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

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.

• What's constant? $g$, $r$, $v$, and angular momentum.. But I'm going to guess that that doesn't help you. Can you help us understand what is troubling you? – garyp May 29 '17 at 20:20
• @garyp How will $v$ be constant? And I think my trouble is fundamentally the concept behind what happens when the mass is near the top. I reasoned that because there are no horizontal forces when the mass is exactly on the top, the mass' horizontal velocity moves it forward and the tension and weight pull it down. I wanted to then look at what changes with respect to a small change in the angle. This is to get a better picture of what's happening when the mass is rotating. – Neev Parikh May 29 '17 at 20:24
• Sorry ... since you specified $v$, I assumed it was given as a constant ... like a Ferris wheel. So with my new understanding: angular momentum is not conserved, but energy is. – garyp May 29 '17 at 22:44

You can calculate the velocity as a function of angle $\theta$ from conservation of energy.
• And then I can differentiate with respect to $\theta$, because velocity will be in terms of $\theta$. I will try it and comment back in a bit. – Neev Parikh May 30 '17 at 4:15