The shape of the lesson
By the end of the lesson, learners can
- The force points to the centre: Swing a ball on a string in a vertical circle.
- A different context: the banked turn: Now a horizontal circle with no string at all.
- Two formulas, one direction: Because the velocity changes direction, a body in uniform circular motion accelerates toward the centre with a = rω² = v² / r.
- Angular speed: ω = 2πf = v / r.
- Centripetal acceleration: a = v² / r = rω².
- Centripetal force: F = mv² / r = mrω².
One picture every idea returns to
A body in uniform circular motion accelerates toward the centre, a = rω² = v²/r, so the resultant force is F = mrω² = mv²/r, also toward the centre.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The force that turns.. Moving in a circle means accelerating toward the centre, even at constant speed. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The force points to the centre | Swing a ball on a string in a vertical circle. Project tab 2 of the lab (Vertical circle) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: A different context: the banked turn | Now a horizontal circle with no string at all. Project tab 4 of the lab (Banked turns: plane, road, bike) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Two formulas, one direction | Because the velocity changes direction, a body in uniform circular motion accelerates toward the centre with a = rω² = v² / r. | Connect-Extend-Challenge |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (8 questions, 88 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Two formulas, one direction, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Centripetal acceleration and force: topic worksheet
8 questions, 88 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A ball of mass 0.15 kg is attached to a string of length 0.80 m and is swung in a vertical circle, as shown in Fig. 2.1.
(a) State the two forces acting on the ball when it is at the top of the circle, and the direction of each. [2]
(b) At the top of the circle the ball moves with speed v.
(i) Show that, for the string to remain taut, the speed at the top must be at least 2.8 m s⁻¹. [3]
(ii) State the name of the force that becomes zero at this minimum speed. [1]
(c) On one turn the ball passes the top of the circle with a speed of 3.5 m s⁻¹.
(i) Calculate the tension in the string at the top of the circle. [2]
(ii) Show that the speed of the ball at the bottom of the circle is 6.6 m s⁻¹. [2]
(iii) Calculate the tension at the bottom of the circle, and state why it is larger than at the top. [3]
A small bob of mass 0.20 kg is attached to a string of length 1.2 m. The bob moves in a horizontal circle so that the string makes a constant angle of 35° with the vertical, as shown in Fig. 3.1.
(a) State the two forces acting on the bob, and the direction of each. [2]
(b) The tension in the string is T. By resolving the forces on the bob, show that tan θ = v² / (rg), where v is the speed of the bob and r is the radius of its circular path. [3]
(c) For this motion:
(i) show that the radius r of the circular path is 0.69 m, [1]
(ii) calculate the speed v of the bob, [2]
(iii) calculate the time taken for one complete revolution. [2]
(d) The bob is then made to move faster. State and explain what happens to the angle θ. [2]
A turntable completes one revolution every 1.8 s. Point P lies 12 cm from the centre and point Q lies 24 cm from the centre.
(a) Calculate the angular speed of the turntable. [2]
(b) Calculate the linear speed of P and of Q. [2]
(c) State and explain how the angular speeds of P and Q compare. [2]
(d) Calculate the angle, in radians, turned through in 0.50 s. [2]
(e) Calculate the centripetal acceleration of Q. [2]
A 0.50 kg mass on a string moves in a horizontal circle of radius 0.80 m on a frictionless table, at a constant speed of 3.0 m s⁻¹.
(a) State what provides the centripetal force. [1]
(b) State the direction of the centripetal force. [1]
(c) Calculate the centripetal acceleration. [2]
(d) Calculate the tension in the string. [2]
(e) The string will break when the tension reaches 20 N. Calculate the maximum speed. [3]
(f) State and explain the direction in which the mass moves immediately after the string breaks. [2]
A car of mass 1200 kg rounds a flat (unbanked) bend of radius 50 m. The maximum frictional force available is 0.60 of the car’s weight.
(a) State what provides the centripetal force on the car. [1]
(b) Write the condition for the car not to skid. [2]
(c) Derive an expression for the maximum cornering speed. [2]
(d) Calculate the maximum speed. [2]
(e) State and explain what happens if the car exceeds this speed. [2]
(f) State one change that would raise the maximum cornering speed. [1]
A bend is banked at an angle θ so that a car can round it at a design speed without relying on friction, as shown in Fig. 8.1.
