Rate of change of tension with respect to theta in vertical circular motion

Question

I was considering vertical circular motion of a mass $m$ tied to a string of length $r$ that has an angular velocity $\omega$, linear speed $v$. The tension in the string is $T$ and is constantly changing.

The components of the weight are $$mg\sin(\theta) \text{ and }mg\cos(\theta)$$ with $mg\sin(\theta)$ acting tangentially and $mg\cos(\theta)$ radially. The speed must constantly be changing with a change in the angle $\theta$, and so the equation below:

$$ T - mg\cos(\theta) = \frac{mv^2}{r} $$

must be changing with respect to theta. If speed changes then $\omega$ must change because $\omega = \frac{v}{r}$.

Here I am confused slightly about the entire motion in general. Does anything stay constant in this scenario? I think I have a conceptual misunderstanding of this type of motion in general. Can someone help me clarify this concept?

I tried to differentiate with respect to $\theta$ thinking that might lead to something:

$$ \frac{d(T - mg\cos(\theta))}{d\theta} = \frac{d (mr(\omega)^2)}{d\theta} $$ $$ \implies \frac{dT}{d\theta} = mr\frac{d(\omega)}{d\theta} - mg\sin(\theta)$$

I don't know what to do with the $\frac{d\omega}{d\theta}$ in the expression. I also am not completely sure what I did is valid in the first place.

Answer 1

You are making this too complicated.

You can calculate the velocity as a function of angle $\theta$ from conservation of energy.

Once you have that, you know the centripetal force needed - which will be the sum of the radial component of gravity and the tension in the string.