- Convert between v, ω, T and f and apply the rule “same v on a belt, same ω on an axle”.
- Find the centripetal acceleration and name the real force that supplies the centripetal force.
- Solve horizontal-circle problems: flat and banked turns and the conical pendulum.
- Find the minimum speed at the top of a vertical circle for a string and for a rod, and the forces at the top and the bottom.
A washing-machine drum spins at 1200 revolutions per minute, a car goes round a roundabout, and a child swings a bucket of water over her head without spilling a drop. All three are circular motion, and one formula — F = mv²/r — explains them. In the CSCA syllabus this topic is the first half of the line “Circular motion and universal gravitation”: here you learn the kinematics and dynamics of turning, while planets and satellites are in the lesson “Universal gravitation and satellites”. From “Newton's laws of motion and their applications” you only need F = ma — except that the acceleration now points to the centre of the circle.
Describing circular motion: v, ω, T, f; belts and shared axles
Motion along a circle with a constant speed. The velocity vector is always tangent to the circle and keeps changing direction, so the velocity is not constant and the body does accelerate (Chinese textbooks: 匀速圆周运动).
Angular velocity ω — the angle (in radians) the radius turns through per second, rad/s. Period T — the time of one full revolution, s. Frequency f (rotational speed n) — revolutions per second, Hz = s⁻¹. Revolutions per minute (rpm) are first divided by 60.
- vlinear speed, m/s (along the tangent)
- ωangular velocity, rad/s
- Tperiod, s
- ffrequency, Hz (revolutions per second)
- rradius (distance of the point from the axis), m
In one revolution the point travels 2πr and the radius turns through 2π rad, both in time T. That gives v = ωr.
1) A fan blade makes 300 revolutions per minute. Find f, T and ω.
2) The tip of the blade is 0.4 m from the axis. What is its linear speed?
3) The minute hand of a clock is 1.5 times longer than the hour hand. What is the ratio of the speeds of their tips?
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2) v = ωr = 10π · 0.4 = 4π m/s ≈ 12.6 m/s.
3) The minute hand has T = 1 h, the hour hand T = 12 h, so their ω are in the ratio 12 : 1. With v = ωr: v₁ : v₂ = 12 · 1.5 : 1 = 18 : 1.
When two wheels are linked by a belt or a chain (or two gears touch without slipping), every point of the belt moves at the same speed, so the rims have the same linear speed v. All points of one rigid body turning about one axle (a disc, a wheel and its small sprocket) have the same angular velocity ω.
- r₁, r₂radii of the wheels or distances of the points from the axis; for gears the radius is proportional to the number of teeth
On a belt the smaller wheel spins faster; on one axle the point farther from the axis moves faster.
1) A bicycle has 48 teeth on the pedal chainring and 16 on the rear sprocket. Aysel pedals at 1 revolution per second, and the rear wheel has a radius of 0.35 m. How fast does the bicycle move?
2) Gear A with 20 teeth meshes with gear B with 60 teeth. A makes 90 revolutions per minute. How many does B make?
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2) Gears: same v, f₁r₁ = f₂r₂ ⇒ fB = 90 · 20/60 = 30 rpm. The bigger gear turns more slowly.
Wheels A and B are linked by a belt that does not slip, and rB = 2r_A. Point C is fixed on wheel B at a distance rA from its axis. What is the ratio of the centripetal accelerations aA : aB : aC of the rim points of A and B and of point C?
A) 1 : 2 : 1 B) 4 : 2 : 1 C) 2 : 1 : 1 D) 2 : 2 : 1
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D is the ratio of the linear speeds and C the ratio of the angular velocities. A comes from using a = ω²r with the same ω for all three points — but wheel A spins faster.
Centripetal acceleration and centripetal force
Even though the speed stays the same, the direction of the velocity changes. Over a short time Δt the change of the velocity vector, Δv, points to the centre of the circle, so the acceleration points to the centre too — it is called the centripetal acceleration. By Newton’s second law, the resultant of the forces toward the centre produces it.
- acentripetal acceleration, m/s², always toward the centre
- Fcentripetal force — the resultant force toward the centre, N
- mmass, kg
The directions of a and F change every instant, so uniform circular motion is not uniformly accelerated motion. The force is perpendicular to the velocity, so it does no work and the kinetic energy stays constant.