(a) Explain why banking allows a higher cornering speed than a flat road. [2]
(b) For a frictionless banked track, show that tanθ = v²/(rg). [3]
(c) A bend of radius 80 m is to be taken at 25 m s⁻¹. Calculate the banking angle. [3]
(d) State and explain what tends to happen if a car travels below the design speed on a frictionless bank. [2]
(e) State the direction of the centripetal force on the car. [1]
A centrifuge spins a 5.0 g sample at a radius of 8.0 cm, rotating at 4000 revolutions per minute. This question links the angular quantities of Q4 and Q5 with the centripetal force ideas of Q6 to Q8.
(a) Calculate the angular speed in rad s⁻¹. [2]
(b) Calculate the linear speed of the sample. [1]
(c) Calculate the centripetal acceleration of the sample. [2]
(d) Express this acceleration as a multiple of g. [2]
(e) Calculate the centripetal force on the sample. [2]
(f) State the direction of this force. [1]
In a fairground "rotor" ride, riders stand against the inside wall of a vertical cylinder of radius 3.0 m. The cylinder spins, the floor drops away, and friction holds the riders up, as shown in Fig. 10.1. The coefficient of friction between clothing and wall is 0.40.
(a) State what provides the centripetal force on a rider. [1]
(b) State the direction of the centripetal force. [1]
(c) Write the condition for a rider not to slip down the wall. [2]
(d) Show that the minimum angular speed is ω = √(g/(μr)). [3]
(e) Calculate the minimum angular speed and the linear speed of a rider at the wall. [3]
(f) State why the rider’s mass does not affect the minimum speed. [1]
Answer key and marking notes
- (a)Weight (mg) vertically downward; tension (T) along the string, also directed downward toward the centre ✓✓
- (b)(i)At minimum speed the tension is zero, so weight provides the centripetal force: mg = mv²/r, giving v² = gr = 9.81 × 0.80, v = 2.80 m s⁻¹ ✓✓✓
- (b)(ii)Tension ✓
- (c)(i)T = m(v²/r − g) = 0.15(3.5²/0.80 − 9.81) = 0.83 N ✓✓
- (c)(ii)Energy: ½mvb² = ½mvt² + mg(2r), so vb² = vt² + 4gr = 3.5² + 4 × 9.81 × 0.80 = 43.6; vb = 6.6 m s⁻¹ ✓✓
- (c)(iii)T = m(vb²/r + g) = 0.15(43.6/0.80 + 9.81) = 9.7 N ✓✓; larger because at the bottom the tension must both provide the centripetal force and support the weight, and the ball moves faster there ✓
- (a)Weight mg vertically downward; tension T along the string, directed up toward the support ✓✓
- (b)Vertical: T cosθ = mg. Horizontal (centripetal): T sinθ = mv²/r. Dividing the two: tanθ = (v²/r)/g = v²/(rg) ✓✓✓
- (c)(i)r = L sinθ = 1.2 × sin 35° = 0.69 m ✓
- (c)(ii)v² = rg tanθ = 0.688 × 9.81 × tan 35° = 4.73; v = 2.2 m s⁻¹ ✓✓
- (c)(iii)t = 2πr/v = 2π × 0.688 / 2.17 = 2.0 s ✓✓
- (d)θ increases ✓; a greater speed needs a greater centripetal force, so the horizontal component T sinθ must increase, which requires a larger angle (the string rises toward the horizontal) ✓
- (a)ω = 2π/T = 2π/1.8 ✓ = 3.5 rad s⁻¹ ✓
- (b)vₚ = rω = 0.12 × 3.49 = 0.42 m s⁻¹ ✓; vₜ = 0.24 × 3.49 = 0.84 m s⁻¹ ✓
- (c)Equal ✓: both complete a revolution in the same time, so they share the same angular speed ✓
- (d)θ = ωt = 3.49 × 0.50 ✓ = 1.7 rad ✓
- (e)a = rω² = 0.24 × 3.49² ✓ = 2.9 m s⁻² ✓
- (a)The tension in the string ✓
- (b)Toward the centre of the circle ✓
