Motion in a Vertical Circle

In short

When an object moves in a vertical circle of radius r, gravity makes the speed vary — fastest at the bottom, slowest at the top. At the top of the loop, the minimum speed to keep the string taut is v_{\text{top}} = \sqrt{gr}. Using energy conservation, the minimum speed at the bottom is v_{\text{bottom}} = \sqrt{5gr}. If the speed drops below these critical values, the string goes slack and the object falls away from the circular path.

Tie a stone to a string and swing it in a vertical circle over your head. If you swing fast enough, the stone traces a smooth loop — around and around, like Diwali sparklers tracing glowing rings in the night sky. But slow down even a little, and the stone fails to make it over the top. The string goes slack, the circle breaks, and the stone drops like a regular projectile.

What decides whether the stone completes the loop or not? It is not just speed — it is the relationship between speed, gravity, and the radius of the circle. And the physics that governs this relationship is a beautiful interplay of Newton's second law and energy conservation.

Think about swinging a bucket of water over your head — a classic trick where the water stays in the bucket even when it is upside down at the top. The bucket needs to move fast enough that the required centripetal acceleration exceeds g. Too slow, and the water falls out. That critical speed is what this article derives.

Why speed changes around the circle

In horizontal circular motion, speed stays constant — gravity acts perpendicular to the motion and does no work. In a vertical circle, gravity is no longer perpendicular. As the object rises from the bottom to the top, gravity opposes the motion and slows it down. As it falls from the top back to the bottom, gravity assists the motion and speeds it up.

The result: the object is fastest at the bottom and slowest at the top. The speed at every point around the circle is different, and so is the tension in the string. This is what makes vertical circular motion fundamentally different from the uniform circular motion you studied earlier.

Forces at each position — the free body diagrams

To understand the tension, you need to draw free body diagrams at three key positions: the bottom of the circle, the top, and the side. At each position, the net radial force provides the centripetal acceleration v^2/r.

The vertical circle with free body diagrams at three key positions. At the top, both tension and weight point toward the centre. At the bottom, tension points toward the centre while weight points away. At the side, only tension provides the centripetal force; weight acts downward (tangentially at this point).

At the top of the circle

At the topmost point, the centre of the circle is directly below the ball. Both the tension T and the weight mg point downward — toward the centre. The net centripetal force is:

T + mg = \frac{mv_{\text{top}}^2}{r} \tag{1}

Why: both forces act in the same direction (toward the centre), so they add. The right side is the centripetal force required for circular motion at speed v_{\text{top}}.

At the bottom of the circle

At the lowest point, the centre is directly above the ball. Tension points upward (toward the centre) and weight points downward (away from the centre). The net centripetal force is:

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

Why: tension and weight oppose each other. The tension must overcome the weight and still provide the centripetal force. This is why the tension is greatest at the bottom.

At the side (level with the centre)

When the ball is at the 3 o'clock or 9 o'clock position, the string is horizontal. Tension points toward the centre (horizontally), and weight acts vertically downward — perpendicular to the string. Only the tension provides the centripetal force:

T = \frac{mv_{\text{side}}^2}{r} \tag{3}

Why: at this position, mg is entirely tangential (it slows the ball down as it rises, or speeds it up as it descends) and contributes nothing to the centripetal direction.

The critical speed at the top

Here is the key question: what is the minimum speed the ball can have at the top and still maintain a circular path?

The string can only pull, not push. So the tension T can be zero (string goes limp) but never negative. The critical condition is T = 0 at the top. Set T = 0 in equation (1):

0 + mg = \frac{mv_{\text{min,top}}^2}{r}

mg = \frac{mv_{\text{min,top}}^2}{r}

Why: when the tension is zero, gravity alone provides the centripetal force. The string is on the verge of going slack — one bit slower and the ball would fall away from the circle.

Cancel m from both sides:

g = \frac{v_{\text{min,top}}^2}{r}

\boxed{v_{\text{min,top}} = \sqrt{gr}} \tag{4}

Why: the mass cancels — the critical speed depends only on the radius of the circle and the gravitational acceleration, not on the mass of the object. A heavy ball and a light ball on strings of the same length both need the same minimum speed at the top.

For a circle of radius 1 m: v_{\text{min,top}} = \sqrt{9.8 \times 1} \approx 3.13 m/s. That is roughly 11 km/h — a brisk jog. Below this speed, the string goes slack and the ball departs from the circular path.

