Uniform circular motion: stopper, coin on a disk and conical pendulum AP Physics 1 · Lesson 5 of 7
Moving in a circle at constant speed means accelerating toward the centre. This lesson explains why, shows you how to find the real force that provides that inward pull, and works through the three set-ups your class uses: the stopper on a string, the coin on a turntable and the conical pendulum.
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In this lesson
This lesson covers an introduction to uniform circular motion, the stopper-on-a-string derivation, the coin on a disk and conical pendulums:
- Introduction to uniform circular motion
- Derivation: stopper on a string
- Coin on a disk
- Conical pendulums
All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.
Uniform circular motion: centripetal acceleration
An object in uniform circular motion moves around a circle at constant speed. Its velocity is not constant, because the direction keeps changing, so it is accelerating. That acceleration, the centripetal acceleration, always points to the centre of the circle:
ac = v²/r v = 2πr/T f = 1/T
T is the period (the time for one revolution, in s) and f is the frequency (revolutions per second, in Hz).
By Newton's second law, an inward acceleration needs an inward net force: ΣFtoward centre = mv²/r. This net force is sometimes called the “centripetal force”, but it is not a new force. It is provided by a real force you can name: tension in a string, friction from the road, gravity, the normal force, or a combination.
Common trap
Never draw “Fc” or “centripetal force” on a free-body diagram. Draw the real forces (tension, gravity, normal, friction), then set the sum of their components toward the centre equal to mv²/r. AP scoring guides take points off for an extra centripetal force arrow.
Method for any circular motion problem
- Draw the free-body diagram with real forces only.
- Find the centre of the circle. Take “toward the centre” as positive.
- Write ΣFin = mv²/r (and ΣF = 0 in the direction perpendicular to the plane of the circle, if it is horizontal).
- Solve, then sense-check: a faster speed should need a bigger inward force.
Worked example 1: ball on a string
A 0.20 kg ball on a 0.80 m string is whirled in a horizontal circle on a frictionless table, making 2.0 revolutions per second. Find its speed, its acceleration and the tension.
- T = 1/f = 0.50 s, so v = 2πr/T = 2π(0.80)/0.50 = 10.1 m/s.
- ac = v²/r = 10.1²/0.80 = 126 m/s² toward the centre.
- On the table, the tension is the only horizontal force: FT = mac = 0.20 × 126 = 25 N.
If the speed is changing as well, there is also a tangential acceleration along the velocity. The total acceleration is the vector sum of the tangential and centripetal parts, so it no longer points exactly at the centre.
Key idea
If the string breaks, the ball flies off along the tangent in a straight line (Newton's first law), not outward along the radius. There is no outward force; the ball simply stops being pulled inward.
Derivation: stopper on a string
The stopper-on-a-string lab is the classic way to test ac = v²/r. A rubber stopper of mass m is whirled in a horizontal circle of radius r above a tube; the string runs down through the tube to a hanging mass M. Swing the stopper at just the speed that keeps the hanging mass still.
The derivation
- The hanging mass is in equilibrium, so the tension is FT = Mg.
- Treat the string to the stopper as horizontal (good if the stopper moves fast). The tension is the net inward force on the stopper: FT = mv²/r.
- So Mg = mv²/r, which gives v² = Mgr/m.
- Using v = 2πr/T: Mg = m(4π²r)/T², so T² = 4π²mr/(Mg).
This is what makes it a good AP lab. Keep m and r fixed and change M: a graph of v² against M (or 1/T² against M) should be a straight line through the origin with gradient gr/m. Keep M and m fixed and change r: a graph of T² against r should be straight with gradient 4π²m/(Mg).
Worked example: the stopper
A 0.025 kg stopper is whirled on a 0.60 m radius while a 0.10 kg mass hangs at rest below the tube. Find the stopper's speed and the period.
- FT = Mg = 0.10 × 9.8 = 0.98 N.
- v = √(FTr/m) = √(0.98 × 0.60/0.025) = √23.5 = 4.85 m/s.
- Period: T = 2πr/v = 2π(0.60)/4.85 = 0.78 s.
