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Calculate angle of swing in a carousel

  1. May 31, 2015 #1
    Hello, this problem has given me headaches for days

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


    The diameter of a carousel is 15 m and it rotates 8,0 rounds per minute. Calculate how large of an angle is formed with respect to the vertical plane by the swing that hangs by the ropes fixed to the edge of carousel. The length of ropes is 2,5 m.

    So I mark:
    [itex]n=8,0 r/min=0,1333... r/s[/itex]
    [itex]r=(15/2) m=7,5 m[/itex]
    [itex]x=2,5 m[/itex]
    Angle between the vertical plane at the edge of carousel and swing is [itex]\theta[/itex]

    2. Relevant equations

    [itex] F=m\frac {v^2}{r}[/itex]
    [itex] F=mg [/itex]
    [itex] \omega=n2\pi[/itex]
    [itex][/itex]
    3. The attempt at a solution

    The angular velocity:

    [itex]\omega=n2\pi=0,1333... r/s*2\pi=0,8377... 1/s[/itex]

    Then I thought that I need the distance of swing from the edge of carousel. I mark this distance as R

    Thus, [itex]R=x*sin\theta[/itex]
    Then the total distance of swing from the middle is [itex]r+R[/itex], when tangential velocity [itex]v=\omega(r+R)[/itex]

    I figured that the angle should be solvable from vertical and horizontal forces which act on the swing:

    [itex]\left\{\begin{array}{cc} Fsin\theta=m\frac{v^2}{r+R}\\Fcos\theta=mg\end{array}\right.[/itex]

    Those divided gives:

    [itex]\frac{sin\theta}{cos\theta}=\frac{v^2}{g(r+R)}=\frac{\omega^2(r+R)^2}{g(r+R)}[/itex]

    [itex]tan\theta=\frac{\omega^2r+\omega^2x*sin\theta}{g}[/itex]

    [itex]g*tan\theta-\omega^2x*sin\theta-\omega^2r=0[/itex]

    and this is where I got stuck.. I tried solving this equation with substituting [itex]tan\theta=\frac{sin\theta}{\pm\surd(1-sin\theta)}[/itex] and [itex]sin\theta=u[/itex] which did not help. So maybe I got some idea wrong and my equation is wrong or I am just missing some trick by how to solve that equation. On the other hand I checked that does the correct angle (32) work on that equation and it gives about 0=0, so my equation should be correct (I think?).
     
  2. jcsd
  3. May 31, 2015 #2

    TSny

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    Yes, solving that exactly looks hard.

    Here's one approach that will lead to a good approximation in just a few steps.

    Rearrange the equation as ##tan \theta = \frac{\omega^2 r}{g} + \frac{\omega^2 x \sin \theta}{g}## .

    Start with a guess for ##\theta##. You can even start with ##\theta = 0## if you want. Plug this value into the right hand side of the equation to calculate a value for ##\tan \theta## and hence a new value for ##\theta##. Now plug this new value into the right hand side to find a new value of ##\tan \theta## and hence ##\theta##. If you repeat this a few times you will see that you will begin to converge to the answer. It only takes about three cycles to get 3 significant figures for this particular equation.
     
  4. Jun 1, 2015 #3
    Thank you for answering

    That method gives me the correct result, which is fantastic! But I don't know how I could have known to use it. Does this 'iteration' work because ##tan \theta## and ##sin \theta## are trigonometric functions (periodic)? Since not all functions have fixed points to which they converge
     
  5. Jun 1, 2015 #4

    TSny

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    I find that this method often converges to the answer. I don't know a general way to tell if it will work before trying it.
     
  6. Jun 1, 2015 #5
    Okay, but thanks for helping me. I'm glad that your method worked and I got an answer to the problem. I was only thinking algebraic methods.. :biggrin:
     
  7. Jun 1, 2015 #6

    haruspex

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    You didn't post what happened when you tried your u substitution. It produces a quartic, no? In principle, you can solve that algebraically.
     
  8. Jun 2, 2015 #7

    ehild

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    In general, an iteration procedure xn+1=f(xn) converges in a region where |df/dx| <1. See picture.
    In the figure, the iteration converges to A but it does not converge to B.
    iteration.JPG
     
  9. Jun 2, 2015 #8
    Yeah and I took it as a warning sign that I must be doing something wrong. I have never solved a quartic function and nobody have taught me that in school. So surely I should not be meant to solve it like that, which is why I stopped trying to solve it. But there really seems to be no good trigonometric identity which I could use to make equation simpler

    Well anyway, after solving it numerically, I got curious about solving that quartic function and found a quartic equation calculator.. so I can show the equation I got and the answers

    [itex]tan\theta=\frac{\omega^2r+\omega^2x*sin\theta}{g}[/itex]

    [itex]g~tan\theta=\omega^2r+\omega^2x*sin\theta[/itex]

    [itex]g\frac{sin\theta}{\pm\surd(1-sin^2\theta)}=\omega^2r+\omega^2x*sin\theta[/itex]

    [itex]g\frac{u}{\pm\surd(1-u^2)}=\omega^2r+\omega^2x~u[/itex]

    [itex]g^2\frac{u^2}{1-u^2}=(\omega^2x~u+\omega^2r)^2[/itex]

    [itex]g^2\frac{u^2}{1-u^2}=\omega^4x^2u^2+2\omega^4~x~r~u+\omega^4r^2[/itex]

    .
    .
    .

    [itex]x^2u^4+2~x~r~u^3+(r^2-x^2+\frac{g^2}{\omega^4})u^2-2~x~r~u-r^2=0[/itex]

    Then I used this site's http://www.1728.org/quartic.htm calculator which gave:

    [itex]x_1=0.5343342065691381[/itex]

    [itex]x_2=-3.05761046744197 +i* 5.553303328218349[/itex]

    [itex]x_3=-3.05761046744197 -i* 5.553303328218349[/itex]

    [itex]x_4=-0.41911327168519863[/itex]

    [itex]x_2[/itex] and [itex]x_3[/itex] can be dismissed since answer should be a real number. [itex]x_1[/itex] is correct, but how to get rid of [itex]x_4[/itex]? Because it is negative? But anyway.. I just used a calculator and certainly this method should not be preferred even though it is interesting.
     
  10. Jun 2, 2015 #9

    ehild

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    You mean u when writing x1...x4, don't you? And these are values for sinθ. How would the swings stand if sinθ was negative? Outward or inward?

    carousel.JPG
     
  11. Jun 2, 2015 #10
    Yes I mean u

    Okay yeah, those values are for [itex]sin\theta[/itex]. So [itex]\theta[/itex] should be positive to make sense physically (because of centripetal force), for swings to be outward. Thus it leaves only [itex]x_1[/itex] ([itex]u_1[/itex])
     
  12. Jun 2, 2015 #11

    ehild

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    Yes, the angle should be outward, so positive, and it also can not be greater than pi/2, so the negative sign in front of your square root cancels.
     
  13. Jun 2, 2015 #12
    I see, so [itex]\frac{sin\theta}{\pm\surd(1-sin^2\theta)}[/itex] needs to be positive → [itex]\frac{sin\theta}{\surd(1-sin^2\theta)}≥0[/itex] because [itex]tan\theta≥0[/itex] and [itex]\theta≤\frac{\pi}{2}[/itex] by definition. So [itex]0≤\theta≤\frac{\pi}{2}[/itex]
     
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