A very nice thing you pointed out there which many people tend to skip...
Actually the fundamental rule of taking components of forces is that the coordinate axes should be perpendicular to the instantaneous direction of velocity or you can say the instantaneous direction of motion.
As the tension force is already perpendicular to the direction of motion, we resolve the $mg$ force (weight) into two components.
Hope this helps!
The answer is clearly b), T = mgcosθ + mv2/r and the reason shines up precisely through a comparison between Tension and Normal Force.
Normal Force is the one exerted by a surface to avoid penetration into it and Tension Force is the one exerted like here by a hanging string to avoid escape from it. One could say that the usually called Normal Force is pushing Normal Force, avoiding that an object breaks in, while Tension is pulling Normal Force, avoiding that the object breaks away. One of the actors in these interactions is often the earth (which pushes on a block or which pulls from a tree, which in turn pulls from a string that pulls from the pendulum’s bob), so we can be positive that the earth can gather sufficient strength to avoid penetration or escape, as long as the material involved has sufficient cohesion. Also in both cases, the Newton 3rd law pair of that force is another Normal Force. The pair of the Normal Force exerted by the earth on a block is the Normal Force exerted by the block on the earth. And the pair of the Tension exerted by the string (indirectly the earth) over the pendulum’s bob is the Tension exerted by the bob over the string (and indirectly the earth). A different thing is that the block and the bob, with their respective masses, have been previously accelerated by Gravity acting over them and this entails that they come to face the earth with a Normal Force that is tainted by their previous interaction, so it happens that if the earth stands up to the task and succeeds in avoiding penetration/escape, it will be because it has built up sufficient force to match its own Gravity Force, the object’s Weight…
In our case, as per formula b), Tension does two things:
In short form: it neutralizes any acceleration in the radial direction whose origin is the radial component of Gravity. In full form: given that the bob has been radially accelerated by Gravity and that consequently it pulls on the string with Tension whatever, the string reacts by applying to the bob Tension whatever, which turns out to totally match and cancel the “plans” of the said radial component of Gravity. This is logical: there is nothing in that radial escape that is compatible with Tension’s job of maintaining the distance bob-pivot, so all is cancelled.
Instead Tension has no problem with Gravity accelerating the bob tangentially and giving it more and more velocity as it descends. Its objection in this respect is only partial: it is not against the modulus, but only against the fact that the bob’s direction tends to remain constant, which would also entail higher separation between pivot and bob. Thus when the string feels that the bob is doing this (trying to escape tangentially) and thus exerting the corresponding Normal Force, the string reacts by applying equal Normal Force aimed at, exclusively, changing the bob’s direction inasmuch as necessary to keep circular motion. In doing this, Tension turns out to frustrate only one element of the “plans“ of the tangential component of Gravity, i.e. the plan that the bob continues in a straight line. At each instant, the velocity modulus faced by Tension is different (because acceleration in modulus is not counter-acted), but that makes no difference: still this term of Tension is mv2/r, even if that v is varying.
Finally a metaphor: it is like if the bob were mounted simultaneously on two vehicles: one suddenly accelerates away radially, but the string holds tight the bob, which falls off onto the second vehicle, so that first one has no impact on the motion; then the second vehicle accelerates tangentially and the string consents the change in modulus but constrains its motion along the circular path, like a sort of geodesic.
Conclusion: b) is right and this is consistent with the nature of Tension as a sort of pulling Normal Force, because the string does what is needed to fight not one, but the two things that are incompatible with maintaining the distance pivot-bob: radial acceleration and constant tangential direction. In both cases there is a pull over the string, i.e. there is Tension exerted by the bob to which the string reacts by applying its own Tension over the bob. And in both cases, thus the “plans” of Gravity over the bob are frustrated or counter-acted upon, either totally or partially.
Best Answer
In general, Newton's second law in polar coordinates is given by
$$\mathbf F=m\left(\ddot r-r\dot\theta^2\right)\hat r+m\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$
In the case of the pendulum, $\dot r=0$ and $\ddot r=0$, and so we are left with
$$\mathbf F=-mr\dot\theta^2\hat r+mr\ddot\theta\hat\theta$$
Also for our pendulum, we have $\mathbf F=\left(mg\cos\theta-T\right)\hat r-mg\sin\theta\,\hat\theta$
Therefore, we can determine from Newton's second law that
$$mg\cos\theta-T=-mr\dot\theta^2$$ $$-mg\sin\theta=mr\ddot\theta$$
At a point in time where $\dot\theta=0$, we must then have $T=mg\cos\theta$ at that point in time as well.
We have circular motion, but not uniform circular motion. The centripetal acceleration varies during the motion of the pendulum, and as we can see is equal to $0$ at the ends of the swings.