# Why is $mg$ split into components instead of the tension in vertical uniform circular motion?

I am a beginner physics student.

I am trying to prove that the magnitude of the tension varies sinusoidally as an object P travels around the circle or something of that sort. Thus, I am evaluating the tension at certain points, or a general $$\theta$$. I am trying to solve for the tension in terms of the other variables, but when I split the tension into components, I got non-customary results. After some research, I found that $$mg$$ is to be split into components instead of the tension. Why so?

It seems that as the object undergoes vertical uniform circular motion, the tension's direction changes, while $$mg$$ does not. How come when solving for tension when there is an angle, mg is split into components instead of the tension? In a horizontal uniform circular motion, the tension was split into components, but what caused the change in the variable being split into components?

Also, here it says that the tension at the bottom is $$6mg$$ more than the tension at the top. Shouldn't it only be a difference of $$2mg$$, because the tension at the top is $$m(\frac{v^2}{r}+g)$$ while the tension at the bottom is $$m(\frac{v^2}{r}-g)$$?

An example for reference:

Image source: here

• Please provide a clear example of vertical uniform circular motion and horizontal uniform circular motion, as the choice of which vectors to resolve into components is usually context dependent. Dec 30 '20 at 21:12
• I have added an example to which I am referring too.
– user275102
Dec 30 '20 at 22:36

You have to consider conservation of energy: $$E_{kin}^{top}+E_{pot}^{top} = E_{kin}^{bottom}+E_{pot}^{bottom}$$ You can then solve for e.g. the velocity of the mass at the top and then compare the forces you obtain: $$|F_{top}|= \dfrac{mv_{top}^2}{r} - mg\qquad\text{and} \qquad |F_{bottom}| = \dfrac{mv_{bottom}^2}{r} + mg$$ I think you can figure it out from there, good luck.

Edit: why does only the $$mg$$ part split into components? Remember that forces are actually vector quantities, i.e. have a direction, then think about which direction the gravitational force has and which the centripetal force has.

• Thank you! Could you elaborate a little the last part where you remark about the direction of gravitation force and centripetal force? I understand that gravity acts downwards, and that the centripetal force can have more than one component, but I am still confused.
– user275102
Dec 30 '20 at 22:38
• Both forces have 3 components, if you choose an appropriate coordinate system, in this case z-axis pointing 'up', then some vectors may only point in the direction of one of your axis, e.g. $\vec{F}_G=-mg\hat{e}_z$. The centripetal force on the other hand always points to the center of rotation. The tension of a string is defined as the magnitude of forces 'pulling' on it. Since shearing forces (i.e. forces acting orthogonal to the string) don't pull on the string, they don't contribute to the tension. Dec 31 '20 at 9:20
• Thanks! Could I also say that because the magnitude of the tension is always changing, it is better to split $mg$ into components because it stays constant?
– user275102
Dec 31 '20 at 21:29
• Let me rehrease that: when analyzing the tension in vertical uniform motion by using a free body diagram, it is better to split $mg$ into components instead of the tension because the magnitude and direction of the tension vary, while $mg$ stays the same. Is this right?
– user275102
Jan 1 '21 at 2:04