# How to know when to divide what components and equate them in dynamics of circular motion? [duplicate]

Here's what I'm trying to say:

Why did we equate $$mg\cos\theta = T$$ instead of:

Why is $$T\cos\theta = mg$$ incorrect?

How do I know when to divide which components? like I could've equated $$T\cos\theta = mg$$ but this incorrect, but how come $$mg\cos\theta = T$$ is correct? what made this true and others false?

• On this site we strongly discourage the use of images for equations. Please use mathjax for all equations. Mathjax/Latex is the site standard. Jul 7 at 2:29
• And as it is, your images are the wrong way up making it hard to read. Jul 7 at 2:30
• This question exists here:-physics.stackexchange.com/questions/648058/…
– ACB
Jul 7 at 2:43
• @ACB The question you linked to is a different scenario. In that case $T$ does equal $mg\cos \theta$. Jul 7 at 2:59
• Does this answer your question? How to understand the ambiguity of vector resolvation?
– ACB
Aug 7 at 7:02

The key here is that the acceleration $$a_t$$ whose magnitude is $$|\frac{dv}{dt}|$$ is non-zero at the instant you snap the string, despite the speed being zero at that instant. (It resembles the situation where a ball at the top of it's trajectory has a velocity of zero ; it still has an acceleration.)
Hence, when you apply the laws of motion in the direction in which you have resolved forces $$Tcos\theta$$ and $$mg$$, you will find that:$$\sum F_y = ma_y => Tcos\theta - mg = m(a_tsin\theta)$$
However when you resolve the forces in the direction along the radius of curvature, at the instant the string snaps, the centripetal acceleration $$a_c$$ whose magnitude is $$v^2/R$$ is necessarily zero because $$v = 0$$ as you wrote. Furthermore, as $$a_t$$ and $$a_c$$ are perpendicular, there is no component of tangential acceleration along the radius. Hence, the laws of motion give you:$$\sum F_r = ma_r => T - mgcos\theta = mv^2/R = 0 => T = mgcos\theta$$