Question

A small bob tied at one end of a thin string of length $1m$ 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 ______ $m/s$. (take$g=10m/s^2$)

Answer 1

Step 1. Given data

Length of string, $l=5m$

Ration between tension $\frac{T_{max}}{T_{min}}=5:1$

Acceleration due to gravity,$g=10m/s^2$

Step 2. Velocity at highest point:

At highest point, the velocity is minimum, given by ${v_{min}}^2=v^2-2g\left(2l\right)$

The corresponding tension is,$T_{min}=\frac{m}{l}\left(v^2-4gl\right)-mg$

At lowest point, velocity is $v$

The corresponding tension is,$T_{max}=\frac{m}{l}{v}^2+mg$

Taking ration of the tensions, $\frac{T_{max}}{T_{min}}=\frac{\frac{m}{l}{v}^2+mg}{\frac{m}{l}\left(v^2-4gl\right)-mg}=\frac{5}{1}$

Putting the values, we get

$\Rightarrow \frac{5}{1}=\frac{v^2+10}{v^2-50}$

$\Rightarrow v^2=65$

$\Rightarrow {v_{min}}^2=65-4\times 10=25$

$v_{min}=5m/s$

Hence, the correct answer is $5$ m/s.