To make it easier to analyze, let's decompose this 'reaction' force on the pivot in directions perpendicular $(x')$ and parallel $(y')$ to the bar (instead of the indicated $x$ and $y$ you indicated). Let's call these components $F_x'$ and $F_y'$.
Equations for translational motion:
$$F_y' - P_y' = 0$$
$$F_x' - P_x' = m*a$$
Now, let's calculate the torque about the center of mass, assuming the pendulum's length is L:
$$T(cm) = -F_x'*L/2 = I(cm)*b$$
Where $b$ is the angular acceleration of the pendulum about the center of mass. $I(cm)$ for a pendulum is equal to $mL²/12$. Hence:
$$F_x' = -mL*b/6$$
The final problem is determining the relationship between $b$ and $a$. That is easily done considering the center of mass rotates in a circular motion about the pivot, hence:
$$a = b*L/2$$
$L/2$ being the radius of the circumference described by the pendulum's center of mass. Making the substitution:
$$-mLb/6 - P_x' = mLb/2$$
Remember that $P_x' = mg*sin(\theta), \theta = angle$ between the pendulum and the vertical direction $y$.
- $mgsin(\theta) = -2mLb/3$
Which is the same differential equation you would obtain considering the torque about the pivot? Don't forget $b$ is the second time derivative of $\theta$.
I hope this clarifies your doubts a bit. As for the nature of these forces, they are of electromagnetic origin (just as any normal force) caused by the bond strength of whichever materials are in contact at the pivot. And yes, I am considering there is no attrition and hence energy dissipation.
To understand the physics here we should first consider the assumptions made, and perhaps try to justify them with some physical arguments.
Assumptions of this simple pendulum:
- Newtonian physics apply. (ie any non-classical effects are negligible)
- The "string", or perhaps more accurately rod, has a fixed length $l$. (That is the string remains under tension but does not stretch)
- The string is massless. (or is negligible next to the mass of the ball)
- The system is frictionless, leaving only gravity (taken to be constant) and tension to act on the ball.
The string keeps the ball at a fixed distance $l$ from the pivot and hence the ball moves to trace out the arc of a circle. Knowing this we use Newtons Laws to resolve the forces involved. The acceleration due to gravity is taken to be $g$ hence the force (acting down) on the ball is $F_g = mg$, where $m$ is the mass of the ball. Decomposing this gravitational force into radial and tangential components we arrive at the expressions given in the diagram: $$F_{\text{radial}} = mg\cos \theta\ \text{ and }\ F_{\text{tangent}} = mg\sin \theta.$$ The tension $T$ in the string acts only radially and since it does not stretch, has magnitude equal to the sum of the centripetal force and the radial component of the gravitational force. The force necessary to keep an object in circular motion in free space (the centripetal force) is given by $$F_c = \dfrac{m v^2}{l},$$ where $v$ is the instantaneous velocity of the ball. Thus $$T = \dfrac{m v^2}{l} + mg\cos\theta.$$
So the tension in the string varies as the ball speeds up and slows down along it's trajectory. It is strongest when $\theta=0$ (at the bottom of the pendulum) since at this point the string is in direct opposition to gravity, and it is weakest when $v=0$ (at the top of the swing) as this is when the centripetal force vanishes and the radial component of gravity is at it's lowest.
It's worth thinking about the assumptions we've made and when they fail (for example is the string really always under tension?). When is this a useful model and how could we adapt it to account for the trickier edge cases?
Best Answer
You are missing concepts.
In a pendulum, the net force at any instant can be resolved into radial and tangential components(They are infact mutually perpendicular to each other at each instant). The radial component is along the string, directed towards the centre of circular motion. It actually provides the required centripetal acceleration . When the string makes angle $\theta$ from the vertical , you can write: $$T-mg cos\theta=\frac {mv^2}{l}$$
Note that at the instant of maximum displacement only, the pendulum instantaneously comes to rest. So at maximum angular displacement $\theta_0$, we can write:
$$T_0=mgcos\theta_0,$$ where $T_0$ is the tension in the string at that instant.
Note that the tension also varies accordingly throughout the motion.
The tangential component of the net force is $mgsin\theta$ which provides the required tangential acceleration (or, restoring torque for oscillation)This is along the motion of the pendulum. This is zero only when the string is vertical.
So now you can see, at the extreme position, only tangential component of force is present.
You can use mechanical energy conservation for further analysis of pendulum system.