A simple pendulum of length l is pulled aside by θ = 60^°…

A simple pendulum of length $l$ is pulled aside by $\theta = 60^\circ$ and released. As it passes through the equilibrium position, its string catches on a peg O located $l/2$ below the point of suspension. By how much does the string tension change when the string hits the peg?

Explanation of Method

Pendulum string catching on a peg.

Step 1: Speed at the bottom

The bob drops by $h = l(1 - \cos 60^\circ) = l/2$: $\tfrac12 mv^2 = mg\,\tfrac{l}{2} \;\Longrightarrow\; v^2 = gl$.

Step 2: Tension just before the peg

At the bottom the string must hold the weight and supply the centripetal force for a circle of radius $l$: \[T_1 = mg + \frac{mv^2}{l} = mg + mg = 2mg.\]

Step 3: Tension just after the peg

The speed does not change at that instant, but the radius drops to $l/2$: \[T_2 = mg + \frac{mv^2}{l/2} = mg + 2mg = 3mg.\]

Step 4: Change

\[\Delta T = T_2 - T_1 = mg\]

The tension grows by half, from $2mg$ to $3mg$.

$60^\circ$ $\text{peg at } \tfrac{l}{2}$ $v = \sqrt{gl}$ $\text{before the peg:}$ $T_1 = mg + \tfrac{mv^2}{l} = 2mg$ $\text{after the peg:}$ $T_2 = mg + \tfrac{mv^2}{l/2} = 3mg$

The bob swings down from 60° along a circle of radius $l$. At the bottom the string catches on the peg, so the bob continues on a circle of half the radius at the same speed, which needs twice the centripetal force.

Result

\[\boxed{\,\Delta T = T_2 - T_1 = mg\,}\]

Remark

Check. The original result is correct. After the peg the bob swings up on the short circle; with $v^2 = gl$ it rises to height $l/2$ again, i.e. exactly level with the peg.