Problem with how to apply pseudo force in this situation [closed]

This is a situation, where both the bodies move with a constant angular velocity. M1 is connected to a rotating rod and M2 is connected to M1 with a string. There is no tangential acceleration. I want to figure out the forces on M2 w.r.t M1. I say, there is

1-mg

2- Tension

3- Pseudo force as M1 is accelerating.

Now,suppose a body is moving with an acceleration a and i want to apply Newton’s laws on B wrt A. I say that the body B is accelerating with an acceleration -a, even if it’s stationary ( wrt earth of course ). This is how we apply pseudo force. M1 is accelerating with v^2/r1.

Now, when i apply this acceleration on M2 as a pseudo force, i get the wrong answer, but if i apply a pseudo force on M2, but by taking the radius r2, i get the right answer. That is, If i apply a pseudo force on M2 = mv^2/r1 ( r1 because i am applying it w.r.t M1 ) i get the wrong answer. But if i apply a force mv^2/r2 on M2, i get the right answer! Why is this happening?

closed as off-topic by knzhou, user259412, CuriousOne, heather, GertAug 4 '16 at 0:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• i don’t understand why people are wanting to close this. Ofcourse this is not a home science question. :/ – Aaryan Dewan Aug 3 '16 at 11:37

Your definition of $r_2$ in the figure and in the text are inconsistent.
I will assume your text is the correct one and let $r_2$ be the distance of $m_2$ from the center.
For an observer at the center, you only need to add the centrifugal force $$m_2 \omega^2 r_2$$
For an observer at $r_1$, it is orbiting and spinning, so you need to add two forces $$m_2 a_1 + m_2 \omega^2 (r_2 - r_1)$$ But $$a_1=\omega^2 r_1$$ So you have $$m_2 \omega^2 r_1 + m_2 \omega^2 (r_2-r_1)=m_2\omega^2 r_2$$ which is the same as the first answer.