I think it is worth doing a quick and dirty tutorial here.
I'll try not to undermine Haruspex's hints too much: you should verify things for yourself.
... the easier approach involves understanding three main things;
1. Understanding kinetic energy:
kinetic energy is usually given by ##K=\frac{1}{2}I\omega^2## ... but you can use the fact that ##L=I\omega## to get a more useful form for these conservation of momentum problems, vis: ##K=L^2/2I## ... this is the rotational equivalent to the linear ##K=p^2/2m##, and it combines the conservation of momentum step with the first half of the conservation of energy step.
That is useful in this case because it allows you to skip the step where you calculated the angular velocity: you know the initial angular momentum and that it is conserved.
There are other cases where alternate forms of the kinetic energy equation are useful - you don't need to memorize them, just know the possibility is there.
The next bit is probably more important...
2. Understanding potential energy:
The gravitational potential energy is the work done to lift the (centre of) mass through a height... it does not matter if all the mass is lifted in one go, or separately: it's the same work. This means that, for the purposes of conservation of energy, we can treat the parts of the rod+ball system separately:
##\Delta U_{tot} = \Delta U_{rod} + \Delta U_{ball}##
This removes the need to calculate the centre of mass, which gave you so much trouble.
Now look at the rod all by itself: when it rotates, how far does the centre of mass move?
So what does that mean about ##\Delta U_{rod}##?
This is actually the place where you would have saved the most time, ferinstance: in an exam.
It's how examiners test and reward your understanding of physics: you are penalized if you just go blindly step-by-step through the equations because that takes longer and you have less time to complete other problems.
3. Math discipline:
It is best practise to do all the algebra before plugging numbers into the equations. eg.
If we say the rod has mass M and length r, and the ball has mass m and initial velocity v ... then:
##L=I_{ball}\omega_{ball} = m(r/2)^2(v/(r/2))=\frac{1}{2}mvr##
##I = I_{ball}+I_{rod}=\frac{1}{4}mr^2 + \frac{1}{12}Mr^2##
##\Delta U = mgh = \frac{1}{2}mgr(1-\cos\theta)## (by geometry)
... and ##\Delta K = \Delta U## (conservation of energy)
So: $$ \frac{(\frac{1}{2}mvr)^2}{2(\frac{1}{4}mr^2 + \frac{1}{12}Mr^2)} = \frac{1}{2}mgr(1-\cos\theta)$$ ... simplify and solve for ##\theta##.
Hopefully you can see how the above symbolic form of the equation is easier to troubleshoot?
Notice also how I commented the lines of equations - this reminds me where i got them from and means others can easily figure out what I did wrong.
(Caveat: there may well be a mistake or two in the equations: do not take my word for it - check it yourself.)
The general principle is to stand back and look at what is going on in terms of the physics instead of just looking at the equations, then choosing the form of the equations to suit the problem. The trick, which you will learn by practise, is to treat the mathematics as a language. Those aren't abstract equations, they are descriptions.
Now that is a lot in one go - take it slow.
