Condition for Completing Vertical Circle

The question

A mass on a string of length $R$ is given speed $v_0$ at the bottom. What is the minimum $v_0$ for the mass to complete a full vertical circle without the string going slack?

Strategy

The critical point is the top of the circle — that is where the speed is minimum and where the string is most likely to go slack.

Minimum condition at the top: $v_{top} = \sqrt{gR}$ (from the previous result).

Now use energy conservation to find what speed at the bottom corresponds to $v_{top} = \sqrt{gR}$.

Derivation

The bottom of the circle is at height $0$. The top is at height $2R$ above the bottom.

Energy conservation from bottom to top:

$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2R)$

$v_{bottom}^2 = v_{top}^2 + 4gR$

Substituting minimum $v_{top}^2 = gR$:

$v_{bottom}^2 = gR + 4gR = 5gR$

$\boxed{v_{bottom} \geq \sqrt{5gR}}$

What happens at exactly $v_{bottom} = \sqrt{5gR}$

The mass just barely completes the circle. At the top: $v = \sqrt{gR}$, $T = 0$. The string is on the verge of going slack.

What happens if $v_{bottom} < \sqrt{5gR}$

The string goes slack somewhere before the top. The mass leaves the circular path and becomes a projectile.

There are three regimes depending on $v_{bottom}$:

The three cases explained

Case 1 ($v_{bottom} \geq \sqrt{5gR}$): Enough energy to reach the top with $v_{top} \geq \sqrt{gR}$. Full circle completed.

Case 2 ($\sqrt{2gR} < v_{bottom} < \sqrt{5gR}$): Mass reaches above the centre (height $> R$) but string goes slack before the top. From that point, parabolic motion.

Case 3 ($v_{bottom} \leq \sqrt{2gR}$): At height $R$ (horizontal level of centre), $v = 0$ — mass cannot even reach the side. It oscillates like a pendulum.