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 first part of this question series is linked here: Average Tension in pendulum string: Understanding the radial $F = ma$ equation
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
My question involves determining the radial component of the weight. Since we intend on utilizing the radial component of the $F = ma$ equation and physically, by inspection, we know that the only forces on the string in the radial direction are the tension $T$ and the radial component of the weight $W_{rad} = -mg$. $W_{rad}$ has been given in the example as $W_{rad} =-mg\cos\theta$ however when calculating this myself I ended up getting $W_{rad} = -\frac{mg}{\cos\theta}$ I have provided a figure below outlining my strategy: Figure $1$ is simply a physical drawing of the system. Figure $2$ is my attempt at listing the forces on a piece of the string. Figure $3$ I have rotated the bottom triangle to make it clearer that I was utilizing right triangle trigonometry (SOHCAHTOA) to determine the value of the hypotenuse. Where I utilized $$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-mg}{W_{rad}}$$ and simple algebra leads me to my result. So, my question is where exactly did I go wrong? And more generally, are there any good resources for determining various components of forces? As I tend to be extraordinarily bad at determining the various components of forces given a physical system.
Best Answer
this is the free body diagram
take the sum of the forces toward $~\mathbf e_T~$ you obtain the tension force $~T$
$$T=m\,L\,\dot\theta^2+m\,g\cos(\theta)$$
and take the sum of the torques about the point A you obtain the equations of motion
$$ m\,L^2\,\ddot\theta+m\,g\,L\sin(\theta)=0\quad\Rightarrow\\ \ddot\theta+\frac gL\sin(\theta)=0$$