Question

Three particles are tied by means of strings of negligible mass and equal lengths. They move on a horizontal circular path of radius ${}^{\prime }{r}^{\prime }$ as shown. Find the tension in the string connecting them.

• A. $\frac{mr{\omega }^2}{2-\sqrt{3}}$
• B. $\frac{mr{\omega }^2}{2\sqrt{3}}$
• C. $\frac{mr{\omega }^2}{2+\sqrt{3}}$
• D. $\frac{mr{\omega }^2}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{mr{\omega }^2}{\sqrt{3}}$

Here by the free body diagram of single mass ,
All the strings will have same tension as all the strings are identical and connected to same mass at same distance.

Here by the FBD, analyzing a single mass will have two tensions at an angle of ${60}^{\text{o}}$ as it forms an equilateral triangle, so by the law of vector addition the net resultant tension$\left(T^{\prime }\right)$ will be

$T^{\prime }=\sqrt{T^2+T^2+2T^2\cos {60}^{\circ }}$
$=\sqrt{3}T$
The resultant tension will provide the necessary centripetal force $\left(m{\omega }^{2}r\right)$.

$T^{\prime }=mr{\omega }^2\Rightarrow \sqrt{3}T=mr{\omega }^2\Rightarrow T=\frac{mr{\omega }^2}{\sqrt{3}}$

Hence option D is the correct answer.