As pointed out in a comment: all expressions of rotational dynamics can be constructed from the expressions of linear dynamics.
The only real difference is that linear dynamics is possible with one degree of spatial freedom, whereas dynamics requires a minimum of two spatial dimensions.
For example, there is the linear mechanics classroom demonstration device called 'air track'. The motion is effectively constrained to one spatial dimension.
For demonstration of circumnavigating motion, around some point of attraction of repulsion, an air table is used.
In the case of circular motion there are two ways of decomposing that motion that have practical use. One way is to decompose the motion along the axes of a rectangular grid. Then circular motion can be represented as a linear combination of two harmonic oscillations, 90 degrees out of phase.
Of course the decomposition used most often is according to polar coordinates: radial distance to the center of attraction/repulsion, and angle with respect to some reference line.
In both cases the motion is decomposed in two components. The component motions are at 90 degrees to each other.
Angular momentum
The concept of angular momentum had a precursor: Kepler's law of areas.
The very first proposition in Newton's Principia is a demonstration that Kepler's law of Areas follows logically from the laws of motion as laid down in the Principia.
I represented that derivation in an answer here on stackexchange to a question titled 'Intuition for angular momentum' The elements that go into that derivation are all from linear mechanics.
Point masses versus extended objects.
In mechanics when the motion of a solid object is modeled with an equation what is being tracked is the motion of the center of mass. Whatever the shape of some object A is, if it isn't touching any other objects (hence the motion is not affected) then that object A is effectively treated as a point mass. That is how it is possible to describe an extended object with a single value for its inertia; you treat it as if it is a point mass.
In angular mechanics there is the concept of Moment of inertia. If you have a wheel you can use the moment of inertia of that wheel in a calculation.
That raises the question: there is no way to treat that wheel as a point mass, it is inherently an extended object. How can you assign a single value to the moment of inertia?
The entities of angular mechanics are constructed by capitalizing on rotational symmetry.
A wheel is (for the purpose of the calculation) rotationally symmetric. For simplicity we count only the mass of the rim (treating the mass of the spokes as negligable). For the magnitude of the moment of inertial we treat all of the mass of the rim as if concentrated in a point at some distance 'r' to the center of rotation. This is valid because all that counts for the magnitude of the moment of inertia is the distance to the center of rotation.
This analysis about the point of contact is known as analyzing about the instantaneous axis of rotation(IAOR). It's a neat trick and you can read up on this using any standard textbook.
The angular accelerations in both COM frame and IAOR frame will be the same. Only the point about which angular acceleration is defined is different for the two cases $\Rightarrow$ the linear acceleration of the same point(s) in the body won't match for these two cases(in their respective frame of references) but it'll match in Laboratory frame $\therefore$ when you analyze the motion in the Laboratory frame using either of the angular accelerations, the results will be identical.
Best Answer
First of all if you want to throw the ball after thirty degree of rotation of arm then according to me you should focus on the tangential velocity on that instant at which you have to release the ball.
I'm going to assume that the torque on the Catapult is going to remain constant all the time because you haven't told that if the arm is attached to the spring which is going to provide the torque to the arm.
So for that as you said torque is 30N.m so according to the relation:-
$τ=Iα$
Where tau is the torque on the arm, I is the moment of inertia of the rod and, alpha will be the angular acceleration.
For the calculation of moment of inertia I'm going to add the mass of the ball and diametrically add the diameter of the ball as well.
$I= \frac{1}{3} (M+m) (L+r)^2$
Taking M as the mass of rod and m as the mass of ball.
So it will come out to be.
$I = 0.16Kgm^2$
So by putting all this in the equation of torque we get:-
$30 = 0.16×α$
From here we get angular acceleration as
$α = 187.5$
Putting this angular acceleration in the equation of circular motion:-
$ω^2-u^2=2αθ$
Initial angular velocity will be zero hence final angular velocity will come out to be after putting theta as pi/6 :-
$ω = 14$
So here you got the angular velocity of the ball at the instant at which the arm is making 30 degree which you were asking for.
And hence you can find tangential velocity of the ball with which the ball will leave the Catapult.
That will come out to be:-
$V = ωr$
$V = 6.72 m/s$
This might happen that there may be some calculation mistakes so please go through the answer carefully and I hope you liked it!