Question

A particle of mass $m$ attached to a string of length $l$ fixed at a point such that it can perform circular motion in a vertical plane. The particle is imparted a velocity of $\sqrt{8gl}$ at the lowest position by a sharp hit. Find the ratio of tension in string at the lowermost position to the tension in the string at the topmost position of the vertical circular path. Take $g=10{\text{m/s}}^2$.

• A. $3:1$
• B. $9:5$
• C. $7:3$
• D. $7:5$

Answer 1

The correct option is A $3:1$

Let tension at lowermost point be $T_l$ and tension at topmost point be $T_t$

Applying equation of dynamics towards centre of circular path from the FBD at both positions:


At lowermost point,
$T_l-mg=\frac{m{u}^2}{l}$
$\Rightarrow T_l=mg+\frac{8mgl}{l}=8mg+mg$
$\therefore T_l=9mg$

Using energy conservation between lowermost and topmost position:
$K_l+U_l=K_t+U_t$
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}_t^2+2mgl$
$\frac{1}{2}m\left(8gl\right)=\frac{1}{2}m{v}_t^2+2mgl$
$\Rightarrow v_t^2=4gl$

At topmost point,
$T_t+mg=\frac{m{v}_t^2}{l}$
$T_t=\frac{4mgl}{l}-mg=4mg-mg=3mg$

$\Rightarrow \frac{T_l}{T_t}=\frac{9mg}{3mg}=3$
$\therefore T_l:T_t=3:1$