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Speed of conical pendulum at angle alpha to horizontal

  1. Jun 18, 2014 #1
    1. The problem statement, all variables and given/known data

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

    picture: http://upload.wikimedia.org/wikipedia/commons/5/53/Conical_pendulum.svg


    2. Relevant equations

    [itex]\text{F}_{\text{net}}=ma[/itex]
    [itex]a=\frac{v^2}{r}[/itex]


    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:

    [itex]\text{F}_{\text{net}} = \text{T}\,\sin(\theta + \alpha)[/itex]

    [itex]ma_{x} = m(a\cos \alpha) = \text{T}\, \sin (\theta + \alpha)[/itex]

    [itex]\implies \text{T} = \frac{ma\cos \alpha}{\sin(\theta + \alpha)} [/itex]

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

    and for the vertical plane i get:

    [itex]\text{F}_{\text{net}} = \text{T}\cos (\theta + \alpha) - mg[/itex]

    [itex]ma_{y} = m(a\sin \alpha) = \text{T} \cos (\alpha + \theta) - mg[/itex]

    and substituting [itex]\text{T}[/itex] into our equation I get;

    [itex]a\sin \alpha = \frac{a \cos \alpha}{\sin (\theta + \alpha)} \cos (\theta + \alpha) - g[/itex]

    since [itex]a = \frac{v^2}{r}[/itex], I substitute and isolate for [itex]v^2[/itex] to get

    [itex]v^2 = \frac{gL\sin (\theta) \sin (\theta + \alpha)}{\cos (\theta + 2\alpha)}[/itex]

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

    [itex]v = \sin \theta \sqrt{\frac{gL}{\cos \theta}}[/itex], 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 [itex]v[/itex] be constant and is my approach correct?
     
  2. jcsd
  3. Jun 18, 2014 #2

    Orodruin

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    If I get your problem correctly you want the circle to have a tilt with respect to the horizontal plane. If this is the case, the speed will not be constant as the energy is being converted from potential to kinetic.
     
  4. Jun 18, 2014 #3
    Yes, that is how I set up the problem, and exactly what I was wondering about the speed as its height from the ground is constantly changing. any thoughts on how to take that into account?
     
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