MathsFurther Mechanics 2 › Motion in a vertical circle

Motion in a vertical circle

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.

Year FMEDEXCEL 9FM0 FM2

Builds on Angular speed and horizontal circular motion and Work, energy and power.

IN THIS TOPIC

  • Combine conservation of energy with the radial equation of motion.
  • Find the condition for a particle on a string to complete a vertical circle.
  • Distinguish the string case from the rod and inside-surface cases.

WHAT YOU PROBABLY THINK

To complete a vertical circle on a string, the particle only needs enough speed to reach the top.

Energy round, forces across

Two separate ideas do the work. Conservation of energy relates the speed at any point to the height, since the tension is always perpendicular to the motion and does none. The radial equation of motion, resolving towards the centre, then gives the tension or reaction at that point:

T - mgcos θ = mv2r

There is a tangential acceleration too, from the component of gravity along the path, which is why the speed varies. It is rarely needed directly: the energy equation has already accounted for it.

A complete vertical circle on a string: at the top gravity alone must be enough, so v² is at least grmgTat the top: T + mg = mv²/rat the bottom: T − mg = mu²/rv² ≥ gr = 7.84so u ≥ 6.26 m/s
FIG. 1A vertical circle with the forces at the top and at the bottom, and the condition that sets the minimum speed.

WORKED EXAMPLE

Completing the circle

A particle on a light string of length 0.8 m is swung in a vertical circle. Find the least speed at the lowest point for a complete circle, taking g = 9.8 m/s².

At the top the string can pull but not push, so T ≥ 0, and the radial equation T + mg = mv²/r needs v² ≥ gr = 7.84.

Energy from bottom to top: u² = v² + 4gr = 7.84 + 31.36 = 39.2.

So u ≥ 6.26 m/s. At that speed the tension at the bottom is m(g + u²/r) = 58.8m newtons, six times the weight.

Three different critical conditions

Whether a circle is completed depends on what is holding the particle. On a string, or on the inside of a circular track, the constraint can only pull inwards, so the critical condition is that the tension or reaction reaches zero at the top: v² = gr there. Reaching the top with less speed than that is impossible, because the string would have gone slack earlier and the particle would have left the circle. That is what the opening claim misses.

On a light rod, or a bead threaded on a wire, the constraint can push as well as pull, so the only requirement is that the particle arrives at the top with any speed at all: v² > 0, giving u² > 4gr. Below the critical speed on a string the particle either oscillates, if it does not pass the horizontal, or leaves the circle and becomes a projectile.

String against rod: a string must still be taut at the top, a rod need only get the particle therestringT = 0 at the topu ≥ 6.26 m/slight rodv > 0 at the topu > 5.60 m/sthe constraint decides the conditiona rod can push outwards; a string cannot
FIG. 2The two critical conditions side by side: zero tension at the top for a string, any speed at all for a rod.

YOUR TURN

String against rod

A particle is attached to a light rod of length 0.8 m and swung in a vertical circle. Find the least speed at the lowest point for a complete circle, and compare it with the string case.

Show the working

A rod can push, so the particle needs only to reach the top with a speed above zero.

Energy: u² > 4gr = 4(9.8)(0.8) = 31.36.

So u > 5.60 m/s, against 6.26 m/s on a string.

The rod is the easier case, and at the critical rod speed the rod is thrusting outwards at the top rather than pulling in.

THE EXAM BIT

  • Use energy for speeds and the radial equation for tensions; do not try to do both with one equation.
  • State the critical condition explicitly: T = 0 at the top for a string, v > 0 for a rod.
  • Measure heights from a single level, usually the lowest point of the circle.
  • If the particle leaves the circle, say so and switch to projectile motion from that point.

CHECK YOURSELF

A particle on a string of length 0.5 m passes the top of a vertical circle. Find the least speed it can have there, taking g = 9.8 m/s².

Show a hint

The tension is zero at the critical speed.

Show the answer

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Energy relates speed to height, since the tension does no work; the radial equation then gives the tension at any point.

A string needs v² ≥ gr at the top, because it cannot push; a rod needs only v > 0 there.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Combine conservation of energy with the radial equation of motion.
  • Find the condition for a particle on a string to complete a vertical circle.
  • Distinguish the string case from the rod and inside-surface cases.

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