-style question
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