A simple pendulum of length l = 0.5 m is pulled aside by α =…

A simple pendulum of length $l = 0.5\ \mathrm{m}$ is pulled aside by $\alpha = 60^\circ$, and the bob is given a speed $v_0 = 3\ \mathrm{m/s}$ downward, perpendicular to the string. Where does the bob get to? Neglect friction.

Step 1: Energy available

Measured from the lowest point, the energy per unit mass is

\[ gl(1 - \cos\alpha) + \tfrac12 v_0^2 = 2.45 + 4.50 = 6.95\ \mathrm{J/kg}, \] enough to lift the bob $6.95/9.81 = 0.709\ \mathrm{m}$, which is more than $l$: the bob swings through the bottom and rises above the suspension point's level. There, however, the string may go slack.

Step 2: Where the string goes slack

At angle $\theta$ from the lowest point, the speed and the string tension are

\[ v^2 = 2\big(6.95 - gl(1 - \cos\theta)\big), \]

\[ \frac{T}{m} = \frac{v^2}{l} + g\cos\theta = 8.19 + 29.43\cos\theta. \]

The tension drops to zero when

\[ \cos\theta = -\frac{8.19}{29.43} = -0.278. \]

\[\theta \approx 106^\circ \ \text{from the lowest point}\]

That is 16° above the horizontal. There the bob is $l(1 - \cos\theta) = 0.64\ \mathrm{m}$ above the lowest point and moves at $v = \sqrt{gl\,|\cos\theta|} \approx 1.17\ \mathrm{m/s}$.

Step 3: After that

With a slack string the bob is a projectile. Its velocity (tangent to the circle) has a vertical part $v\sin 74^\circ \approx 1.12\ \mathrm{m/s}$, so it rises another $1.12^2/(2g) \approx 0.064\ \mathrm{m}$:

\[h_{\max} \approx 0.70\ \mathrm{m} \ \text{above the lowest point}\]

That is about 0.20 m above the suspension point O. It then falls back inside the circle until the string snaps taut again.