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

deancodemo

## 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?

Last edited:

nickjer
Draw a free body diagram of the forces acting on the mass at P. You will see both ropes on either side of the mass pulling in opposite directions. That is why your 2nd set of equations is correct.