Self-studying Classical Mechanics right now and was working through an example until I got to a point where I felt certain questions need be addressed. I will list the problem statement, the example up to the point in which I decided I need to clarify some things and then my question below.
The second part of this question is listed here: Average Tension in the string of a pendulum: Determining radial component of weight
Problem Statement
Is the average (over time) tension in the string of a pendulum larger or smaller than $mg$? By how much? As usual, assume that the angular amplitude $A$ is small.
Relevant portion of the example
Let $\cal{l}$ be the length of the pendulum. Then the angle $\theta$ depends on time like: $$\theta(t) = A\cos(\omega t)$$ Where $\omega = \sqrt{\frac{g}{\cal{l}}}$
Since $T$ is the tension in the string the radial $F = ma$ equation is: $$T – mg = m\cal{l}\dot{\theta^2}$$
My Question
The radial component of force in polar coordinates is given by $$F_r = m(\ddot{r} – r\dot{\theta^2})$$ Now I understand that we don't have an $m \ddot{r}$ term in our radial $F=ma$ equation since the radius $r$ in this case is the pendulum length $\cal{l}$ and the pendulum length is fixed hence $$\cal{l} = \text{constant} \implies \dot{\cal{l}} = 0 \implies \ddot{\cal{l}} = 0$$. But why is it that the term $m\cal{l}\dot{\theta^2}$ is positive above? It was my understanding that when setting up $F = ma$ equations the LHS is occupied by external forces (in this case the tension $(T)$ and the radial component of weight $(-mg\cos\theta)$ and the RHS by the mass times the acceleration of the system where the radial component of acceleration for polar coordinates is given by ($\cal{\ddot l} – \cal{l}\dot \theta^2$). The only thing I can think of is that the $m(\cal{\ddot l} – \cal{l}\dot \theta^2)$ term is also on the LHS of the $F = ma$ equation and then is moved to the RHS by addition. Then my understanding of setting up $F = ma$ equations is clearly flawed and I would appreciate any clarification on the matter!
Best Answer
You have your directions mixed up. If we pick the origin of our coordinate system to be the top of the pendulum (as you naturally do), this implies the positive $\hat{r}$ direction points outwards. So, resolving the forces with outwards being positive, we have:
$$mg\cos\theta - T = -ml\dot{\theta}^2$$
which is equivalent to the equation they give you.