What happens below the critical speed?

If the ball's speed at the top drops below \sqrt{gr}, equation (1) would require a negative tension to maintain circular motion. But a string cannot push — it can only pull. So the string goes slack (T = 0), the radial constraint disappears, and the ball becomes a projectile. It follows a parabolic trajectory under gravity alone until the string becomes taut again (if it does at all).

This is why swinging a bucket of water over your head requires confidence and speed. If you hesitate at the top and slow down, the water falls out — and possibly onto you.

Connecting top and bottom speeds with energy conservation

You now know the minimum speed at the top. But the ball is launched from the bottom — so you need the minimum speed at the bottom that guarantees the ball reaches the top with at least \sqrt{gr}.

This is where energy conservation enters. Take the bottom of the circle as the reference level for potential energy (h = 0). The top of the circle is at height h = 2r (a full diameter above the bottom).

Assumptions: The string is massless and inextensible. No air resistance. No energy is lost to friction.

Apply conservation of mechanical energy between the bottom and the top:

\frac{1}{2}mv_{\text{bottom}}^2 + 0 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)

Why: at the bottom, all energy is kinetic (PE = 0 at the reference level). At the top, energy is split between kinetic and gravitational potential. Total mechanical energy is conserved because only gravity and the tension do work — and the tension, being perpendicular to the velocity, does zero work.

Step 1. Simplify. Cancel \frac{1}{2}m from the kinetic energy terms:

v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr \tag{5}

Why: this is the fundamental relationship connecting speeds at the top and bottom. The 4gr term is the "gravitational tax" — the kinetic energy the ball must surrender to climb through a height of 2r.

Step 2. Substitute the critical condition v_{\text{top}} = \sqrt{gr} (i.e., v_{\text{top}}^2 = gr):

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

\boxed{v_{\text{min,bottom}} = \sqrt{5gr}} \tag{6}

Why: the minimum speed at the bottom must provide enough kinetic energy for two things: (1) climbing through a height 2r, which costs 4gr worth of v^2, and (2) still moving at \sqrt{gr} at the top to maintain the circular path. Adding these gives 5gr.

For a circle of radius 1 m: v_{\text{min,bottom}} = \sqrt{5 \times 9.8 \times 1} = \sqrt{49} = 7 m/s — about 25 km/h. That is the speed of a fast cyclist. Anything slower, and the ball will not make it over the top.

Speed at the side

At the 3 o'clock position (level with the centre), the height above the bottom is r (one radius). Using energy conservation between the bottom and the side:

\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{side}}^2 + mgr

v_{\text{side}}^2 = v_{\text{bottom}}^2 - 2gr \tag{7}

At the critical condition (v_{\text{bottom}}^2 = 5gr):

v_{\text{side}}^2 = 5gr - 2gr = 3gr

v_{\text{side}} = \sqrt{3gr}

Why: the ball has climbed only half the total height (r instead of 2r), so it has lost less speed than at the top. The speed at the side is intermediate between the bottom speed (\sqrt{5gr}) and the top speed (\sqrt{gr}).

Tension at each position — the full picture

Now you can find the tension at every key point. Use the Newton's second law equations (1), (2), (3) with the speeds from energy conservation.

Tension at the bottom

From equation (2): T_{\text{bottom}} - mg = mv_{\text{bottom}}^2/r

T_{\text{bottom}} = mg + \frac{mv_{\text{bottom}}^2}{r}

At the critical speed (v_{\text{bottom}}^2 = 5gr):

T_{\text{bottom}} = mg + \frac{m(5gr)}{r} = mg + 5mg = 6mg \tag{8}

Why: the tension at the bottom is a full six times the weight. The string must support the weight and provide the centripetal force for a ball moving at \sqrt{5gr}. This is the maximum tension anywhere in the loop — the string is most likely to break at the bottom, not the top.

Tension at the top

From equation (1): T_{\text{top}} = mv_{\text{top}}^2/r - mg

At the critical speed (v_{\text{top}}^2 = gr):

T_{\text{top}} = \frac{m(gr)}{r} - mg = mg - mg = 0

Why: this confirms the critical condition. At the minimum speed, the tension at the top is exactly zero — gravity alone provides the centripetal force.

