Maths › Further 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.
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.
- Say what happens when the circle is not completed.
COMMON MISCONCEPTION
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 stays 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:
There is a tangential acceleration too, from the component of gravity along the path, and that is what makes the speed vary. You rarely need it directly, because the energy equation has already accounted for it.
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
What holds the particle decides the test. On a string, or on the inside of a circular track, the constraint acts inwards only: a string pulls in and a track pushes in. Neither can act outwards, so the critical condition is that the tension or reaction reaches zero at the top, giving v² = gr there. Arriving at the top any slower is impossible, because the string would already have gone slack lower down and the particle would have left the circle.
On a light rod, or a bead threaded on a wire, the constraint can push as well as pull. The only requirement is that the particle arrives at the top with some speed, so v² > 0 and u² > 4gr at the bottom.
GUIDED PRACTICE
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 instead of pulling in.
When the circle is not completed
For a particle on a string of length r projected with speed u from the lowest point, there are three outcomes and the boundaries are worth memorising. If u² ≥ 5gr the circle is completed. If u² ≤ 2gr the particle never rises past the horizontal through the centre, so the string stays taut and it swings back and forth like a pendulum.
Between those two, 2gr < u² < 5gr, the string goes slack somewhere above the horizontal and the particle leaves the circle. Setting T = 0 in the radial equation locates that point, and from there the particle is a projectile under gravity alone until the string snaps taut again. Say all of that explicitly in an answer; the marks are for identifying the case, not only for the arithmetic.
ASSESSMENT FOCUS
- Use energy for speeds and the radial equation for tensions. One equation will not do both.
- 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, find where T = 0, and switch to projectile motion from that point.
- Remember the 2gr and 5gr boundaries; questions often ask only which case applies.
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
With T = 0 the radial equation gives mg = mv²/r, so v² = gr = 4.9 and v = 2.21 m/s.
Energy relates speed to height, since the tension does no work; the radial equation then gives the tension at any point.
A string or an inside track can act inwards only, so it needs v² ≥ gr at the top; a rod or a wire can also push, and needs only v > 0.
From the lowest point, u² ≥ 5gr completes the circle and u² ≤ 2gr oscillates; anything between leaves the circle above the horizontal.
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.
Or read them with their worked answers on the motion in a vertical circle questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- 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.
- Say what happens when the circle is not completed.
Open the full revision checklist to see every objective in the course in one place.