Answer
$$\frac{m_1}{m_2}=4$$
Work Step by Step
The formula for centripetal force is given by $$F_c=\frac{mv^2}{r}$$
However, we also know that $v=\frac{2\pi r}{T}$. Therefore, $$F_c=\frac{m}{r}\times\frac{4\pi^2r^2}{T^2}=\frac{4\pi^2mr}{T^2} (1) $$
If we take the distance from the center of rotation to particle 1 to be $r$, then the distance from the center of rotation to particle 2 will be $2r$. The period of rotation $T$ in both points, nevertheless, is the same.
Finally, we need to find $F_c$ at 2 points where the particles are located.
- At the end of the rod where particle 2 is located, the tension in the outer section provides the centripetal force, so $F_{To}=F_{c2}$
- At the middle of the rod where particle 1 is located, the tension in the inner rod $F_{Ti}$should provide the centripetal force, too. However, $F_{Ti}$ makes up the tension for both $m_1$ and $m_2$, so to find $F_c$ only for particle 1, we need to take out $F_{To}$: $$F_{Ti}-F_{To}=F_{c1}$$ $$F_{c1}=3F_{To}-F_{To}=2F_{To}$$
Therefore, $F_{c1}=2F_{c2}$
Now, using equation (1), we start to compare $$\frac{F_{c1}}{F_{c2}}=\frac{\frac{4\pi^2m_1r}{T^2}}{\frac{4\pi^2m_22r}{T^2}}=\frac{m_1}{2m_2}$$ $$\frac{m_1}{2m_2}=2$$ $$\frac{m_1}{m_2}=4$$