A-level Further Maths Year 13 · Topic 6
Further mechanics
A level further mechanics covers motion in circles and with changing forces.
We will model horizontal and vertical circular motion and find centres of mass by integration.
Then we use variable forces, oblique collisions and simple harmonic motion.
- 5 subtopics
- 250 questions
- Free: 6.1 notes and 10 questions
In this topic
- 6.1Circular motion: horizontal and vertical circlesFree below · 51 questions
- 6.2Centres of mass: particles, laminas and solids by integrationOpen in Full Marks · 41 questions
- 6.3Variable forces and further kinematicsOpen in Full Marks · 51 questions
- 6.4Oblique collisions and impulse vectorsOpen in Full Marks · 53 questions
- 6.5Simple harmonic motion in mechanics: springs and pendulumsOpen in Full Marks · 54 questions
6.1 · Free notes
Circular motion: horizontal and vertical circles
Key points
- For motion in a circle of radius r with angular speed ω: v = rω, and the acceleration towards the centre is v²/r = rω².
- Newton's second law towards the centre: F = mv²/r = mrω², where F is the resultant of forces such as tension, friction, normal reaction or a component of weight.
- Horizontal circles: conical pendulum (T cos θ = mg, T sin θ = mrω²); car on a banked track; particle on a rough turntable (friction provides the force).
- Vertical circles: speed changes; use conservation of energy for the speed at each point and F = mv²/r for tension or reaction. A particle on a string completes the circle if the tension at the top is ≥ 0 (v_top² ≥ gr); on a rigid rod it only needs v_top ≥ 0.
In more detail
- Worked example: a 0.5 kg ball on a 1 m string has speed 6 m/s at the bottom (g = 9.8). At the top: v² = 36 − 4g = 36 − 39.2 < 0, so it cannot reach the top; it leaves the circle before then.
- Banking without friction: tan θ = v²/(rg), the speed at which no sideways friction is needed.
- Exam tip: resolve towards the centre for the circular equation, and vertically (or along the radius) for the other.
Practice
Circular motion: horizontal and vertical circles: 10 questions
Pick an answer to check it, or open the answer for the explanation.
1. What is the acceleration of a particle moving in a circle of radius 2 m at 6 m/s?
- 3 m/s²
- 18 m/s² towards the centre
- 12 m/s²
- 72 m/s²
Show answer
B: 18 m/s² towards the centre. v²/r = 36/2.
2. Convert angular speed: v = 10 m/s, r = 5 m. What is ω?
- 50 rad/s
- 0.5 rad/s
- 15 rad/s
- 2 rad/s
Show answer
D: 2 rad/s. ω = v/r.
3. What provides the centripetal force for a car turning on a flat road?
- Weight
- Air resistance
- Friction
- Normal reaction alone
Show answer
C: Friction. It acts towards the centre.
4. In a conical pendulum, which equation is correct vertically?
- T cos θ = mg
- T sin θ = mg
- T = mg
- T = mrω²
Show answer
A: T cos θ = mg. No vertical acceleration.
5. For a particle on a string in a vertical circle, what is the condition to complete the circle?
- Speed at the top > 0
- Tension at the top ≥ 0
- Speed at the bottom = 0
- Tension at the bottom = 0
Show answer
B: Tension at the top ≥ 0. Otherwise the string goes slack.
6. What is the minimum speed at the top of a vertical circle of radius r on a string?
- 0
- √(2gr)
- gr
- √(gr)
Show answer
D: √(gr). mg = mv²/r when tension is zero.
7. A 2 kg mass moves in a horizontal circle of radius 0.5 m at ω = 4 rad/s. What is the resultant force?
- 8 N
- 4 N
- 16 N
- 32 N
Show answer
C: 16 N. mrω² = 2 × 0.5 × 16.
8. Why use energy in vertical circles?
- The speed changes with height
- The speed is constant
- There is no gravity
- Momentum is conserved
Show answer
A: The speed changes with height. KE converts to GPE.
9. A particle on a rigid rod moves in a vertical circle. What is needed to complete the circle?
- Tension at the top ≥ 0
- Speed at the top > 0 (just reaching it)
- v_top² ≥ gr
- Nothing
Show answer
B: Speed at the top > 0 (just reaching it). A rod can push as well as pull.
10. On a banked track with no friction, what provides the centripetal force?
- Weight
- Friction
- Air resistance
- The horizontal component of the normal reaction
Show answer
D: The horizontal component of the normal reaction. R sin θ = mv²/r.
Keep going in Full Marks
Full Marks has 51 questions on Circular motion: horizontal and vertical circles (41 more than here), set out in practice levels. You also get notes and 3D models for all 5 subtopics of Further mechanics, plus games, timed exams and past papers.