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Motion in a vertical circle questions
Now the speed changes as well as the direction, because gravity does work on the way round. Energy handles the speed and the radial equation handles the force, and the interesting question is whether the circle is completed at all.
6 original questions · 23 marks · the motion in a vertical circle notes · Further Mechanics 2
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Explain why conservation of energy can be used for a particle on a string in a vertical circle, even though the tension varies.
Worked answer
The tension always acts along the string, which is perpendicular to the direction of motion at every instant, so it does no work. Only gravity does work, and it is conservative, so the total of kinetic and potential energy is constant. B1 for the tension doing no work, B1 for gravity being the only force that does work.A particle on a string of length 1.2 m is to complete a vertical circle. Find the least speed it can have at the top, taking g = 9.8 m/s².
Worked answer
At the critical speed the string is just taut, so T = 0 and the only radial force is the weight: mg = mv²/r. Hence v² = gr = 9.8 × 1.2 = 11.76, so v = 3.43 m/s. M1 for setting T = 0, A1 for mg = mv²/r, A1 for the least speed.A particle on a string of length 1.2 m is to complete a vertical circle. Taking g = 9.8 m/s², find the least speed it can have at the lowest point.
Worked answer
Conserving energy from the bottom to the top, a rise of 2r, gives ½u² = ½v² + g(2r), so u² = v² + 4gr. The critical case has v² = gr, so u² = 5gr = 5 × 9.8 × 1.2 = 58.8 and u = 7.67 m/s. M1 for conserving energy over a rise of 2r, A1 for u² = v² + 4gr, M1 for using v² = gr, A1 for the least speed. The result u² = 5gr is worth remembering, but the derivation carries the marks and a quoted formula alone will not score them.A bead slides from rest on the inside of a smooth circular track of radius 2 m, starting level with the centre. Find its speed at the lowest point and the reaction there, per unit mass.
Worked answer
It falls a height equal to the radius, so v² = 2gr = 2(9.8)(2) = 39.2 and v = 6.26 m/s. At the lowest point the radial equation is R − mg = mv²/r, so per unit mass R = 9.8 + 39.2/2 = 29.4 N per kg, three times g. M1 for the energy equation over a drop of r, A1 for the speed, M1 for the radial equation at the lowest point, A1 for the reaction. The reaction is three times the weight at the lowest point, and the extra 2mg is what the circular motion demands.State the condition for a complete circle when the particle is attached to a light rod rather than a string, and explain the difference.
Worked answer
A rod can push outwards as well as pull inwards, so the constraint never fails and the particle only has to reach the top still moving. That needs v > 0 at the top and so u² > 4gr at the bottom, here u > 6.86 m/s. A string can only pull, so its tension must stay non-negative all the way round, forcing v² ≥ gr at the top and u² ≥ 5gr at the bottom. B1 for the rod being able to push, B1 for needing only v > 0 at the top, B1 for the string needing u² ≥ 5gr. A bead threaded on a wire or inside a smooth tube behaves like the rod, since the tube can push inwards or outwards.A particle slides from rest at the top of a smooth sphere of radius r. Show that it leaves the surface where the radius makes an angle θ with the upward vertical satisfying cos θ = 2/3, and find its speed at that instant in terms of g and r.
Worked answer
At angle θ the radial equation is mg cos θ − R = mv²/r. Energy from the top, having dropped r(1 − cos θ): v² = 2gr(1 − cos θ). The particle leaves when R = 0, so g cos θ = v²/r = 2g(1 − cos θ). Hence 3cos θ = 2 and cos θ = 2/3, that is θ = 48.2°.
Substituting back, v² = 2gr(1 − 2/3) = 2gr/3, so the speed is √(2gr/3).
M1 for the radial equation, A1 for mg cos θ − R = mv²/r, M1 for the energy equation, A1 for v² = 2gr(1 − cos θ), M1 for setting R = 0, A1 for cos θ = 2/3, A1 for the speed.
The result is independent of r and of the mass, so every smooth sphere sheds a particle at the same place. Two equations are needed and each carries marks. The radial equation must resolve the weight along the radius as mg cos θ, and the energy equation must use the vertical drop r(1 − cos θ) rather than an arc length.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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