Pendulum Circular Motion query Using two different approaches, I appear to receive contradictory information about the tension force in a simple pendulum.
Under the idea of a centripetal force, the Tension - component of $mg$ in that direction(i.e. line of action of tension force) $=$ centripetal force to provide the 'circular motion' for small $\theta$ in a simple pendulum
However, resolving forces in that direction and using $F= ma$, since the string length is fixed, the bob at the end of the string doesn't change 'height' in direction of the tension force so surely
Tension $=$ Component of $mg$ in that direction
Any help would be appreciated!
 A: You can simply imagine a circular motion.
In a circular motion the distance of a particle describing the circle from the centre remains constant. This distance is the string of the pendulum (which doesn't slack) but still you have some acceleration towards the centre.
What happens is that the particle constantly tries to fall towards centre but is taken back due to a velocity tangentially at any given point. So instead of simply falling to the centre it rotates around.
Similarly, in the case of a pendulum though the length remains same. It still has some net force towards the centre and Bob does try to fall towards the pivoting point but instead of that, it undergoes the pendulum motion due to the presence of a tangential component of $mg$ (perpendicular to the component which is along the direction of tension) which helps the Bob to describe a kind of circular shape.
So stick with the first one , i.e. : $T$ - component  of   $mg$
in that direction= centripetal force and
$F_{net}$ is not equal to zero (along $T$)
$\implies T$ is not equal to the component of $mg$ along this direction
A: Just because there is no change in length of the string doesn't mean there is no acceleration in the radial direction. If you understand circular motion properly, you'll see that the particle undergoing circular motion is always accelerated towards the centre, even if the radius remains constant. The acceleration is such that the direction of velocity changes, as a tangent to the circular path. Thus when you apply $F=ma$, you have to consider not only the tensile and gravitational forces, but also the centripetal acceleration.
