Question

A particle tied to a string describes a vertical circular motion of radius $R$ continuously. If it has velocity $\sqrt{3gR}$ at the highest point, then the ratio of respective tensions in the string at the highest and lowest point is

• A. $4:3$
• B. $5:4$
• C. $1:4$
• D. $3:2$

Answer 1

The correct option is C $1:4$

Solving from the frame of reference of the particle, there will be a centrifugal force acting radially outward.
At top most point ($A$):


$T_{Top}+mg=\frac{m{v}^2}{R}$
$\Rightarrow T_{Top}+mg=\frac{m}{R}\left(3gR\right)$
(given $v=\sqrt{3gR}$ at the top)
$\Rightarrow T_{Top}=3mg-mg=2mg$
$\Rightarrow T_{Top}=2mg.......\left(i\right)$


Step - 2
Applying law of conservation of energy
$U_A+K{E}_A=U_B+K{E}_B$
$\Rightarrow mg\left(2R\right)+\frac{1}{2}m\left(3gR\right)=\frac{1}{2}m{v}_B^2$
$(\because U_B=0)$
$\Rightarrow v_B=\sqrt{7gR}$

Step - 3
Now at bottom most point

$T_{Bottom}-mg=\frac{m{v}_B^2}{R}$
$T_{Bottom}=mg+\frac{m}{R}\times 7gR$
$T_{Bottom}=8mg.......\left(ii\right)$

From ($i$) & $\left(ii\right)$
$\frac{T_{Top}}{T_{Bottom}}=\frac{2mg}{8mg}=\frac{1}{4}$