Question

A small sphere $B$ of mass $1\text{kg}$ is released from rest in the position shown and swings freely in a vertical plane, first about $O$ and then about the peg $A$ after the cord comes in contact with the peg. The tension in the cord just after it comes in contact with the peg is
[Take $g=10{\text{m/s}}^2$]

• A. $30\text{N}$
• B. $20\text{N}$
• C. $15\text{N}$
• D. $25\text{N}$

Answer 1

The correct option is D $25\text{N}$


Suppose, the speed of the small sphere is $v$ just after it comes in contact with the peg.
From law of conservation of mechanical energy,
Decrease in potential energy = Increase in kinetic energy
$\Rightarrow mgh=\frac{1}{2}m{v}^2$
$\Rightarrow 10\times 0.8\times \sin {30}^{\circ }=\frac{1}{2}\times v^2$
$\Rightarrow 10\times 0.8\times \frac{1}{2}=\frac{1}{2}\times v^2$
$\Rightarrow v^2=8\text{m/s}$ ... (1)

After coming in contact with the peg, its radius of vertical circular motion becomes $r=0.8-0.4=0.4\text{m}$
So, for rotation about $A$,
$T=\frac{m{v}^2}{r}+mg\cos {60}^{\circ }$
$\Rightarrow T=\frac{1\times 8}{0.4}+10\times \frac{1}{2}$
$\Rightarrow T=25\text{N}$