# Identifying the forces acting on masses in uniform circular motion.

## Homework Statement

Two particles of masses 2kg and 1kg are attached to a light string at distances 0.5m and 1m respectively from one fixed end. Given that the string rotates in a horizontal plane at 5 revolutions/second, find the tensions in both portions of the string.

## The Attempt at a Solution

First I convert the angular velocity from rev/sec to rad/sec.
$$\omega = 5$$ rev/sec
$$\omega = 10\pi$$ rad/sec

Now, let the 2kg mass be P, the 1kg mass be Q and the fixed point be O. I understand that the tension in PQ < tension in OP. Actually, the tension in OP should equal the centripetal force acting on P plus the tension in PQ. Right?
I get confused when identifying the forces acting on P. By using R = ma:
$$R = ma$$
(Tension in OP) + (Tension in PQ) $$= m r \omega^2$$
(Tension in OP) = $$m r \omega^2 -$$ (Tension in PQ)

This is incorrect!

The only solution would be:
$$R = ma$$
(Tension in OP) - (Tension in PQ) $$= m r \omega^2$$
(Tension in OP) = $$m r \omega^2 +$$ (Tension in PQ)

Which makes sense, but this means that the tension forces acting on P in the different sections of string act in opposite directions. Is that right?

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