If the ball moves faster than the critical speed, v_{\text{top}} > \sqrt{gr}, the tension at the top becomes positive — the string pulls the ball inward beyond what gravity provides.

Tension at the side

From equation (3): T_{\text{side}} = mv_{\text{side}}^2/r

At the critical speed (v_{\text{side}}^2 = 3gr):

T_{\text{side}} = \frac{m(3gr)}{r} = 3mg

Why: at the side, the weight is tangential and does not contribute to the centripetal direction, so the tension alone must provide all the centripetal force. The tension at the side (3mg) is between the top (0) and the bottom (6mg).

Summary of the critical-speed results:

Position	Height above bottom	Speed	Tension
Bottom	0	\sqrt{5gr}	6mg
Side	r	\sqrt{3gr}	3mg
Top	2r	\sqrt{gr}	0

The tension changes by 6mg from bottom to top — a dramatic variation. This is why a ball on a string in a vertical circle is so different from one in a horizontal circle, where the tension stays constant.

Worked examples

Example 1: Ball on a string in a vertical circle

A ball of mass 200 g is tied to a string of length 1 m and swung in a vertical circle. Find: (a) the minimum speed at the top, (b) the minimum speed at the bottom to complete the loop, and (c) the tension in the string at the bottom when the ball just barely completes the loop.

Take g = 9.8 m/s².

The vertical circle for Example 1. At the top, the ball barely maintains the loop with T = 0 and speed 3.13 m/s. At the bottom, the string bears 11.76 N — six times the ball's weight.

Given: m = 0.2 kg, r = 1 m, g = 9.8 m/s².

(a) Minimum speed at the top.

At the top, the critical condition is T = 0, so gravity alone provides the centripetal force:

v_{\text{min,top}} = \sqrt{gr} = \sqrt{9.8 \times 1} = \sqrt{9.8}

\boxed{v_{\text{min,top}} \approx 3.13 \text{ m/s}}

Why: the mass does not appear — any ball on a 1 m string needs the same minimum speed at the top. The string is on the verge of going slack.

(b) Minimum speed at the bottom.

Use equation (6):

v_{\text{min,bottom}} = \sqrt{5gr} = \sqrt{5 \times 9.8 \times 1} = \sqrt{49}

\boxed{v_{\text{min,bottom}} = 7 \text{ m/s}}

Why: the ball must have enough kinetic energy to climb through 2r = 2 m while retaining \sqrt{gr} at the top. The factor of 5 comes from adding the 4gr cost of climbing to the gr needed at the top.

(c) Tension at the bottom.

From equation (8):

T_{\text{bottom}} = 6mg = 6 \times 0.2 \times 9.8

\boxed{T_{\text{bottom}} = 11.76 \text{ N}}

Why: 6mg means the tension is six times the weight. The ball's weight is mg = 0.2 \times 9.8 = 1.96 N, and the tension is 11.76 N — quite a lot for a small ball. If the string has a breaking strength below 11.76 N, it will snap at the bottom, not the top.

Verification. Check energy conservation: KE at bottom = \frac{1}{2}(0.2)(7)^2 = 4.9 J. KE at top = \frac{1}{2}(0.2)(3.13)^2 \approx 0.98 J. PE gained = mg(2r) = 0.2 \times 9.8 \times 2 = 3.92 J. KE at top + PE = 0.98 + 3.92 = 4.9 J. Energy is conserved.

What this shows: A ball on a 1 m string needs at least 7 m/s at the bottom to complete the vertical circle. The tension swings from zero at the top to nearly 12 N at the bottom. The string is under the most stress at the lowest point, not the highest.

Example 2: Toy car on a loop-the-loop track

A toy car rolls down a frictionless ramp and enters a circular loop of radius r = 20 cm (0.2 m). What is the minimum height h of the ramp above the bottom of the loop for the car to complete the loop without losing contact with the track?

The loop-the-loop setup. The car starts from rest at height $h$ and must reach the top of the loop (height $2r$) with enough speed to maintain contact. The minimum ramp height turns out to be $h = 2.5r$.

Key difference from the string problem: This is a track, not a string. The track can push (normal force) but cannot pull. At the top of the loop, the car is on the inside of the track, so the normal force N points downward (toward the centre), just like the tension did. The critical condition is the same: N = 0 at the top.

Step 1. Apply Newton's second law at the top of the loop.