- (c)a = v²/r = 3.0²/0.80 ✓ = 11 m s⁻² ✓
- (d)F = mv²/r = 0.50 × 11.25 ✓ = 5.6 N ✓
- (e)v = √(Fr/m) = √(20 × 0.80 / 0.50) ✓✓ = 5.7 m s⁻¹ ✓
- (f)Along a straight line, tangent to the circle ✓: with no resultant force it continues at constant velocity (Newton’s first law) ✓
- (a)Friction between the tyres and the road ✓
- (b)The friction must supply the centripetal force: μmg ≥ mv²/r ✓✓
- (c)At the limit μmg = mv²/r, so v_max = √(μrg) ✓✓
- (d)v = √(0.60 × 50 × 9.81) ✓ = 17 m s⁻¹ ✓
- (e)The required centripetal force exceeds the maximum friction ✓, so the car skids outward and leaves the bend ✓
- (f)A larger radius, a higher coefficient of friction, or banking the road ✓
- (a)The horizontal component of the normal contact force can supply the centripetal force ✓, so the car does not depend on friction, which can be exceeded ✓
- (b)Horizontally N sinθ = mv²/r ✓; vertically N cosθ = mg ✓; dividing gives tanθ = v²/(rg) ✓
- (c)tanθ = 25²/(80 × 9.81) = 625/784.8 = 0.796 ✓✓; θ = 39° ✓
- (d)It tends to slide down the bank, toward the centre ✓, because the horizontal force component now exceeds the centripetal force needed ✓
- (e)Horizontal, directed toward the centre of the circular path ✓
- (a)ω = 4000 × 2π/60 ✓ = 4.2 × 10² rad s⁻¹ ✓
- (b)v = rω = 0.080 × 418.9 = 34 m s⁻¹ ✓
- (c)a = rω² = 0.080 × 418.9² ✓ = 1.4 × 10⁴ m s⁻² ✓
- (d)a/g = 1.4 × 10⁴ / 9.81 ≈ 1.4 × 10³ g ✓✓
- (e)F = ma = 0.0050 × 1.40 × 10⁴ ✓ = 70 N ✓
- (f)Toward the axis of rotation (the centre) ✓
- (a)The normal (contact) force from the wall ✓
- (b)Horizontally, toward the axis of the cylinder ✓
- (c)The friction must at least support the weight: μN ≥ mg, with N = mrω² ✓✓
- (d)μ(mrω²) ≥ mg ✓; ω² ≥ g/(μr) ✓; so ω_min = √(g/(μr)) ✓
- (e)ω = √(9.81/(0.40 × 3.0)) = √8.18 = 2.9 rad s⁻¹ ✓✓; v = rω = 3.0 × 2.86 = 8.6 m s⁻¹ ✓
- (f)The mass cancels: both the friction needed and the weight are proportional to m ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
An object moves in a circle at constant speed. The direction of its acceleration is:
- Atoward the centre
- Balong the direction of motion
- Caway from the centre
- Dzero, because the speed is constant
The centripetal force on a body moving in a circle is best described as:
- Athe resultant of the real forces, directed toward the centre
- Ban extra outward force on the body
- Ca separate force called centrifugal force
- Dalways equal to the weight of the body
A 0.50 kg ball moves in a circle of radius 0.80 m at 4.0 m s⁻¹. The centripetal force F = mv²/r is:
- A10 N
- B2.5 N
- C8.0 N
- D20 N
Keeping the mass and radius fixed, doubling the speed changes the centripetal force to:
- Afour times as large
- Btwice as large
- Chalf as large
- Dunchanged
Quiz key
The velocity keeps changing direction, so there is an acceleration even at constant speed; for uniform circular motion it points to the centre, a = v²/r.
Centripetal force is just the name for the net inward force. It is provided by tension, friction or gravity, not by an additional force.
F = mv²/r = 0.50 × 4.0² / 0.80 = 0.50 × 16 / 0.80 = 10 N.
F = mv²/r is proportional to v², so doubling the speed multiplies the force by four.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.