Lab tip
In reality the string to the stopper sags slightly below horizontal, so only the horizontal part of the tension points to the centre, and friction in the tube affects the tension. These are the sources of error an AP LAB question expects you to name.
Coin on a disk (turntable)
Put a coin on a spinning turntable. The coin goes round in a circle, so something must pull it inward. The normal force and gravity are vertical. The only horizontal force is static friction, and it points toward the centre.
Vertical: FN = mg Inward: Ff = mv²/r ≤ μsmg
The coin stays on as long as the friction it needs is no more than the maximum static friction. At the point of slipping, mv²/r = μsmg, so the mass cancels and
vmax = √(μsgr)
For a turntable spinning with period T, every coin has the same period, but coins further out move faster (v = 2πr/T), so they need more friction. The coin furthest from the centre slips first.
Worked example 1: will it stay?
A coin sits 0.12 m from the centre of a turntable that turns at 45 revolutions per minute. μs = 0.30. Does the coin slip?
- f = 45/60 = 0.75 Hz, so T = 1.33 s and v = 2π(0.12)/1.33 = 0.565 m/s.
- Friction needed per kg: v²/r = 0.565²/0.12 = 2.66 m/s². Available: μsg = 0.30 × 9.8 = 2.94 m/s².
- 2.66 < 2.94, so the coin stays on, just.
Worked example 2: furthest position
On the same turntable (45 rpm, μs = 0.30), how far from the centre can the coin be placed without slipping?
- μsg = v²/r = (2πr/T)²/r = 4π²r/T².
- r = μsgT²/(4π²) = 0.30 × 9.8 × 1.33²/39.5 = 0.13 m.
Key idea
The same physics applies to a car turning on a flat road: static friction between the tyres and road provides the inward force, and vmax = √(μsgr). On ice, μs is small, so the safe speed is much lower.
Conical pendulums
A conical pendulum is a mass on a string that swings in a horizontal circle, so the string traces out a cone. The string makes angle θ with the vertical, and the radius of the circle is r = L sinθ.
Only two forces act on the bob: tension and gravity. Split the tension into vertical and horizontal (inward) parts:
Vertical: FT cosθ = mg Inward: FT sinθ = mv²/r
Divide the second equation by the first: tanθ = v²/(rg), so v² = gr tanθ. The mass cancels. Faster swinging needs a bigger angle, and the tension always exceeds mg because only its vertical part holds the bob up.
Worked example 1: conical pendulum
A 0.50 kg bob on a 1.2 m string swings in a horizontal circle with the string at 30° to the vertical. Find the tension, the speed and the period.
- Tension: FT = mg/cosθ = 4.9/0.866 = 5.66 N.
- r = L sin 30° = 0.60 m; v = √(gr tanθ) = √(9.8 × 0.60 × 0.577) = 1.84 m/s.
- Period: T = 2πr/v = 2π(0.60)/1.84 = 2.05 s.
Worked example 2: what angle?
A child on a swing ride (chains 5.0 m long) moves in a circle at 6.0 m/s. The ride's centre post and chain attachments are small enough to ignore. Find the angle of the chains.
- v² = gr tanθ with r = L sinθ gives v² = gL sinθ tanθ.
- 36 = 49 sinθ tanθ, so sinθ tanθ = 0.735. Trying values: θ = 45° gives 0.707 and θ = 46° gives 0.745.
- θ ≈ 46° from the vertical.
Key equations
| Centripetal acceleration | ac = v²/r |
|---|---|
| Speed, period and frequency | v = 2πr/T, f = 1/T |
| Newton's second law, toward the centre | ΣFin = mv²/r |
| Stopper on a string | Mg = mv²/r |
| Coin on a disk, about to slip | vmax = √(μsgr) |
| Conical pendulum | FT cosθ = mg, v² = gr tanθ |
Practice problems
17 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.
Hint
Constant speed in a circle still means an acceleration toward the centre.
Worked solution
- ac = v²/r = 144/30 = 4.80 m/s², toward the centre.
Hint
Velocity is a vector: is its direction constant?
Worked solution
- The direction of the velocity changes continuously, so there is an acceleration.
- For constant speed it is entirely centripetal: toward the centre, at 90° to v. The answer is C.