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

At the critical condition (N = 0): v_{\text{top}}^2 = gr.

Why: this is identical to the string result. Whether it is tension or normal force providing the constraint, the critical speed at the top is \sqrt{gr}.

Step 2. Apply energy conservation from the ramp top to the loop top.

The car starts from rest (v = 0) at height h. At the top of the loop, the car is at height 2r with speed v_{\text{top}}.

mgh = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)

Why: all the initial potential energy converts to kinetic energy plus the remaining potential energy at the top of the loop. No friction means no energy loss.

Step 3. Substitute v_{\text{top}}^2 = gr and solve for h.

mgh = \frac{1}{2}m(gr) + mg(2r)

Cancel m:

gh = \frac{1}{2}gr + 2gr = \frac{5}{2}gr

h = \frac{5}{2}r = 2.5r

\boxed{h = 2.5r = 2.5 \times 0.2 = 0.5 \text{ m}}

Why: the ramp height must be 2.5 times the loop radius. The factor 2.5 comes from two contributions: 2r for climbing to the top of the loop, plus \frac{1}{2}r for arriving with the critical speed \sqrt{gr} at the top.

Result: The ramp must be at least 0.5 m (50 cm) high — that is 2.5 times the loop radius.

What this shows: The minimum ramp height h = 2.5r is a universal result — it does not depend on the car's mass or on g. A heavy car and a light car both need the same ramp height to complete the same loop. This is why toy loop-the-loop tracks work reliably: the manufacturer just needs to make the ramp 2.5 times taller than the loop radius.

Common confusions

"The string is most likely to break at the top." This is backwards. The tension is maximum at the bottom (6mg at critical speed) and minimum at the top (zero at critical speed). If the string is going to break, it will break at the bottom. The top is where the string goes slack, not where it snaps.
"The ball slows down because of tension." No — tension is always along the string, which is along the radius. Since the velocity is tangent to the circle and the radius is perpendicular to the tangent, the tension is always perpendicular to the velocity. A force perpendicular to the velocity does zero work and cannot change the speed. It is gravity that changes the speed — gravity has a component along the direction of motion (tangential component) that does work on the ball.
"At the top of the loop, the ball is weightless." Not quite. The ball's weight mg still acts downward — gravity does not switch off. What happens at the critical speed is that mg alone provides the centripetal force, and the string tension is zero. The ball is in a state similar to free fall — it "feels" weightless in the sense that there is no contact force — but gravity is very much present. It is the same sensation astronauts feel in orbit.
"The formula v = √(5gr) works at the top." No — v = \sqrt{5gr} is the critical speed at the bottom. At the top, the critical speed is v = \sqrt{gr}. These two are connected by energy conservation, but they apply at different points.
"A bead on a track and a ball on a string have the same critical speed." Only at the top. At the bottom, the physics is the same because the constraint force (tension or normal force) acts the same way. But there is a subtle difference: a string can go slack (lose contact) only by failing to pull, while a bead on a track can lose contact only by failing to push. For the inside of a loop-the-loop, the analysis is identical. But for the outside of a curve (like the top of a hill), the normal force reversal means the critical condition is different — see the Going Deeper section.

If you came here to find the critical speed, derive the tension at each point, and solve problems, you have everything you need. What follows is for readers who want to explore the general angle formula, the bead-on-a-track variant, and the case where the circle is not completed.

Tension as a function of angle

Let \theta be the angle measured from the bottom of the circle (so \theta = 0 at the bottom, \theta = \pi at the top). At a general angle \theta, the height of the ball above the bottom is h = r - r\cos\theta = r(1 - \cos\theta).

Using energy conservation between the bottom and angle \theta:

\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv^2 + mgr(1 - \cos\theta)

v^2 = v_{\text{bottom}}^2 - 2gr(1 - \cos\theta) \tag{9}

Why: the speed at any angle depends on how much height the ball has gained. At \theta = 0, v = v_{\text{bottom}}. At \theta = \pi, v^2 = v_{\text{bottom}}^2 - 4gr, matching equation (5).

