Question

A bob of mass $m$ is suspended from point $O$ by a massless string of length $l$ as shown. At the bottommost point it is given a velocity $u=\sqrt{12gl}$ for $l=1m$ and $m=1kg$, match the following two columns when string becomes horizontal $(g=10m{s}^{-2})$

Answer 1

Given : $u=\sqrt{12gl}$ $m=1kg$ $l=1$ m $g=10m/s^2$

Let the velocity of the bob be $v$ when the string becomes horizontal.

Using work-energy theorem : $W=\Delta K.E$

$\therefore$ $-mgl=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

OR $-2gl=v^2-\left(12gl\right)$ $⟹v=\sqrt{10gl}$

$\therefore$ $v=\sqrt{10\times 10\times 1}=10$ m/s

Radial acceleration, $a_r=\frac{v^2}{l}=\frac{{10}^2}{1}=100$ $m/s^2$

Tension in the string, $T=m{a}_r=1\times 100=100$ $N$

Tangential acceleration, $a_t=mg=1\times 10=10$ $N$

$\therefore$ Total acceleration, $a=\sqrt{a_r^2+a_t^2}=\sqrt{(100{)}^2+(10{)}^2}=10\sqrt{101}=100.5$ $m/s^2$