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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.enter image description here

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?

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

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

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – knzhou, user259412, CuriousOne, heather, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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