A simple pendulum of length l = 1 m is pulled aside by α =…

A simple pendulum of length $l = 1\ \mathrm{m}$ is pulled aside by $\alpha = 60^\circ$ and given a speed $v_0$ downward, perpendicular to the string. What must $v_0$ be for the bob to go all the way round?

Explanation of Method

Speed for a pendulum to go all the way round.

Step 1: Condition at the top

At the top the string may just go slack ($T = 0$), so gravity alone provides the centripetal force: $mg = \frac{mv_{\text{top}}^2}{l} \;\Longrightarrow\; v_{\text{top}}^2 = gl$.

Step 2: Energy

Heights above the lowest point: start $l(1 - \cos\alpha) = l/2$, top $2l$: $\tfrac12 v_0^2 + g\,\frac{l}{2} = g\,(2l) + \tfrac12 gl \;\Longrightarrow\; v_0^2 = 4gl$.

\[v_0 = 2\sqrt{gl} \approx 6.26\ \mathrm{m/s}\]

$v_0$ $v_{\text{top}} = \sqrt{gl}$ $60^\circ$ $\tfrac12 v_0^2 + gl(1 - \cos\alpha)$ $= 2gl + \tfrac12 gl$ $v_0 = 2\sqrt{gl} \approx 6.26\ \mathrm{m/s}$

To go all the way round, the bob must still move at $\sqrt{gl}$ at the top, so that gravity alone supplies the centripetal force there (string just taut).

Result

\[\boxed{\,v_0 = 2\sqrt{gl} \approx 6.26\ \mathrm{m/s}\,}\]

Remark

About the original solution. The number 6.2 m/s is about right, but the formula it wrote, $v_0^2 = 2gl + g^2$, is wrong (it is not even dimensionally consistent, and it would give 10.8 m/s). The correct relation is $v_0^2 = 4gl$, i.e. $v_0 = 6.26\ \mathrm{m/s}$.