向心力)Not a new kind of force but a role: tension, friction, gravity, the normal force or their resultant plays it. Never draw it as an extra arrow in a force diagram — draw the real forces, then set their sum toward the centre equal to mv²/r.
1) On a smooth horizontal table a 0.2 kg puck is tied to a pin by a 0.5 m string and makes 2 revolutions per second. Find the acceleration and the tension.
2) A coin lies 0.1 m from the centre of a turntable; the coefficient of static friction is μ = 0.4 (g = 10 m/s²). What is the largest angular velocity at which the coin does not slide?
3) The speed doubles at the same radius; the radius doubles at the same ω; the radius doubles at the same v. How does the acceleration change each time?
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2) The force toward the centre is static friction, and it cannot exceed μmg: μmg = mω²r ⇒ ω² = μg/r = 4/0.1 = 40 ⇒ ω = 2√10 rad/s ≈ 6.3 rad/s. The mass cancels.
3) a = v²/r ⇒ 4 times larger; a = ω²r ⇒ 2 times larger; a = v²/r ⇒ 2 times smaller.
If the available force is smaller than the required mv²/r, the body leaves the circle and moves away from the centre — this is centrifugal motion (离心运动). A spin dryer works this way: the force holding a water drop on the clothes is not enough to provide mω²r, so the drop flies off along the tangent. No “centrifugal force” pushes it out.
Horizontal circles: turns, banked roads, the conical pendulum
- 1Draw the real forces
Gravity mg, the normal force or tension, friction — nothing else.
- 2Find the centre of the circle
The centre lies in the plane of the motion (not at the suspension point!); the radius is the horizontal distance from the body to the axis.
- 3Vertical direction: ΣF = 0
The body does not move up or down, so the vertical forces balance.
- 4Toward the centre: ΣF = mv²/r
Add the horizontal components, set them equal to mv²/r = mω²r and solve.
- μcoefficient of static friction
- θbank angle of the road, or the angle of the string to the vertical
- Llength of the string, m
- hheight of the cone h = L cos θ — vertical distance from the pivot to the plane of the circle
On a banked turn and in the conical pendulum the resultant of mg and the normal force (tension) is horizontal: F = mg tan θ, so a = g tan θ.
1) A flat bend has a radius of 80 m, and μ = 0.5 between the tyres and the asphalt (g = 10 m/s²). What is the highest speed at which a car can take the bend without skidding?
2) A banked bend of radius 120 m is designed for 30 m/s with no friction needed. What is the bank angle? What happens if a car goes faster?
3) A 0.1 kg ball hangs on a string 0.5 m long and moves in a horizontal circle with the string at 60° to the vertical. Find ω, the period and the tension.
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2) tan θ = v²/(gr) = 900/1200 = 0.75 ⇒ θ = 37°. A faster car tends to slide outward, so friction acts down the slope; for a slower car it acts up the slope. On a railway the outer rail is laid higher (
外轨高于内轨); a train faster than the design speed is pushed inward by the outer rail on its wheel flanges.3) h = L cos 60° = 0.25 m, ω = √(g/h) = √40 = 2√10 rad/s ≈ 6.3 rad/s, T = 2π/ω = π/√10 ≈ 1.0 s. Vertically: F cos θ = mg ⇒ F = mg/cos 60° = 2 N.
Two balls of different masses hang from the same point on strings of different lengths and move in circles in the same horizontal plane. Which quantity is the same for both balls?
A) linear speed B) angular velocity C) centripetal acceleration D) tension in the string
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A: v = ωr and the radii differ. C: a = g tan θ and the angles differ. D: F = mg/cos θ depends on the mass and the angle. The longer string “looks slower” but it sweeps a larger circle — that is the trap.
Vertical circles: string or rod?
In a vertical circle gravity does work, so the speed changes and the motion is not uniform. CSCA usually asks only about the highest and the lowest points: there all forces are vertical and their sum toward the centre equals mv²/r. The speeds at the two points are linked by conservation of mechanical energy (lesson “Work, power and energy”): vbottom² = vtop² + 4gr.
