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Further Mechanics

Vertical loops and centripetal force calculations

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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.

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Key term

Why do passengers feel thrown outwards when a vehicle turns a sharp corner?

Questions from this subtopic

Each one with the model answer underneath.

  1. 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. 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. 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. 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. 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. 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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