Question

A small bob tied at one end of a thin string of length $1\text{m}$ is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio $5:1$. The velocity of the bob at the highest position is ______$\text{m/s}$.
$($Take $g=10{\text{m/s}}^2)$

Answer 1

Here, $T_1$ is maximum tension and $T_2$ is minimum tension.

From Law of Conservation of Energy,
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}^2+2mgR$

$\Rightarrow u^2=v^2+4gR...\left(1\right)$

From Newton's Second Law of Motion,

At bottommost point, $T_1-mg=\frac{m{u}^2}{R}$

$\Rightarrow T_1=\frac{m}{R}(v^2+4gR)+mg=\frac{m{v}^2}{R}+5mg$
$[$From $\left(1\right)]$

At topmost point, $T_2+mg=\frac{m{v}^2}{R}$

$\Rightarrow T_2=\frac{m{v}^2}{R}-mg$

Now,

$\frac{T_1}{T_2}=\frac{\frac{m{v}^2}{R}+5mg}{\frac{m{v}^2}{R}-mg}$

$\Rightarrow \frac{5}{1}=\frac{\frac{m{v}^2}{R}+5mg}{\frac{m{v}^2}{R}-mg}$

On solving this, we get,

$v=\sqrt{\frac{5gR}{2}}=\sqrt{\frac{5\times 10\times 1}{2}}=5\text{m/s}$