Visual walkthrough — Vertical circular motion — minimum speed conditions

This is the visual companion to the parent topic. Read that first if you want the words; read this if you want the pictures.

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Step 1 — What "moving in a circle" demands

WHAT. Look at the figure. A ball sits somewhere on a circle. The red arrow points from the ball straight to the center.

WHY. To turn — to keep curving instead of flying off in a straight line — an object always needs a net force pointing toward the center. Straight-line motion needs no sideways force; curving does. That inward-pointing need is the whole story of this chapter.

PICTURE. Notice the red inward arrow rotates as the ball moves: at the top it points down, at the bottom it points up, at the side it points sideways. Same rule, different direction. Keep this in mind — it is the single fact that makes the top and bottom behave so differently.

See Centripetal force and acceleration for where $\frac{m{v}^2}{r}$ comes from.

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Step 2 — The forces at the top (draw them, don't guess)

WHAT. The ball is at the very top. The center is below it. So the string, running from the ball to the center, pulls downward. Gravity also pulls downward. Both real forces point the same way.

WHY draw it first? Because the most common mistake (see the parent's mistake box) is to assume the string pulls up at the top out of habit. The picture kills that habit: the center is below, so the pull is down.

PICTURE. Two black arrows leave the red ball, both pointing down: the short one is $T$, the long one is $mg$. Their combined length is the total inward force available at the top.

This is just Newton's Second Law — net force form with "inward" chosen as positive.

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Step 3 — Squeeze the speed down and watch tension vanish

WHAT. Rearrange Step 2 to isolate the tension: $T=\frac{m{v}_{\text{top}}^2}{r}-mg$

WHY. We want to know what happens as the ball goes slower at the top. Slower means smaller $v_{\text{top}}$, which shrinks $\frac{m{v}_{\text{top}}^2}{r}$. Since $mg$ is fixed, the only thing that can shrink to match is $T$.

PICTURE. In the figure, three copies of the ball at the top for three speeds: fast (long $T$ arrow), slower (short $T$), and the critical speed (no $T$ arrow at all — only gravity). The red arrow is the tension, and you literally watch it shorten to nothing.

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Step 4 — The edge: set the tension to exactly zero

WHAT. Put $T=0$ into Step 2: $0+mg=\frac{m{v}_{\text{top,min}}^2}{r}$

WHY. At this instant, gravity alone supplies the entire inward force — nothing left over, nothing missing.

PICTURE. One red gravity arrow, and it is exactly as long as the inward requirement bracket beside it. Gravity and the "need" are equal — a perfect match.

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Step 5 — Getting to the top: the energy bridge

WHAT. From bottom to top the ball rises a height of $2r$ (bottom of the circle to top of the circle is one diameter).

WHY energy and not forces? Forces at the top and bottom point in different directions and vary all around the loop — tracking them mid-loop is painful. Energy only cares about height and speed, ignoring direction. It is the right tool for "how does speed change with height?"

PICTURE. The circle with a vertical ruler on the side: bottom at height $0$, top at height $2r$, side at height $r$. The red ball climbs the ruler; a dashed line marks the $2r$ it must gain.

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Step 6 — Solve for the bottom speed

WHAT. Cancel $\frac{1}{2}m$ everywhere (again mass and the half drop out): $v_{\text{bot}}^2=v_{\text{top}}^2+4gr$ Now feed in the Step 4 result $v_{\text{top}}^2=gr$: $v_{\text{bot}}^2=gr+4gr=5gr\Rightarrow \boxed{v_{\text{bot,min}}=\sqrt{5gr}}$

WHY the $4gr$? Multiplying $mg(2r)$ back through the cancelled $\frac{1}{2}m$ turns the height cost $2r$ into a $v^2$ cost of $2\cdot 2gr=4gr$. That is the "energy tax" for the climb.

PICTURE. A bar chart in the theme colors: the bottom bar (height $5gr$) splits into the top bar ($gr$, red) plus the climbing tax ($4gr$). You can see where the 5 comes from: $1+4$.

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Step 7 — The bottom tension, and the famous $6mg$

WHAT. At the bottom the center is above the ball, so tension points up while gravity points down. Net upward force = requirement: $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r}$

WHY the minus sign? Because here the two real forces oppose: up minus down. Compare Step 2 (top) where they added. Same law, opposite geometry — exactly the "where is the center?" rule from Step 1.

PICTURE. The ball at the bottom: a long red $T$ arrow up, a shorter $mg$ arrow down; the net up arrow equals the inward requirement.

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Step 8 — The degenerate case: a rigid rod

WHAT. With the rod, the only surviving requirement at the top is that the ball actually gets there: $v_{\text{top}}\ge 0$. So the minimum top speed is $0$.

WHY it matters. The parent's mnemonic ($5$–$3$–$1$, $6mg$) is a string/track result. Change the constraint, change the answer. Never memorize $\sqrt{gr}$ as a law of nature — it is a consequence of "no pushing."

PICTURE. Two panels. Left: a string at the top with $v_{\text{top}}=\sqrt{gr}$, a taut red line. Right: a rod at the top with $v_{\text{top}}\to 0$, the rod pushing up (red arrow reversed) to hold the ball on the circle.

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The one-picture summary

WHAT. One figure gathers the whole loop: the circle with the ball at bottom, side, top; the squared-speed labels $5gr$, $3gr$, $gr$; the tensions $6mg$ and $0$; the inward arrows at each point; and the height ruler $0$, $r$, $2r$. This is the entire derivation compressed.

Recall Feynman retelling — say the whole walkthrough in plain words

To curve, you always need a pull toward the middle. At the top of a loop, the middle is below you, so gravity is already helping pull you inward. If you go slowly, the circle only needs a gentle inward pull — gentler than gravity — so the string would have to push to hold you back, and strings can't push. The slowest you can go is when gravity's pull is exactly what the circle needs, and the string does nothing at all: that speed is $\sqrt{gr}$. To arrive at the top that fast, you must start faster at the bottom, because climbing the height of two radii steals speed — energy bookkeeping turns that climb into an extra $4gr$, so the bottom speed squared is $1+4=5$ times $gr$. At the bottom the string must both hold your weight and provide the big inward pull, so it strains at $6mg$ while at the top it strains at nothing. Swap the string for a stiff rod and the "no pushing" rule vanishes, so you can crawl over the top at zero speed — the magic $\sqrt{gr}$ was never a law, just a string's honest limit.

Recall The five numbers, from memory

Squared speeds $5gr,3gr,1gr$ at bottom / side / top. Tensions $6mg$ at bottom, $0$ at top. Rod: $2\sqrt{gr}$ at the bottom.

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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