Hint
One revolution covers a distance of 2πr in one period.
Worked solution
- v = 2πr/T = 2π(0.50)/0.40 = 7.85 m/s.
Hint
The tension is the only horizontal force, so it is the net inward force.
Worked solution
- FT = mv²/r = 0.30 × 16/0.75 = 6.40 N.
Hint
Once the string breaks, what horizontal forces act?
Worked solution
- No horizontal force acts, so the ball keeps its velocity (Newton's first law).
- That velocity was along the tangent. The answer is B.
Hint
Is “centripetal force” a separate force, or the role played by a real force?
Worked solution
- Friction is the real force providing the inward net force.
- Adding a separate centripetal force counts it twice. The answer is C.
Hint
Tension = Mg, and tension provides mv²/r for the stopper.
Worked solution
- Mg = mv²/r, so v = √(Mgr/m).
- v = √(0.98 × 0.60/0.025) = 4.85 m/s.
Hint
Mg = mv²/r, so v² ∝ M.
Worked solution
- Doubling M doubles v².
- So v increases by a factor of √2. The answer is A.
Hint
Start from Mg = m(4π²r)/T².
Worked solution
- Rearranging: T² = (4π²m/(Mg))r, so T² is proportional to r.
- A graph of T² against r is a straight line through the origin. The answer is B.
Hint
At the limit, maximum static friction provides mv²/r.
Worked solution
- μsmg = mv²/r, so v = √(μsgr) = √(0.40 × 9.8 × 0.15) = 0.767 m/s.
Hint
Both have the same period. Compare the inward acceleration they need: a = 4π²r/T².
Worked solution
- Same period, so the needed acceleration is proportional to r.
- The outer coin needs twice the friction, so it reaches the limit μsg first. The answer is B.
Hint
Find v from the period, then set μsg = v²/r.
Worked solution
- v = 2π(0.10)/0.80 = 0.785 m/s.
- μs = v²/(gr) = 0.617/(9.8 × 0.10) = 0.629.
Hint
Only the vertical part of the tension balances the weight.
Worked solution
- FT cos 30° = mg, so FT = 4.9/0.866 = 5.66 N.
Hint
v² = gr tanθ, with r = L sinθ.
Worked solution
- r = 1.2 sin 30° = 0.600 m.
- v = √(9.8 × 0.600 × tan 30°) = 1.84 m/s.
Hint
FT cosθ = mg, and tanθ = v²/(rg).
Worked solution
- A faster bob needs a bigger inward force, so the string tilts further: θ increases.
- FT = mg/cosθ grows as cosθ shrinks. The answer is A.
Hint
Find v from v² = gr tanθ, then T = 2πr/v. (Or show T = 2π√(L cosθ/g).)
Worked solution
- r = 0.80 sin 40° = 0.514 m; v = √(9.8 × 0.514 × tan 40°) = 2.06 m/s.
- T = 2πr/v = 1.57 s. This matches 2π√(L cosθ/g).
Hint
FT = mv²/r.
Worked solution
- With v fixed, FT ∝ 1/r.
- Halving r doubles the tension. The answer is C.
Free-response practice
Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.
Mathematical Routines (MR) style · 8 points · about 12 minutes
A small coin of mass m is placed on a flat turntable at distance r from its centre. The coefficient of static friction between the coin and the turntable is μs. The turntable starts from rest and its rate of rotation is increased very slowly.
- (a) Draw a free-body diagram for the coin while it rotates with the turntable at constant speed.
- (b) Derive an expression for the period T of the turntable at which the coin starts to slip, in terms of r, μs and physical constants.
- (c) A second, identical coin is placed at 2r. Using your expression, explain which coin slips first as the turntable speeds up.
- (d) After the outer coin starts to slip, describe its path as seen from above, relative to the ground.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Dot with three forces: weight mg down, normal force up (same length), static friction horizontally toward the centre of the turntable. No separate centripetal force.
Part (b)
At slipping, μsmg = mv²/r with v = 2πr/T. So μsg = 4π²r/T² and T = 2π√(r/(μsg)).
