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## 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.

## Homework Equations

## The Attempt at a Solution

First I convert the angular velocity from rev/sec to rad/sec.

[tex]\omega = 5[/tex] rev/sec

[tex]\omega = 10\pi[/tex] 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:

[tex]R = ma[/tex]

(Tension in OP) + (Tension in PQ) [tex]= m r \omega^2[/tex]

(Tension in OP) = [tex]m r \omega^2 - [/tex] (Tension in PQ)

This is incorrect!

The only solution would be:

[tex]R = ma[/tex]

(Tension in OP) - (Tension in PQ) [tex]= m r \omega^2[/tex]

(Tension in OP) = [tex]m r \omega^2 + [/tex] (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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