Vertical Circular Motion
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Direct answer
A body whirled in a vertical circle of radius r must satisfy, at the highest point, mv^2/r ≥ mg — gravity is allowed to supply the whole centripetal force — so the minimum (critical) speed at the top is v_top = √(gr), about 7.0 m/s for a 5 m circle with g = 9.8 m/s^2. For a string-or-track loop where the body just completes the circle, energy conservation over the diameter then fixes the bottom speed at √(5gr), roughly 2.24 times the top speed. The forces are T_top = mv^2/r − mg and T_bottom = mv^2/r + mg, and for the just-completing case these work out to 0 and 6mg, a difference worth memorising outright. Water stays in a swung bucket, and a pilot blacks out at the bottom of a dive rather than the top, for exactly these reasons.
What you must remember
- Critical speed at the top: v_min = √(gr) for a string or the inside of a track, where gravity alone can provide the centripetal force; any slower and the string goes slack (the body leaves circular path into projectile motion).
- Bottom speed for just completing the loop: v_bottom = √(5gr), from ½mv_bottom^2 = ½mv_top^2 + mg(2r).
- Force equations: at the top, T + mg = mv^2/r; at the bottom, T − mg = mv^2/r. Keep weight towards the centre at the top, away from it at the bottom.
- The 6mg result: in the minimum-speed case, T_bottom = 6mg and T_top = 0, so T_bottom − T_top = 6mg — one of the most reused numbers in NEET rotational problems.
- Rod versus string: a rigid rod can also push, so a body attached to a rod can complete the circle with v_top = 0 (the rod supports it); the string cannot. NEET tests this distinction often.
- Energy bookkeeping: between any two points, ½mv_1^2 + mgh_1 = ½mv_2^2 + mgh_2; the speed difference across the vertical diameter is purely gravitational, independent of the string.
- Outside-a-track case: for a ball rolling on the outside of a sphere, leaving occurs when N = 0, i.e. at cos θ = 2/3 down from the top — a classic result appearing in NCERT Exemplar-level problems.
A worked loop-the-loop number
Take a 1 kg block sliding on the inside of a frictionless vertical loop of radius r = 2.0 m, entering at the bottom. What is the least entry speed so it maintains contact throughout? Set the top-speed condition first: v_top^2 = gr = 9.8 × 2 = 19.6 m^2/s^2. Now climb the diameter with energy: ½mv_bottom^2 = ½mv_top^2 + mg(4), so v_bottom^2 = 19.6 + 2 × 9.8 × 4 = 19.6 + 78.4 = 98, giving v_bottom = √98 ≈ 9.9 m/s. If the block arrives with exactly this speed, the track force at the top is zero (pure gravity steering it) while at the bottom the track pushes with mv^2/r + mg = (98/2) + 9.8 = 58.8 N, i.e. 6mg as advertised. If instead the block arrives at 9.0 m/s, v_top^2 = 81 − 78.4 = 2.6, and the required centripetal acceleration mv^2/r = 1.3 m/s^2 falls short of g = 9.8 — the needed inward force exceeds what gravity supplies at the top, meaning the block would have needed a pull it cannot get, so it leaves the loop earlier. This single arithmetic chain — top condition, energy climb, force check — is the complete exam algorithm.
Where students slip
The most common error is writing T = mv^2/r at the top with gravity forgotten, which undercounts the centripetal force; gravity always acts and at the top it helps the circle, so the correct budget is T + mg = mv^2/r. The second slip is assuming the 6mg tension difference holds universally — it is specific to the minimum-speed case; for a general speed the difference is m(v_bottom^2 − v_top^2)/r + 2mg = 6mg only because v_bottom^2 − v_top^2 = 4gr in every loop, which is actually the cleaner way to remember it. Finally, when the question says "just completes the vertical circle", read it as the critical condition v_top = √(gr), not v_top = 0; the zero-speed option belongs to the rod, and examiners bank on candidates mixing the two.
Frequently asked questions
What is the minimum speed at the highest point of a vertical circle?
For a string or the inside of a track, v_top(min) = √(gr), the speed at which gravity alone supplies the entire centripetal force and tension drops to zero.
Why is the bottom speed √(5gr) for a just-completing loop?
Energy conservation between bottom and top gives ½mv_b^2 = ½mv_t^2 + mg(2r); substituting v_t^2 = gr yields v_b^2 = 5gr.
Why can a rod complete the circle with zero top speed but a string cannot?
A rod exerts push as well as pull, supporting the body at the top, whereas a string can only pull and goes slack the moment the required force reverses.
Why does water not fall from a bucket swung vertically?
At the top, if the bucket moves at least at √(gr), gravity itself provides the centripetal acceleration needed to bend the water's path; the water is already "falling" along the circle, so it exerts no tendency to leave the bucket.
What is the difference between the tensions at the bottom and the top?
T_bottom − T_top = m(v_b^2 − v_t^2)/r + 2mg = 6mg for any loop completed in a string, since v_b^2 − v_t^2 = 4gr from energy conservation.