Motion in a vertical circle

Swing a bucket of water in a fast vertical loop and not a drop spills, even when the bucket is upside down at the top. A roller-coaster car does the same trick around a loop-the-loop. Both rely on one idea from Unit 3, Laws of Motion: a body moving in a circle needs a net centre-directed force $mv^2/r$ at every instant, and in a vertical circle gravity itself supplies or opposes part of that force depending on where the body is. Unlike a horizontal circle, the speed keeps changing — gravity does work as the body rises and falls around the loop, so energy conservation has to join hands with Newton's second law to describe the motion fully. This article builds the two force equations at the top and bottom of the loop, combines them with energy conservation to find the famous "just completes the loop" condition, and works through the string-tension calculations that ISC boards ask most often.

The core insight

In a vertical circle the centripetal force equation changes with position because gravity adds to the required force at the top and subtracts from it at the bottom, so a string can go slack unless the speed at the top is at least $\sqrt{gr}$ .

The core idea

Picture a small body of mass $m$ on a string of length $r$ (or a bead on a wire, or a car on a circular track), moving in a vertical circle. At every point, the net force directed toward the centre must equal $mv^2/r$ — that is simply Newton's second law applied to circular motion, and it holds everywhere, not just at special points. What changes as the body moves around the loop is which forces point toward the centre and which oppose it, because gravity always points straight down.

At the top of the loop, the centre is below the body, so both the string tension $T$ and the weight $mg$ point toward the centre (downward) and add together to supply the centripetal force:

T + mg = \frac{mv_{top}^2}{r}

where: $T$ is the string tension (or track's normal reaction) at the top, $m$ is the mass, $v_{top}$ is the speed at the top, and $r$ is the radius of the circle.

At the bottom of the loop, the centre is above the body, so tension points toward the centre (upward) while weight points away from the centre (downward) — they oppose:

T - mg = \frac{mv_{bottom}^2}{r}

A string can only pull, never push, so physically $T \ge 0$ at every point of the loop. That single constraint, applied at the top where gravity is already helping supply the centripetal force, is what produces the classic minimum-speed condition below.

Figure

Force diagram of a ball on a string at the top and bottom of a vertical circle showing tension and weight vectors

Deriving the minimum speed to complete the loop

The goal of this derivation is to combine the top-of-loop force equation with energy conservation between the bottom and top of the loop so as to find the two critical speeds — the minimum speed at the top below which the string cannot stay taut, and the corresponding minimum speed at the bottom needed to reach it, both expressed purely in terms of $g$ and the radius $r$ .

Step 1. Start from the critical condition at the top. A string can only pull, so $T \ge 0$ . The slowest possible motion that still keeps the string taut is the boundary case $T = 0$ , where gravity alone supplies the entire centripetal force:

mg = \frac{mv_{top,\,min}^2}{r} \implies v_{top,\,min} = \sqrt{gr}

If $v_{top}$ were any smaller, the equation would demand $T < 0$ — a push from the string — which is impossible, so the body leaves the circular path before reaching the top.

Step 2. Bring in energy conservation to relate this top speed to the speed at the bottom. Taking the bottom of the loop as the reference height, the top is higher by $2r$ (the diameter). With no friction or air resistance, mechanical energy is conserved between bottom and top:

\frac12 mv_{bottom}^2 = \frac12 mv_{top}^2 + mg(2r)

v_{bottom}^2 = v_{top}^2 + 4gr

Step 3. Substitute the critical top speed from Step 1 into the energy relation from Step 2. Since $v_{top,\,min}^2 = gr$ , the minimum speed the body needs at the bottom to just complete the loop is:

v_{bottom,\,min}^2 = gr + 4gr = 5gr \implies v_{bottom,\,min} = \sqrt{5gr}

Step 4. Read the result physically. Completing a vertical circular loop needs two separate conditions checked together — a dynamics condition ( $T \ge 0$ at the top) and a kinematics/energy condition (enough speed at the bottom to still be moving fast enough at the top). $v_{bottom,\,min} = \sqrt{5gr}$ is the number every problem eventually reduces to, and it is always larger than $v_{top,\,min} = \sqrt{gr}$ because the body must climb height $2r$ against gravity on the way up.

