Question

Particles each of mass ${}^{\prime }{m}^{\prime }$ are tied by means of strings of equal length as shown in figure. If the system is rotated with constant angular velocity ${}^{\prime }{\omega }^{\prime }$ in horizontal plane, find the tension in the string.
[Given ${}^{\prime }{r}^{\prime }$is the radius of the circle]

• A. $\frac{mr{\omega }^2}{\sqrt{2}}$
• B. $\frac{mr{\omega }^2}{\sqrt{5}}$
• C. $\frac{mr{\omega }^2}{2}$
• D. $\frac{mr{\omega }^2}{2\sqrt{2}}$

Answer 1

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

Free body diagram of single mass $\left(D\right)$ ,


From FBD, a single mass will have two tensions at an angle of ${90}^{\text{o}}$ as it forms a square, so by the law of vector addition the net resultant tension$\left(T^{\prime }\right)$ will be

$T^{\prime }=\sqrt{T^2+T^2+2T^2\cos {90}^0}=T\sqrt{2}$ and the resultant tension $T^{\prime }$will provide the necessary centripetal force.
Hence $T^{\prime }=mr{\omega }^2$
$\Rightarrow T\sqrt{2}=mr{\omega }^2\Rightarrow T=\frac{mr{\omega }^2}{\sqrt{2}}$

Hence option A is the correct answer.