Vertical circular motion

In a vertical circle the speed is no longer constant: the particle slows as it climbs and speeds up as it falls. So we cannot just plug into one formula — we use two tools, in order.

Step 1 — Energy conservation (find the speed). On a smooth track or a light inextensible string there is no friction, so mechanical energy is conserved:

$\tfrac{1}{2}mv_1^2 + mgh_1 = \tfrac{1}{2}mv_2^2 + mgh_2$

The mass cancels, leaving a relation between speeds and heights. This gives the speed $v$ at any point of the circle.

Step 2 — Newton's second law toward the centre (find the force). At the point of interest, resolve the real forces along the radius (toward the centre) and set the resultant equal to the centripetal term:

$F_{\text{toward centre}} = \frac{mv^2}{r}$

The weight contributes only its radial component $mg\cos\theta$, where $\theta$ is the angle of the radius from the vertical.

Which tool for which quantity?

• Asked for a speed at a point → use energy conservation (Step 1).
• Asked for a tension or normal reaction → first find the speed (Step 1), then use the radial equation (Step 2).

Cambridge tip. Almost every vertical-circle question needs both steps. Find the speed with energy, then the force with the radial equation — in that order.

The direction 'toward the centre' changes as the particle moves, so the radial equation looks different at each position. Always ask: where is the centre relative to the particle?

At the lowest point the centre is directly above, so 'toward the centre' is up. Tension acts up, weight acts down:

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

Here the tension is large — it must both support the weight and provide the centripetal force.

At the highest point the centre is directly below, so 'toward the centre' is down. Both the tension and the weight act down:

$T + mg = \frac{mv^2}{r}$

Here the tension is small (the weight already helps pull the particle toward the centre).

At the sides (string horizontal) the centre is horizontal from the particle, so 'toward the centre' is horizontal. The weight is vertical — perpendicular to the radius — so it contributes nothing to the centripetal equation:

$T = \frac{mv^2}{r}$

At a general angle $\theta$ from the downward vertical, the radial component of the weight is $mg\cos\theta$ (pointing away from the centre on the lower half):

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

Cambridge tip. Before writing the radial equation, draw the radius and mark where the centre is. Then decide whether the weight helps ($+$) or opposes ($-$) the motion toward the centre.

For a particle to travel all the way round a vertical circle, the connection must never fail. The highest point is the hardest place to maintain — the speed is least there and the weight is pulling the particle away from the circle.

On a string (or inside a track). A string can only pull, so the tension cannot be negative: $T \geq 0$. At the top:

$T + mg = \frac{mv_{\text{top}}^2}{r} \quad\Longrightarrow\quad T = \frac{mv_{\text{top}}^2}{r} - mg \geq 0$

This gives the critical condition:

$v_{\text{top}}^2 \geq gr$

The minimum case ($T = 0$ at the top) combined with energy conservation over the rise of $2r$ gives the famous result for the lowest point:

$v_{\text{low}}^2 \geq 5gr$

On a rigid rod. A rod can also push, so it can hold the particle in place at the top even at zero speed. The particle need only just reach the top ($v_{\text{top}} = 0$):

$v_{\text{low}}^2 \geq 4gr$

The 'inside of a track' behaves like a string: the track can only push the particle inward (reaction $\geq 0$), so the condition is the same as for a string — leaving requires $R = 0$ at the top.

Summary table:

Connection	Can it push?	Top condition	Bottom condition
String / inside a track	No (pull only)	$v_{\text{top}}^2 \geq gr$	$v_{\text{low}}^2 \geq 5gr$
Rigid rod	Yes	$v_{\text{top}} \geq 0$	$v_{\text{low}}^2 \geq 4gr$

Cambridge tip. Read the words carefully. "String" or "inside a smooth tube/track" → $5gr$. "Rigid rod" → $4gr$. Quoting the wrong one is a guaranteed lost answer.

A particle sliding on the outside of a smooth sphere is pressed against the surface by gravity, but as it speeds up and the surface curves away it eventually leaves the sphere. The crucial idea:

A particle leaves a surface at the instant the normal reaction $R$ becomes zero.

(The speed is generally not zero there — a common confusion.)

Setting it up. Let $\theta$ be the angle the radius makes with the upward vertical, with the particle starting from rest at the top of a sphere of radius $a$.

Energy conservation (height dropped $= a(1 - \cos\theta)$):

$v^2 = 2ga(1 - \cos\theta)$

Radial equation toward the centre. On the outside the centre is inward; the weight's radial component $mg\cos\theta$ points inward, the reaction $R$ points outward:

$mg\cos\theta - R = \frac{mv^2}{a}$

At the moment of leaving, $R = 0$:

$mg\cos\theta = \frac{mv^2}{a} \quad\Longrightarrow\quad v^2 = ga\cos\theta$

Equating the two expressions for $v^2$:

$2ga(1 - \cos\theta) = ga\cos\theta \quad\Longrightarrow\quad \cos\theta = \tfrac{2}{3}$

So the particle leaves the sphere where $\cos\theta = \frac{2}{3}$ ($\theta \approx 48.2°$ from the top) — a result independent of $a$, $g$ and $m$.

Cambridge tip. Set $R = 0$ — never the speed — to find where contact is lost. Then solve the two $v^2$ expressions simultaneously.