At a general angle, the radial direction points toward the centre. The component of weight along the radial direction (toward the centre) is mg\cos\theta when \theta is measured from the bottom. Newton's second law in the radial direction gives:

T - mg\cos\theta = \frac{mv^2}{r}

T = \frac{mv^2}{r} + mg\cos\theta \tag{10}

Substitute v^2 from equation (9):

T = \frac{m}{r}\left[v_{\text{bottom}}^2 - 2gr(1 - \cos\theta)\right] + mg\cos\theta

T = \frac{mv_{\text{bottom}}^2}{r} - 2mg(1 - \cos\theta) + mg\cos\theta

T = \frac{mv_{\text{bottom}}^2}{r} - 2mg + 2mg\cos\theta + mg\cos\theta

T = \frac{mv_{\text{bottom}}^2}{r} - 2mg + 3mg\cos\theta

\boxed{T = \frac{mv_{\text{bottom}}^2}{r} + mg(3\cos\theta - 2)} \tag{11}

Why: this single formula gives the tension at any position on the circle, as a function of the speed at the bottom and the angle. At \theta = 0: T = mv_{\text{bottom}}^2/r + mg, which matches equation (2). At \theta = \pi: T = mv_{\text{bottom}}^2/r - 5mg, and setting v_{\text{bottom}}^2 = 5gr gives T = 5mg - 5mg = 0, confirming the critical condition.

The tension decreases as the ball rises (as \cos\theta decreases from 1 to -1) and reaches its minimum at the top. This equation tells you that the tension varies linearly with \cos\theta — the 3mg\cos\theta term is the dominant angular variation.

The bead on a frictionless circular track

A bead threaded on a vertical circular wire (or a car on a loop-the-loop track) differs from a ball on a string in one important way: the track can exert a normal force in either direction — inward or outward. The string can only pull (inward).

For the bead on the inside of the loop (like a car on the inside of a loop-the-loop), the analysis is identical to the string: the normal force replaces the tension, both point toward the centre, and the critical condition is N = 0 at the top with v_{\text{top}} = \sqrt{gr}.

But consider a bead on the outside of a circular hump — like a car going over the top of a hill. At the top of the hump, the weight mg points downward (toward the centre), while the normal force N points upward (away from the centre):

mg - N = \frac{mv^2}{r}

Here the critical condition is N = 0, which gives v_{\text{critical}} = \sqrt{gr} — the maximum speed at which the car can stay on the road. Beyond this speed, the car would need a negative normal force (the road would need to pull it down), which a road cannot do. The car lifts off.

This is exactly what happens when you drive fast over a speed bump — your stomach drops because the normal force temporarily decreases. Drive fast enough, and the car leaves the ground entirely.

Where the string goes slack — the departure point

If the ball's speed at the bottom is less than \sqrt{5gr}, it will not complete the circle. The string goes slack at some angle \theta_s where the tension first reaches zero. Setting T = 0 in equation (11):

0 = \frac{mv_{\text{bottom}}^2}{r} + mg(3\cos\theta_s - 2)

\cos\theta_s = \frac{2}{3} - \frac{v_{\text{bottom}}^2}{3gr}

Why: this gives the exact angle at which the string goes slack. If v_{\text{bottom}}^2 = 5gr, then \cos\theta_s = 2/3 - 5/3 = -1, meaning \theta_s = \pi — the string just barely stays taut all the way to the top. For lower speeds, \theta_s < \pi — the string goes slack before the top.

After the string goes slack, the ball is in free fall (projectile motion) until the string becomes taut again. The trajectory during the slack phase is a parabola, not a circle. This is the "messy" physics that happens below the critical speed — and it is why the critical speed calculation matters: it tells you the boundary between clean circular motion and complicated projectile motion.

Energy approach for a general starting height

For a ball released from rest at an angle \theta_0 from the bottom (height r(1 - \cos\theta_0)), the analysis generalises cleanly. Energy conservation gives the speed at any other angle, and the tension formula (11) still applies with v_{\text{bottom}}^2 replaced by the speed at the bottom (found from energy conservation). JEE Advanced problems often start the ball from an arbitrary position, not from the bottom — but the method is the same. Find the speed at the point of interest using energy conservation, then use Newton's second law in the radial direction to find the constraint force.

Where this leads next

Centripetal Force — the foundational concept of radial acceleration and the force required for circular motion.
Work and Energy — the conservation law that connects speeds at different points in the circle.
Conical Pendulum and Banking of Roads — circular motion in a horizontal plane, where gravity is perpendicular to the motion.
Centrifugal Force and Non-Inertial Frames — analysing the same problem from the rotating frame of the ball.