Worked example

A small ball of mass $0.5\ \text{kg}$ is attached to a string of length $0.8\ \text{m}$ and whirled in a vertical circle. Find (a) the minimum speed at the top so the string does not go slack, (b) the corresponding minimum speed at the bottom, and (c) the tension at the bottom if the ball actually moves at $5\ \text{m/s}$ there.

Step 1. Minimum speed at the top, using $v_{top,\,min} = \sqrt{gr}$ with $g = 9.8\ \text{m/s}^2$ and $r = 0.8\ \text{m}$ :

v_{top,\,min} = \sqrt{9.8 \times 0.8} = \sqrt{7.84} = 2.8\ \text{m/s}

Step 2. Minimum speed at the bottom, using $v_{bottom,\,min} = \sqrt{5gr}$ :

v_{bottom,\,min} = \sqrt{5 \times 9.8 \times 0.8} = \sqrt{39.2} \approx 6.26\ \text{m/s}

Step 3. Now suppose the ball is actually whirled with $v = 5\ \text{m/s}$ at the bottom (a mass $m = 0.5\ \text{kg}$ this time is not needed for parts a and b, but is needed here). Using the bottom equation $T - mg = mv^2/r$ :

T = mg + \frac{mv^2}{r} = (0.5)(9.8) + \frac{(0.5)(5)^2}{0.8} = 4.9 + \frac{12.5}{0.8} = 4.9 + 15.625 = 20.525\ \text{N}

The ball needs at least $2.8\ \text{m/s}$ at the top and $6.26\ \text{m/s}$ at the bottom to complete the loop; at an actual bottom speed of $5\ \text{m/s}$ the string tension there is $20.5\ \text{N}$ , well above the ball's weight of $4.9\ \text{N}$ , since the string must also supply the centripetal force.

Competitive stretch

If instead of a string the ball sits in a smooth circular groove (a track that can push as well as pull), the "string goes slack" restriction disappears — the track's normal reaction can point either toward or away from the centre. The body can then complete the loop at speeds below $\sqrt{5gr}$ at the bottom, because the track (unlike a string) is able to push outward on the body if needed. This is why loop-the-loop roller-coasters use rigid rails, not cables.

Common traps

Treating the speed as constant around the loop, as it is in horizontal circular motion — gravity does work as height changes, so speed varies continuously in a vertical circle.
Forgetting the string can only pull ( $T \ge 0$ ), so the critical condition at the top is $T = 0$ , not some other special value of $T$ .
Getting the sign of gravity's contribution backwards — at the top gravity acts toward the centre (adds to $T$ in the force equation); at the bottom it acts away from the centre (subtracts from $T$ ).
Confusing $v_{top,\,min} = \sqrt{gr}$ (the critical speed at the top) with $v_{bottom,\,min} = \sqrt{5gr}$ (the critical speed at the bottom needed to reach that top speed) — they refer to different points on the loop and are linked only through energy conservation.

Ask the tutor: Want to work through a roller-coaster or bucket-of-water version of this problem, or see what changes when the track can push as well as pull?

Why does the string go slack near the top and not near the bottom?

At the bottom, tension and gravity oppose each other in the force equation ( $T = mg + mv^2/r$ ), so $T$ is always comfortably positive — it only grows as speed increases. At the top, gravity already points toward the centre and does part of the job, so the equation becomes $T = mv^2/r - mg$ . If the speed there is too small, this can go negative, which a string cannot supply, so it simply goes slack instead. The top is always the critical point to check.

Does the mass of the ball affect the minimum speed needed to complete the loop?

No. Both $v_{top,\,min} = \sqrt{gr}$ and $v_{bottom,\,min} = \sqrt{5gr}$ come from equations where $m$ cancels out — it appears on both sides of $mg = mv^2/r$ and of the energy-conservation equation. The minimum speeds depend only on $g$ and the radius $r$ of the loop, not on how heavy the ball, bucket, or car is. Mass does matter, however, when you calculate the actual tension or normal force at a given speed, since those equations keep an explicit $m$ .

Is the vertical-circle case just uniform circular motion with an extra force added?

Not quite — the key difference is that the motion is no longer uniform. In horizontal circular motion the speed stays constant because gravity acts perpendicular to the plane of motion (balanced by a normal force) and does no work. In a vertical circle, gravity has a component along the direction of motion for most of the loop, so it continuously does positive or negative work, changing the speed. That is why you need energy conservation, not just the centripetal force equation, to fully solve vertical-circle problems.