- Fforce of the string, rod or track on the body, taken positive toward the centre
- rradius of the circle, m
A string can only pull: at the top F ≥ 0, so mg ≤ mv²/r. A rod can pull or push: the ball can pass the top at any v ≥ 0; at v = √(gr) the rod exerts no force.
| Model | Minimum speed at the top | Force at the top |
|---|---|---|
| string; inside of a loop track | √(gr) | pulls toward the centre (down); 0 at v = √(gr) |
| light rod; ball inside a tube | 0 | v < √(gr): pushes up; v = √(gr): 0; v > √(gr): pulls down |
| car on top of a hump-backed bridge | — (maximum speed √(gr), faster it leaves the road) | N = mg − mv²/r < mg |
A 0.5 kg ball is fixed to a light rod 0.4 m long and turns in a vertical circle. At the highest point its speed is 1 m/s (g = 10 m/s²). What force does the rod exert on the ball there?
A) 3.75 N, pulling down B) 6.25 N, pulling down C) 3.75 N, pushing up D) 1.25 N, pushing up
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A: right size, wrong direction (v < √(gr) = 2 m/s, so the rod pushes). B: adds mg + mv²/r. D: takes mv²/r as the rod’s force.
1) A 0.2 kg ball on a string 0.9 m long moves in a vertical circle. What is the minimum speed at the top? In that case, what are the speed and the tension at the bottom?
2) A 1000 kg car passes over the top of a hump-backed bridge of radius 50 m at 10 m/s. What force does the bridge exert on the car? What would it be at the bottom of a dip of the same radius at the same speed?
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2) At the top: mg − N = mv²/r ⇒ N = 10 000 − 1000 · 100/50 = 8000 N — the car is “lighter” (
失重). At √(gr) = √500 ≈ 22 m/s, N would be 0 and the car would leave the road. In the dip: N = mg + mv²/r = 12 000 N — “heavier” (超重).How CSCA asks about this
The CSCA physics paper has 48 questions in 60 minutes — about 75 seconds each. Circular-motion items are short, 1–3 steps. The typical patterns:
- rpm or a period is given ⇒ find ω, v, a (convert first);
- belts, gears and shared axles: ratios of v, ω, T, a for three points;
- “which force provides the centripetal force?” and “which statement is correct?” about uniform circular motion;
- flat turns (μ), banked roads and railway curves (outer/inner rail), conical pendulums (same height);
- vertical circles: minimum speed at the top (string or rod), the force at the top or the bottom, humps and dips, often together with energy;
- centrifugal motion: spin dryers, a car skidding outward on a bend.
| Term | 中文 | Pinyin |
|---|---|---|
| uniform circular motion | 匀速圆周运动 | yúnsù yuánzhōu yùndòng |
| linear speed | 线速度 | xiànsùdù |
| angular velocity | 角速度 | jiǎosùdù |
| period / frequency | 周期 / 频率 | zhōuqī / pínlǜ |
| rotational speed (revolutions per second) | 转速 | zhuànsù |
| centripetal acceleration | 向心加速度 | xiàngxīn jiāsùdù |
| centripetal force | 向心力 | xiàngxīnlì |
| centrifugal motion | 离心运动 | líxīn yùndòng |
| belt drive / same axle | 皮带传动 / 同轴转动 | pídài chuándòng / tóngzhóu zhuàndòng |
| conical pendulum | 圆锥摆 | yuánzhuībǎi |
| arch (hump-backed) bridge | 拱形桥 | gǒngxíngqiáo |
| light string / light rod | 轻绳 / 轻杆 | qīngshéng / qīnggān |
| highest / lowest point | 最高点 / 最低点 | zuìgāodiǎn / zuìdīdiǎn |
Key points
- v = ωr, ω = 2π/T = 2πf; divide rpm by 60 first.
- Belts, chains and meshing gears share v; points on one axle share ω.
- a = v²/r = ω²r points to the centre; F = mv²/r is the resultant of the real forces, never an extra force.
- Flat turn: vmax = √(μgr); banked turn and conical pendulum: a = g tan θ; conical pendulum: ω = √(g/h).
- Vertical circle: top F + mg = mv²/r, bottom F − mg = mv²/r; the minimum top speed is √(gr) for a string and 0 for a rod.
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