Part (c)
From (b), the period at which slipping starts is proportional to √r. The coin at 2r slips at a longer period, √2 times longer, which is reached first as the turntable speeds up (the period gets shorter as it speeds up). So the outer coin slips first.
Part (d)
Once it slips, the friction is kinetic and too small to keep it in the circle, so it moves along a path that curves less than the circle, drifting outward along roughly the tangent direction, not straight out along the radius.
Your mark: 0 of 8
Qualitative/Quantitative Translation (QQT) style · 7 points · about 13 minutes
A small ball of mass m hangs from a light string of length L. It moves in a horizontal circle at constant speed, so the string makes a constant angle θ with the vertical.
- (a) A student claims that a heavier ball, on the same string and at the same angle, would need to move faster. Without equations, explain whether the student is correct.
- (b) Derive an expression for the speed of the ball in terms of L, θ and g.
- (c) Explain how your expression supports or contradicts your answer to (a).
- (d) Calculate the speed and the tension for m = 0.2 kg, L = 1.0 m and θ = 35°.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
Incorrect. A heavier ball has a larger weight, so the tension's vertical part must be larger, and at the same angle the horizontal part (the inward force) grows in the same proportion. The inward force needed, mv²/r, also grows in proportion to m at the same speed. Both scale with mass, so the speed is the same.
Part (b)
FT cosθ = mg and FT sinθ = mv²/r. Dividing, tanθ = v²/(rg). With r = L sinθ, v = √(gL sinθ tanθ).
Part (c)
The expression contains no m, so the speed does not depend on the mass, which supports (a).
Part (d)
r = 1.0 sin 35° = 0.574 m; v = √(9.8 × 0.574 × tan 35°) = 1.98 m/s. FT = mg/cosθ = 1.96/0.819 = 2.39 N.
Your mark: 0 of 7
Common mistakes
- Drawing a separate “centripetal force” on the free-body diagram. The inward net force comes from real forces.
- Drawing an outward “centrifugal force”. In AP Physics 1 (an inertial frame), there is no outward force; the feeling of being thrown outward is your inertia.
- Saying an object in uniform circular motion is not accelerating because its speed is constant.
- Using the string length L as the radius of a conical pendulum. The radius is L sinθ.
- Thinking an object flies radially outward when released. It moves along the tangent.
- Mixing up period (seconds per revolution) and frequency (revolutions per second), or forgetting to convert rpm to revolutions per second.
Frequently asked questions
Is centripetal force a real force?
It is a name for a role, not a separate force. Whatever real force (or combination) points toward the centre, such as tension, friction, gravity or the normal force, is providing the centripetal force. That is why it never appears as an extra arrow on a free-body diagram.
What about centrifugal force?
In the frame of the ground (which AP Physics 1 uses), there is no centrifugal force. When a car turns and you slide toward the door, nothing pushes you outward; your body tends to continue in a straight line and the car turns under you. Then the door pushes you inward.
Why does the mass cancel for a coin on a turntable and a conical pendulum?
The force that supplies the inward pull (friction, or the horizontal part of the tension) is proportional to the weight, and the inward force needed, mv²/r, is proportional to the mass. Both scale with m, so m cancels.
Does a satellite in orbit need a force to keep moving forward?
No. It needs a force to keep turning, and gravity provides it, pulling toward Earth's centre. Its forward motion continues by inertia. This is Lesson 6.
Do I need calculus or radians for AP Physics 1 circular motion?
No calculus. Angles are usually in degrees for forces. For rotation (Unit 5) you will use angular speed ω in radians per second, where v = rω, but the problems in this lesson only need v = 2πr/T.
Can an object move in a circle with changing speed?
Yes. Then it has a tangential acceleration along the velocity as well as the centripetal acceleration toward the centre. The total acceleration is their vector sum and points somewhere between the two.
Why is the tension in a conical pendulum always more than mg?
Only the vertical part, FT cosθ, holds up the weight. Since cosθ is less than 1 for any angle, FT must be bigger than mg.
Keep going
Related guides and practice:
- Centripetal force explained
- AP Physics 1 forces summary
- How to draw free-body diagrams
- AP Physics 1 forces practice problems
All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers