Question

A stone of mass $1\text{kg}$ tied to a light string of length $\frac{10}{3}\text{m}$ is whirled in a vertical circle. If the ratio of the maximum tension to minimum tension is 4 and $g=10{\text{m/s}}^2$, then the speed of the stone at the highest point of the circle is

• A. $0\text{m/s}$
• B. $20\text{m/s}$
• C. $10\text{m/s}$
• D. $5\text{m/s}$

Answer 1

The correct option is C $10\text{m/s}$


Let $v_o$ & $v_t$ be the velocities at the bottom and top of the circle.
Tension will be maximum at the bottom of the circle.
$\Rightarrow T_{max}=\frac{m{v}_o^2}{r}+mg$ ------(1)
Tension will be minimum at the top of the circle.
$\Rightarrow T_{min}=\frac{m{v}_t^2}{r}-mg$ -------(2)

$\frac{T_{max}}{T_{min}}=4$ (given)
$\Rightarrow \frac{\frac{m{v}_o^2}{r}+mg}{\frac{m{v}_t^2}{r}-mg}=4$ -----(3)

Applying conservation of energy at points $A$ & $B$, taking $A$ as a reference point
$K{E}_A+P{E}_A=P{E}_B+K{E}_B$
$\frac{1}{2}m{v}_o^2+0=mg\left(2r\right)+\frac{1}{2}m{v}_t^2$
$\frac{v_o^2-v_t^2}{2}=2gr$ -------(4)

From equation (3)
$\frac{m{v}_o^2}{r}+mg=\frac{4m{v}_t^2}{r}-4mg$
$\Rightarrow 5mg=\frac{4m{v}_t^2}{r}-\frac{m{v}_o^2}{r}$
$4v_t^2-v_o^2=5gr$ ---------(5)
From equation (4)
$v_o^2-v_t^2=4gr--------\left(6\right)$

By solving Equation (5) & (6)
$3v_t^2=9gr$
$v_t^2=3gr$
$v_t=\sqrt{3gr}$
$v_t=\sqrt{3\times 10\times \frac{10}{3}}=10\text{m/s}$