Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity $u=\sqrt{10gl}$, where $l$ is the length of the pendulum. Then, the tension in the string when it is horizontal is $xmg$ where $x=$

Answer 1

Let the velocity at B be $v_b$ .
The velocity at A is $u=\sqrt{10gl}$ (given)
Applying energy conservation at $A$ & $B$, taking $A$ as reference.
$(KE{)}_A=(KE{)}_B+(PE{)}_B$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_b^2+mgl$
$\Rightarrow \frac{1}{2}m\left(10gl\right)=\frac{1}{2}m{v}_b^2+mgl$
$\Rightarrow 8mgl=m{v}_b^2$
$\Rightarrow v_b^2=8gl$ ---------(1)

So, tension in the string at horizontal position ($B$)
$T=\frac{m{v}_b^2}{l}=\frac{m\left(8gl\right)}{l}$
$\therefore T=8mg$