Vertical circular motion — minimum speed conditions: Question bank

Setup you must agree on first (read before the questions)

Definition Sign convention — "inward is positive"

Because circular motion needs a net force pointing toward the centre, we choose the radial axis pointing inward (toward the centre) as positive. The tangential axis points along the direction of motion (perpendicular to the radius); we use it only when a force speeds the ball up or slows it down.

Why inward-positive? Newton's second law along this axis reads $F_{net, inward} = m a_{centripetal} = \frac{m v ^{2}}{r}$ . Making inward positive means the required side is $+ \frac{m v ^{2}}{r}$ , so any force pointing inward enters with a $+$ sign and any force pointing outward with a $-$ sign. No guessing.

Apply it once and forever:

At the top, the centre is below, so "inward" = down. Gravity (down) is $+ m g$ , tension (down, toward centre) is $+ T$ : $T + m g = \frac{m v ^{2}}{r}$ .
At the bottom, the centre is above, so "inward" = up. Tension (up, toward centre) is $+ T$ , gravity (down, away from centre) is $- m g$ : $T - m g = \frac{m v ^{2}}{r}$ .

Figure — Vertical circular motion — minimum speed conditions

The free-body diagrams above are the whole game. Notice there is no "centripetal" arrow drawn — only the real forces ( $m g$ and $T$ ). The quantity $\frac{m v ^{2}}{r}$ lives on the other side of the equation as the required inward net.

$\frac{1}{2} m v_{bot}^{2} = \frac{1}{2} m v_{top}^{2} + m g (2 r)$

Multiply through by $2/ m$ : $v_{bot}^{2} = v_{top}^{2} + 4 g r$ . That " $+ 4 g r$ " is the climbing tax you'll meet again and again below. (Energy conservation applies because tension is perpendicular to the motion and does zero work.)

True or false — justify

True or false: At the top of the loop, tension and gravity point in opposite directions.

False. The centre of the circle is below the ball at the top, so the string pulls downward toward it — same direction as gravity. Using inward-positive (inward = down here), both are positive and they add:

T + m g = \frac{m v ^{2}}{r}

True or false: Centripetal force is an extra arrow you must draw on the free-body diagram.

False. On the free-body diagram you draw only real forces you can name — here just gravity

m g

and tension

T

(look at figure s01). The quantity

\frac{m v ^{2}}{r}

is not a force; it is the required inward net that those real forces must add up to, so it sits on the other side of the equation, never as its own arrow.

True or false: The minimum speed to complete the loop is

g r

False as stated.

g r

is the minimum speed at the top only. To complete the whole loop you must enter the bottom with

5 g r

: the top needs

g r

, and the climb adds

4 g r

v^{2}

, so

v_{bot}^{2} = g r + 4 g r = 5 g r

True or false: A string can supply either a push or a pull to keep the ball circling.

False. A string can only pull, so

T \geq 0

. The whole "

g r

" minimum comes from this one-sided constraint: the moment the equation would demand

T < 0

(a push), the string goes slack instead.

True or false: For a ball on a rigid rod, the minimum speed at the top is still

g r

False. A rod can push as well as pull, so

T

may be negative; the slack condition

T \geq 0

vanishes. The rod's only requirement is

v_{top} \geq 0

, giving minimum top speed

0

True or false: The tension difference

T_{bot} - T_{top} = 6 m g

only holds at the minimum speed.

False — it holds for any speed. From the two setup equations:

T_{bot} = \frac{m v _{bot}^{2}}{r} + m g

and

T_{top} = \frac{m v _{top}^{2}}{r} - m g

. Subtracting, the two

\pm m g

terms combine to

+ 2 m g

, and

\frac{m}{r} (v_{bot}^{2} - v_{top}^{2}) = \frac{m}{r} (4 g r) = 4 m g

using the climbing tax. Total:

2 m g + 4 m g = 6 m g

, independent of speed.

True or false: At the minimum-speed top, the ball is momentarily weightless relative to the string.

True. With

T = 0

, the string exerts nothing; gravity alone curves the ball, so it feels no support force — like free fall along a curve. This is exactly why the water doesn't spill.

True or false: Doubling the radius doubles the minimum top speed.

False.

v_{top,min} = g r

, so doubling

r

multiplies the speed by

2

, not

2

. Speed scales with the square root of radius.

Spot the error

"At the top, net force

= T - m g = \frac{m v ^{2}}{r}

, so minimum speed sets

T = 0

giving

- m g = \frac{m v ^{2}}{r}

The sign of

T

is wrong. With inward-positive at the top (inward = down), tension points down toward the centre, so it enters as

+ T

, not

- T

T + m g = \frac{m v ^{2}}{r}

. Setting

T = 0

gives

m g = \frac{m v ^{2}}{r}

, hence

v = g r

"Energy:

\frac{1}{2} m v_{bot}^{2} = \frac{1}{2} m v_{top}^{2} + m g r

, since the top is a height

r

above the bottom."

The top of a loop of radius

r

is a height

2 r

(a full diameter) above the bottom, not

r

— see figure s02. The correct term is

m g (2 r)

, which produces

v_{bot}^{2} = 5 g r

, not

3 g r

"At minimum speed the ball has zero speed at the top, so it just barely reaches it."

That's the rod case. For a string, the ball still moves at

g r

at the top — zero speed would mean the string went slack long before. "Just reaches" (zero top speed) needs a rod or track that can push.

