Minimum Speed at Top of Vertical Circle

The situation

A mass $m$ on a string of length $R$ moves in a vertical circle. At the top of the circle, what is the minimum speed needed to keep the string taut?

The critical condition

At the top of the circle, both gravity ($mg$ downward) and string tension ($T$ downward, since the string pulls toward the centre which is now below) act downward — both provide centripetal force toward the centre.

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

The string goes slack when $T = 0$. At that moment:

$mg = \frac{mv^2}{R}$

$v^2 = gR$

$\boxed{v_{min} = \sqrt{gR}}$

If $v < \sqrt{gR}$ at the top, the string goes slack and the body leaves the circular path.

What happens when the string goes slack

When $T = 0$, the only force is gravity. The body becomes a projectile from that point — it follows a parabolic path, not a circular one.

At the minimum speed, $T = 0$

At exactly $v = \sqrt{gR}$: tension is zero. The body is on the verge of leaving the path. Gravity alone provides the centripetal force.

This is the minimum — any slower and the body cannot maintain the circular path.

For a body on the inside of a track (roller coaster loop)

Same analysis applies. At the top of the loop, the track pushes downward (normal force $N$ downward) and gravity acts downward. The condition $N = 0$ gives the same minimum speed $v_{min} = \sqrt{gR}$.