# speed of conical pendulum at angle alpha to horizontal

1. The problem statement, all variables and given/known data

Assuming we know the length of the string $L$, radius of the swept out circle $r$, angle formed by string and centre of circle, $\theta$, and angle the swept out circle is to the horizontal, $\alpha$, what is the speed, $v$, of the mass if it is constant?

picture: http://ift.tt/1imelD8

2. Relevant equations

$\text{F}_{\text{net}}=ma$
$a=\frac{v^2}{r}$

3. The attempt at a solution

So in trying to solve this, I broke it down into x and y components and in the horizontal plane, I got:

$\text{F}_{\text{net}} = \text{T}\,\sin(\theta + \alpha)$

$ma_{x} = m(a\cos \alpha) = \text{T}\, \sin (\theta + \alpha)$

$\implies \text{T} = \frac{ma\cos \alpha}{\sin(\theta + \alpha)}$

The angle, $\theta + \alpha$ arrises due to some geometry where if i tilt my axis so that $\text{F}_{g}$ points directly down on the negative y-axis and $\text{T}$ points into the first quadrant, the angle from the positive x-axis is $\frac{\pi}{2}-(\theta + \alpha)$

and for the vertical plane i get:

$\text{F}_{\text{net}} = \text{T}\cos (\theta + \alpha) – mg$

$ma_{y} = m(a\sin \alpha) = \text{T} \cos (\alpha + \theta) – mg$

and substituting $\text{T}$ into our equation I get;

$a\sin \alpha = \frac{a \cos \alpha}{\sin (\theta + \alpha)} \cos (\theta + \alpha) – g$

since $a = \frac{v^2}{r}$, I substitute and isolate for $v^2$ to get

$v^2 = \frac{gL\sin (\theta) \sin (\theta + \alpha)}{\cos (\theta + 2\alpha)}$

I am just wondering if I did this properly? I mean if I let $\alpha = 0$, then I end up with

$v = \sin \theta \sqrt{\frac{gL}{\cos \theta}}$, which is what I got when I was working with a normal conical pendulum, however, I feel that as the mass starts to approach the ground then it should speed up. Is it even possible to let $v$ be constant and is my approach correct?

http://ift.tt/1nhlgd5