"Tension does work on the ball as it rises, so we can't use energy conservation."

Tension is always perpendicular to the motion (it points along the radius, the ball moves tangentially), so it does zero work. That's precisely why energy conservation is valid here.

"At the bottom,

T_{bot} = \frac{m v ^{2}}{r}

because that's the centripetal force."

Gravity is missing. At the bottom, inward = up: tension is

+ T

, gravity is

- m g

, so

T_{bot} - m g = \frac{m v ^{2}}{r}

, giving

T_{bot} = m g + \frac{m v ^{2}}{r}

— larger, not equal.

"On a roller-coaster loop, the seat pushes the rider up at the top."

The track/seat is outside the car, so at the top it can only push inward = downward (a normal force

N \geq 0

toward the centre). Together with gravity it supplies the centripetal force; it never pushes outward.

Why questions

Why does gravity "help" at the top but "fight" at the bottom?

The centripetal requirement always points inward. At the top the centre is below, so gravity (down) points inward and enters as

+ m g

— it helps. At the bottom the centre is above, so gravity (down) points outward, entering as

- m g

— tension must overcome it.

Why must the bottom be faster than the top by a fixed energy amount?

Climbing from bottom to top raises the ball by height

2 r

(figure s02), converting kinetic energy into gravitational potential energy. That "tax" is

m g (2 r)

; dividing the energy equation by

\frac{m}{2}

shows it adds

4 g r

v^{2}

v_{bot}^{2} = v_{top}^{2} + 4 g r

Why does the water not fall out of an upside-down bucket at the top?

At (or above) minimum speed, gravity provides at most the centripetal force needed, so the bucket bottom accelerates downward at least as fast as the water would fall. The water and bucket fall together — the bucket never pulls away, so nothing spills.

Why can't we just set

\frac{m v ^{2}}{r} = 0

to find the string's minimum speed at the top?

Follow the logic in order. (1) The physical limit is the string going slack, i.e.

T = 0

— not zero required force. (2) Put

T = 0

into

T + m g = \frac{m v ^{2}}{r}

: the left side is still

m g

, because gravity hasn't vanished. (3) So the circle must still demand

\frac{m v ^{2}}{r} = m g

, giving

v = g r

. Setting

\frac{m v ^{2}}{r} = 0

instead would wrongly assume gravity switches off.

Why is the minimum-speed condition a string/track rule and not a law of physics?

It comes entirely from the constraint

T \geq 0

(strings and outer tracks can only pull/push inward). Change the connector — a rod, a tube, or an inner track — and the constraint changes. It's about the connector, not the motion.

Why does the required centripetal force shrink as the ball slows at the top?

\frac{m v ^{2}}{r}

depends on

v^{2}

. Slower

v

→ smaller required inward force. Since gravity

m g

is fixed, tension must drop to make up the difference; once the requirement falls below

m g

, tension would need to go negative — impossible for a string.

Edge cases

What happens if the ball at the top is faster than

g r

The circle demands more than

m g

, so the string provides the extra:

T = \frac{m v ^{2}}{r} - m g > 0

. The string stays taut — completely fine, just tighter.

What happens at exactly

v_{top} = g r

Tension is exactly zero — the string is on the verge of going slack but still traces a perfect circle, with gravity alone as the centripetal force. This is the boundary between "loops fine" and "falls out".

What if the ball is on a rigid rod and moves slower than

g r

at the top?

Perfectly allowed. The rod simply pushes outward (

T < 0

) to reduce the net inward force to

\frac{m v ^{2}}{r}

. Down to zero top speed the rod keeps it on the circle.

What is the minimum bottom speed for the rod case, and why is it smaller?

v_{bot,min} = 2 g r

(from

v_{bot}^{2} = 0 + 4 g r

), because the rod only needs the ball to reach the top (

v_{top} = 0

), paying just the

4 g r

climbing tax — no extra

g r

for staying taut.

What happens for a ball inside a smooth tube (walls on both sides)?

A tube can push both inward and outward, so like the rod it removes the slack condition: minimum top speed is

0

, and

v_{bot,min} = 2 g r

. Compare with the outer-track/string case where only inward push exists.

If the speed at the bottom is between

2 g r

and

5 g r

(string case), where exactly does the string go slack?

At the angle

θ

above the horizontal (measured from the centre) where tension first hits zero. Let

v_{θ}^{2} = v_{bot}^{2} - 2 g r (1 + sin θ)

(height above bottom is

r (1 + sin θ)

). Setting

T = 0

in the radial equation

T + m g sin θ = \frac{m v _{θ}^{2}}{r}

gives the slack condition

sin θ = \frac{v _{bot}^{2} - 2 g r}{3 g r}

. For example with

v_{bot}^{2} = 4 g r

this gives

sin θ = \frac{2}{3}

(about

4 2^{\circ}

above horizontal); beyond that point the ball leaves the circle as a projectile.

At the horizontal side (height

r

), is any of the ball's weight part of the centripetal force?

No. At the side the centre is horizontal, so gravity (vertical) is entirely tangential — it changes the ball's speed, not its direction. Tension alone provides the centripetal force there:

T = \frac{m v ^{2}}{r}

Connections

Centripetal force and acceleration
Newton's Second Law — net force form
Conservation of mechanical energy
Tension and constraint forces
Normal force in circular tracks
Banked curves and horizontal circular motion
Free-body diagrams