Would like to start by saying I know how to solve this problem - now I just want to know why one of my original methods was wrong.
Fixed points A, B and C are in a vertical line with A above B and B above C. A particle Q of mass 3kg is joined to A, to B and to a particle P of mass 2kg, by three light rods where the length of rod AQ is 2m and the length of rod QP is 1m. Particle Q moves in a horizontal circle with centre B. Particle P moves in a horizontal circle with centre C at the same constant angular speed $\omega$ as Q, so hat A, B, P and Q are coplanar. Rods AQ and QP both make an angle of $30^{\circ}$ with the downward vertical, and rod BQ is horizontal.
Calculate the angular speed $\omega$.
See diagram:
My original method:
Find the tension in the rod AQ:
$$T\:=\:\frac{5g}{cos\left(30\right)}$$
We also know that
$$T\:sin\left(30\right)\:=\:5\:\cdot 2\sin \left(30\right)\cdot \omega ^2$$
$$\omega =\sqrt{\frac{5g\cdot sin\left(30\right)}{sin\left(60\right)\cdot 5\cdot 2\sin \left(30\right)}}=2.38\:rad\:s^{-1}$$
This is the wrong answer. But I get the right answer when I change the radius from 2sin(30) to 3sin(30) - which doesn't make sense as then I'm looking beyond rod AQ?
The second equation is not right. The wording says that Q is attached to B by a rod, so, your analysis for the particle at Q must include the force the rod from B to Q exerts outward over the particle in Q. To know the intensity of this force we need to know the acceleration (two equations for Q and three unknowns), so is, the angular speed.
We are forced to analyze the particle at P as we have only two bodies interacting with it, the Earth (gravity force) and the rod from Q to P. Your equations are ok if you put 2 kg for the mass and $3\sin30º$ for the radius.
$T=\dfrac{2 kg}{\cos30º}$
$T\cdot \sin30º=2 kg\cdot 3 m\cdot \sin30º\cdot \omega ^2$