Question

A string attached to a light thread is whirled in a vertical circle with a constant velocity $v$. The radius of the circle is $10\text{m}$. The maximum and minimum tension of the string is in the ratio $4:1$. What is the value of velocity $v$?
(Take $g=10{\text{m/s}}^2$)

• A. $15.1\text{m/s}$
• B. $13.5\text{m/s}$
• C. $12.9\text{m/s}$
• D. $10.9\text{m/s}$

Answer 1

The correct option is C $12.9\text{m/s}$

The tension in the string is highest at the lowest point during the rotation and the tension in the string is lowest at the highest point in the rotation.

Tension at the lowest point = $T_{max}=\frac{m{v}^2}{r}+mg$
Tension at the highest point =
$T_{min}=\frac{m{v}^2}{r}-mg$
$\frac{T_{max}}{T_{min}}=\frac{\frac{m{v}^2}{r}+mg}{\frac{m{v}^2}{r}-mg}=\frac{4}{1}\Rightarrow \frac{v^2+rg}{v^2-rg}=\frac{4}{1}\Rightarrow 4v^2-4rg=v^2+rg\Rightarrow v^2=\frac{5rg}{3}\Rightarrow v=\sqrt{\frac{5rg}{3}}=\sqrt{\frac{5\times 10\times 10}{3}}=12.9m/s$