Question

Three point masses each of mass m are joined together to form an equilateral triangle of side 'd. The system is placed over a smooth horizontal surface and rotated with a constant angular velocity $\omega$ about a vertical axis passing through the centroid. The tension in each string will be.

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

Answer 1

The correct option is C $\frac{ma{\omega }^2}{\sqrt{3}}$

Distance of each mass from the centroid is $r=\sqrt{3}a$
So, the tension in the strings will be
$T=mr{\omega }^2$
$=m\left(\sqrt{3}a\right){\omega }^2$
Tension in each string will be
$T^{\prime }=\frac{T}{3}=\frac{\sqrt{3}ma{\omega }^2}{3}$
$=\frac{ma{\omega }^2}{\sqrt{3}}$.