Circular motion: radians, angular velocity and inward force
Key idea: Connect the geometry and timing of circular motion to a vector model in which the resultant acceleration and force point towards the centre.
Before you start: Angular Displacement & VelocityNewton's Laws, Vectors & Problem Solving
By the end, you can
- Express signed angular displacement in radians and use s = rθ.
- Use angular velocity and connect period, frequency and tangential speed through v = rω.
- Explain why uniform circular motion has an inward acceleration although its speed is constant.
- Use a = rω², a = v²/r, F = mrω² and F = mv²/r in radial force models.
Starting-point self-check
1. Check your starting point
Attempt all three groups without notes and mark the first angle conversion, motion relation or radial-force decision you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Centripetal acceleration and radial force models 7(d)–(f)
Question 1
An 800 kg car travels at 15 m s⁻¹ around a level bend of radius 50 m. Find its acceleration and required resultant force, and identify the force direction.
Check the model response
a = v²/r = 15²/50 = 4.50 m s⁻². F = ma = 800(4.50) = 3.60 × 10³ N towards the centre. Static friction supplies this inward resultant.
Question 2
Explain why a body moving at constant speed around a circle is accelerating.
Check the model response
Velocity is a vector. Its direction changes continuously even though its magnitude is constant, so there is an acceleration perpendicular to the instantaneous velocity and towards the centre.
repair
2. Repair the six common breaks
Use only the repair matching an error, then redraw the angle, velocity or free-body model before retrying.
Centripetal acceleration and radial force models 7(d)–(f)
Check this idea
Misconception: The velocity and acceleration of a body in uniform circular motion both point towards the centre.
Repair: Velocity is tangent to the path. Centripetal acceleration is perpendicular to that velocity and points towards the centre.
Check this idea
Misconception: Centripetal force is an extra force that must be added to weight, tension or friction.
Repair: Centripetal force names the inward resultant. Resolve the real forces radially and set their inward resultant equal to mv²/r or mrω².
Check this idea
Misconception: An inward force must increase a body's speed because it causes acceleration.
Repair: In uniform circular motion the inward force is perpendicular to velocity, so it changes direction without doing work or changing speed.
worked example
3. Follow three worked models
Follow how each solution fixes the angle unit, converts the rotation rate and identifies the inward resultant before calculating.
Centripetal acceleration and radial force models 7(d)–(f)
Model 1
A 0.30 kg mass on a string moves in a vertical circle of radius 0.80 m. Find the string tension at the top when its speed is 5.0 m s⁻¹ and at the bottom when its speed is 7.0 m s⁻¹.
Check the model response
At the top, inward is down: T + mg = mv²/r, so T = 0.30(5.0²)/0.80 − 0.30(9.81) = 6.43 N. At the bottom, inward is up: T − mg = mv²/r, so T = 0.30(7.0²)/0.80 + 0.30(9.81) = 21.3 N.
guided practice
4. Guided practice
Use each hint only to choose the signed angle, angular relation or inward resultant.
Centripetal acceleration and radial force models 7(d)–(f)
Question 1
A 1200 kg car takes a level bend of radius 60 m at 20 m s⁻¹. Find its acceleration and the frictional force required.
Hint: Static friction is the real force providing the horizontal inward resultant.
Check the model response
a = v²/r = 20²/60 = 6.67 m s⁻². The required inward friction is F = ma = 1200(6.67) = 8.00 × 10³ N.
independent practice
5. Independent practice
Solve without repair notes and show the angle conversion, rotation-rate relation and radial equation.
Centripetal acceleration and radial force models 7(d)–(f)
Question 1
A frictionless road is banked at 14° for a bend of radius 65 m. Derive the design-speed relation and calculate the speed for which no sideways friction is needed.
Check the model response
Resolve the normal contact force: N cos θ = mg and N sin θ = mv²/r. Dividing gives tan θ = v²/(rg), so v = √(rg tan θ) = √[65(9.81)tan 14°] = 12.6 m s⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Centripetal acceleration and radial force models 7(d)–(f)
Question 1
A 0.50 kg mass moves at 4.0 m s⁻¹ at the top of a vertical circle of radius 0.80 m. Find its acceleration and string tension, and state the directions of its velocity and acceleration.
Check the model response
a = v²/r = 4.0²/0.80 = 20 m s⁻² towards the centre, which is downward at the top. T + mg = mv²/r gives T = 0.50(20) − 0.50(9.81) = 5.10 N. Velocity is tangent to the circle and perpendicular to the downward acceleration.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Centripetal acceleration and radial force models 7(d)–(f)
Question 1
A 1000 kg car travels around a level circular bend of radius 80 m at 24 m s⁻¹. Find its acceleration and inward resultant force. State how both change if the speed doubles.
Check the model response
a = v²/r = 24²/80 = 7.20 m s⁻² and F = ma = 7.20 × 10³ N, supplied by friction towards the centre. Since both are proportional to v², doubling speed makes each four times as large.