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Further Mechanics
Vertical loops and centripetal force calculations
- 1 thing to know
- Papers 1 and 3
Overview
Recall questions on the forces at the top and bottom of a vertical loop, the minimum speed condition, and what happens when the centripetal force is removed.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
A car passes over the top of a vertical loop on a track. Which forces contribute to the centripetal force there?
Model answer
Both the weight and the normal contact force from the track act downwards towards the centre, so mg + N = mv²/r.
- 2
What is the condition for the minimum speed at the top of a vertical loop, and what is that speed for a loop of radius 8.0 m?
Model answer
The minimum occurs when the track force falls to zero so that the weight alone provides the centripetal force, giving v = √(gr) = √(9.81 × 8.0) = 8.9 m s⁻¹.
- 3
Why does a passenger feel heaviest at the lowest point of a loop?
Model answer
The seat must push up hard enough both to support the weight and to provide the upward centripetal force, so the contact force is N = mg + mv²/r, which is what the passenger senses.
- 4
A ball of mass 0.20 kg is whirled at 4.0 m s⁻¹ on a string of length 0.80 m. What centripetal force is needed?
Model answer
F = mv²/r = 0.20 × 4.0² ÷ 0.80 = 4.0 N.
- 5
A centrifuge rotor of radius 0.15 m spins at 50 revolutions per second. What is the centripetal acceleration at its rim?
Model answer
The angular velocity is ω = 2π × 50 = 314 rad s⁻¹, so a = rω² = 0.15 × 314² = 1.5 × 10⁴ m s⁻².
- 6
A car of mass 1200 kg rounds a bend of radius 45 m at 15 m s⁻¹. What friction force must the tyres provide?
Model answer
F = mv²/r = 1200 × 15² ÷ 45 = 6.0 × 10³ N directed towards the centre of the bend.
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