Question

A particle is whirled in a vertical circle of radius $1.0m$ using a string with one end fixed. If the ratio of maximum to minimum tensions in the string is $\frac{5}{3}$, then the minimum velocity (in $m/s)$ of the particle during circular motion is

Answer 1

Step1: Draw the rough diagram for given situation


Step2: Find ratio of maximum and minimum tension.
Given, $\frac{T_{\text{max}}}{T_{\text{min}}}=\frac{5}{3}$
At the lower most point,
$T_{\text{max}}-mg=\frac{m{v}_{max}^2}{R}\dots \left(i\right)$
At the upper most point,
$mg+T_{\text{min}}=\frac{m{v}_{min}^2}{R}\dots \left(ii\right)$
Using equation $\left(i\right),\left(ii\right)$

$\frac{T_{\text{max}}}{T_{\text{min}}}=\frac{\left(\frac{m{v}_{\text{max}+mg}^2}{R}\right)}{(\frac{m{v}_{\text{min}}^2}{R}-mg)}\dots \left(iii\right)$
step3: Find minimum velocity of the particle.

Apply conservation of energy at lower most point and upper most point,
$\frac{1}{2}m{v}_{\text{max}}^2=mg\left(2R\right)+\frac{1}{2}m{v}_{\text{min}}^2$
$\Rightarrow v_{\text{max}}^2=2g\left(2R\right)+v_{\text{min}}^2\dots \left(iv\right)$

Using equation $\left(iii\right)$ and $\left(iv\right)$

$\frac{5}{3}=\frac{\frac{v_{\text{min}}^2+2.g.2R}{R}+g}{\frac{V_{\text{min}}^2}{R}-g}$

$\Rightarrow v_{\text{min}}=10m/s$