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S5 Physics | Circular motion prep guide (for tutors) | angular speed, centripetal force, turning
Students think centripetal force is an extra arrow, so they draw a fourth force. This lesson shows it is a job taken by an existing force.
Objectives — Students can: (1) convert a period into angular speed and linear speed and explain v = omega r; (2) use a = v squared / r to compute centripetal acceleration and state that it points to the centre; (3) name the real force supplying it in three contexts and find the minimum speed at the top of a vertical circle.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Speed and velocity | velocity has direction | changing direction is acceleration |
| Newton second law | F = ma | F = 37.0 N |
| Vector subtraction | tip to tail | delta v points inward |
02 Lesson flow (50 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Angle and radian | one turn = 2 pi rad | omega = 7.85 rad/s |
| 8-18 min | Linear speed | r = 1.5 m, T = 0.8 s | v = 11.78 m/s |
| 18-30 min | Centripetal acceleration | v and r | a = 92.5 m/s2 |
| 30-40 min | Who supplies the force | m = 0.4 kg | F = 37.0 N |
| 40-46 min | Top of a vertical circle | r = 1.5 m | v_min = 3.84 m/s |
| 46-50 min | Exit check | three questions collected | independent work |
03 Teaching script (read it out)
Scene 1: Angular speed: how fast one turn takes
With T = 0.8 s the frequency is 1.25 Hz, that is 1.25 turns per second or 75 rpm. Angular speed is omega = 2 pi / T = 7.85 rad/s, meaning 7.85 radians swept per second.
Radians are not decoration: arc length s = r theta and v = omega r only hold in radians. Substituting degrees makes v too small by a factor of about 57, because one radian is 57.3 degrees. Make students write one turn equals 2 pi radians before touching the period.
Radians are not decoration: arc length s = r theta and v = omega r only hold in radians. Substituting degrees makes v too small by a factor of about 57, because one radian is 57.3 degrees. Make students write one turn equals 2 pi radians before touching the period.
Student task: convert T = 0.8 s into omega, frequency and rpm, and say what each number means.
Scene 2: Linear speed and acceleration: direction is the hard part
v = omega r = 7.85 times 1.5 = 11.78 m/s. The speed is constant but the direction keeps changing, so the velocity changes, so there is an acceleration. a = v squared / r = 92.5 m/s2, also written omega squared r.
The classic error is drawing a along the tangent. Delta v is new velocity minus old velocity, tip to tail, and the result points at the centre. Note the size too: 92.5 m/s2 is about 9.4 g, far more than gravity, which is why passengers feel pressed sideways in a turn.
The classic error is drawing a along the tangent. Delta v is new velocity minus old velocity, tip to tail, and the result points at the centre. Note the size too: 92.5 m/s2 is about 9.4 g, far more than gravity, which is why passengers feel pressed sideways in a turn.
Student task: draw velocity vectors at two points, subtract them, and mark the direction of delta v.
Scene 3: Centripetal force: ask which force is doing the job
Centripetal force is not a new kind of force, it is the inward resultant required. F = ma = 0.4 times 92.5 = 37.0 N. Then always ask who supplies it: tension for a whirling mass, static friction for a turning car, gravity for a satellite, gravity plus a normal force on a roller coaster. A car turning a 40 m radius bend at 14 m/s needs a = 4.9 m/s2, so the friction coefficient must reach 0.50; wet roads fall short and the car slides out.
If a student writes that centrifugal force pushes them out, correct it: in the inertial frame there is no outward force, the body simply wants to travel in a straight line. Vertical circle: mg = mv squared / r gives v_min = 3.84 m/s.
If a student writes that centrifugal force pushes them out, correct it: in the inertial frame there is no outward force, the body simply wants to travel in a straight line. Vertical circle: mg = mv squared / r gives v_min = 3.84 m/s.
Student task: for a whirling mass, a turning car and a satellite, name the force and compute the value.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Speed from a period | r = 1.5 m, T = 0.8 s | omega = 2 pi / T, v = omega r, a = v squared / r | v = 11.78 m/s, a = 92.5 m/s2 |
| Force from rpm | m = 0.4 kg at 75 rpm | omega = 2 pi f, F = m omega squared r | F = 37.0 N |
| Minimum friction on a bend | v = 14 m/s, r = 40 m | mu = v squared / (r g) | mu = 0.50 |
| Top of a vertical circle | r = 1.5 m | mg = mv squared / r | v_min = 3.84 m/s |
Sanity check (teach this): An acceleration of 92.5 m/s2, about 9.4 g, is reasonable for a 1.5 m radius at 0.8 s. An answer of 9.2 or 924 means a decimal slip or degrees instead of radians. The minimum speed 3.84 m/s must be below the actual speed 11.78 m/s, otherwise the string would already be slack. A friction coefficient of 0.50 between 0 and 1 is realistic; above 1 suggests bad data.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| r = 0.5 m, T = 0.4 s: find v and a. | v = 7.85 m/s, a = 123.4 m/s2 | omega = 15.71 rad/s first |
| m = 0.2 kg at 120 rpm on a 0.3 m radius: find F. | about 9.5 N | f = 2 Hz, omega = 12.57 |
| Vertical circle r = 0.8 m: minimum speed at the top? | 2.80 m/s | square root of 9.81 x 0.8 |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Centripetal force drawn as an extra arrow | free-body diagram shows both tension and centripetal force | It is not an extra force; ask which existing force points to the centre. |
| Constant speed means no acceleration | writing a = 0 because speed is unchanged | Speed is constant but direction changes, so velocity changes and a is not zero. |
| Degrees used for omega | omega = 360 / T | One turn is 2 pi radians, not 360, in the v = omega r formula. |
| Acceleration drawn along the tangent | a arrow parallel to v | Delta v points inward, so a points inward; the tangent is the velocity. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index (motion, forces, waves, electricity) — do circular motion before gravitation
- Secondary prep guides overview — all subjects
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| T = 0.5 s: what is omega? | 12.57 rad/s | wrong means redo act one |
| Which way does centripetal acceleration point? | towards the centre | wrong means redo act two |
| Why does a ball fall off at the top of a loop? | v is below root gr, normal force cannot act | two wrong means back to act three |
09 Tutor FAQ
Which year is this for?
S5, the circular motion part of the HKDSE forces and motion unit.
How many formulas to memorise?
Four: omega = 2 pi / T, v = omega r, a = v squared / r, F = ma.
Is there paired practice?
The homework links to the physics chapter index, which also lists gravitation and waves.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
