Problem regarding equations in different frame of reference

Question

Previously I asked this question and got the following answer.

I'm asking for reason why you can split Tension(first case) but not mg(second case)

When you say 'splitting,' you really mean projecting a force along some axis.

First let's cover why you 'split' the tension as you did in your first approach.

You know that the net force in the y-direction is zero; finding the y-component of tension helps you solve the problem. That's why you bother to 'split.'

Q. Why can't I split the force of gravity?

A. Of course you can! The question, is, however, along what axes?

From what I understood, you decided to calculate the component of gravity force that lies on the same axis as the tension force (or the string). There is nothing wrong with that. However, the true equation you'd get would be:

$$0 = T - mg\cos\theta + F_{\rm fictitious}$$

The reference frame you chose is non inertial. Though there is no motion along the string, the string itself is accelerating. Thus, there will be what we can call a fictitious force.

The whole reason why we 'split' along x- and y-axis in this problem is because the frame is inertial, so we can solve for the tension.

Here user256872 Writes the equation

$$0 = T - mg\cos\theta + F_{\rm fictitious}$$

and says that it is written in non inertial frame of reference. I don't understand how this equation is related to frame of reference.

Can anybody explain me properly.

Answer 1

Let's forget about non-inertial reference frames and fictitious forces. The key thing to note is that $\frac {v^2} R$ is not a horizontal force, it is a horizontal acceleration. So when we apply Newton’s second law the $\frac {v^2} R$ terms are on the opposite side of the $F=ma$ equation to the forces.

Because the acceleration is horizontal, it has zero vertical component, so if we are working in vertical/horizontal axes we can say

$T_F \cos \theta - mg = m \times \text{ vertical acceleration} = 0 \\T_F \sin \theta = m \times \text{ horizontal acceleration} = m \frac {v^2} R$

Alternatively, we can work in axes that are parallel and perpendicular to the string. The weight of the bob has components along both axes, but the tension in the string has no component perpendicular to the string. The horizontal acceleration can also be resolved along parallel and perpendicular axes.

If we take the positive direction of the parallel axis as being upwards along the string and the positive direction of the perpendicular axis as being downwards and inwards towards the axis of the cone then we have

$T_F - mg \cos \theta = m \times \text { parallel acceleration} = m \frac {v^2} R \sin \theta \\ mg \sin \theta = m \times \text { perpendicular acceleration} = m \frac {v^2} R \cos \theta$

Note that $T_F - mg\cos\theta$ is not zero because although the length of the string is constant, its direction is changing. So there is a non-zero acceleration parallel to the string. You could, if you wanted, work in a non-inertial reference frame rotating with the bob, in which the direction of the string is also constant, but then you would have to introduce fictitious forces, which I think creates confusion.

With a little algebra you can show that the second two equations are equivalent to the first two equations above. So you get the same answer whether you use vertical/horizontal axes or whether you use parallel/perpendicular axes.