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The Swing Problem:How can you go completely around?

  1. Jul 13, 2004 #1
    Every kid has asked the question on a swing, could I go all the way around? How fast would I have to go? I wanted to find the answer.I’ve taken the first year of calculus, and I have been having trouble with the math of this problem :confused: . Im not sure there is an algebraic solution to this problem. :surprise:

    You are on a park swing with radius R. What must your minimum speed be at the bottom of your circular trajectory such that you perfectly traverse the circular path created by the rotating swing?

    Note: There are several parts to this problem. First you must calculate the rate of change of theta. You must then integrate the deceleration of gravity over the swing’s circular path. At all times the upward vector of centrifugal force, (V^2)/R must be greater than the acceleration of gravity so that you stay on your circular path.

    Thanks, Elias
     
  2. jcsd
  3. Jul 13, 2004 #2

    Doc Al

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    The easy way is to examine what happens at the top of the motion, where the speed is lowest. Find the minimum speed needed at the top to maintain tension in the swing (and thus maintain the circular path). Then use energy conservation to find the corresponding speed at the bottom.
     
  4. Jul 13, 2004 #3
    use simple circular motion physics, you dont need any integration. The main fores actiong on you are gravity and centripital acceleration. At the top of the swing is the most miportant part, since the gravity and centripital acceleration are in opposition of eachother, so if you want the kid to have a circular path and make it over the actual swing, you must set the two forces equal to eachother, solve for v, and this will be your MINIMUM speed. Like so:

    mg = mv^2/r (m's cancel out)
    g = v^2/r
    v = (sqrt(gr))
    therefore the minimum speed of the swing must be: v = (sqrt(gr))
     
  5. Jul 13, 2004 #4
    v=(sqrt(gr)) is ONLY the minimum velocity at the apex of the trajectory. The original starting velocity is NOT v=(sqrt(gr)).

    Mathematically, this is wrong because you neglect that V varies according to the gravitational vector you also neglect that V varies according to the varying rate of change of theta.

    Experimentally we can also see this if we solve for a 10 foot swing. V=(sqrt(32X10))
    That means at 12 MPH youve got enough speed to go all the way around. Not True. Try it yourself.
     
  6. Jul 14, 2004 #5
    The forces acting are the magnetic force and the string tension. Not the centr. acc.

    Besides, you can't write
    [tex]\sum \vec F = - m\frac{v^2}{R} \hat \rho[/tex]

    Because there this is NOT a uniform circular motion since the speed ain't constant.

    for E_Man:
    Follow this
     
  7. Jul 14, 2004 #6

    Doc Al

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    Magnetic force?? And, true, the speed isn't constant, but so what?

    This is an easy one. I thought I had explained how to do it in my earlier post. Here are the details:

    At the top of the motion, we need the minimum speed so that the swing remains taut (assuming a flexible chain). From Newton's 2nd Law we get:[itex] F = mg = mv^2/r[/itex], so the minimum KE at the top of the swing is [itex]KE_{top} = mgR/2[/itex]. From conservation of energy, the KE at the bottom is [itex]KE_{bot} = mgR/2 + 2mgR = 5/2 mgR[/itex], so the minumum speed at the bottom must be: [itex]v_{min} = \sqrt{5gR}[/itex].
     
  8. Jul 14, 2004 #7

    Galileo

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    Doc Al's right. I think the easiest way to do it is to find a relation between the speed at the bottom and the speed at the top by means of energy conservation.
    Setting the gravitational potential energy zero at the bottom we find for the energy of the kid at the bottom:
    [tex]E = \frac{1}{2}mv_{bottom}^2[/tex]

    And at the top (at height 2R):

    [tex]E = \frac{1}{2}mv_{top}^2+2mgR[/tex]

    Therefore (equating these and simplifying):
    [tex]v_{bottom}^2=v_{top}^2+4gR[/tex]

    You want [tex]v_{top}^2/R \geq g[/tex], so you must have:

    [tex]v_{bottom}^2 \geq gR+4gR=5gR[/tex]
    And so the minimum speed at the bottom must be [tex]v_{min}=\sqrt{5gR}[/tex]
     
    Last edited: Jul 14, 2004
  9. Jul 14, 2004 #8
    Like your answer Doc but you neglected the fact that g is not constant ,and his mass changes with speed -- if the guy neglected this he might have a bad accident. Ray.
     
  10. Jul 14, 2004 #9

    Doc Al

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    Give me a break. :rolleyes:
     
  11. Jul 14, 2004 #10
    Sorry, I meant gravitational force, I was thinking of something else.

    And the fact that velocity ain't constant means that there is a tangentical acceleration as well, so you have to take that into account in Newton's 2nd Law.
     
  12. Jul 14, 2004 #11

    BobG

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    I think this was the mistake my daughter made. She tied herself to the swing by the waist, which obviously made things safe enough to let go the chains.

    Oooh :eek: One broken arm later, plus a physics lesson learned........
     
  13. Jul 14, 2004 #12

    LURCH

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    Have you ever shortened the chain on a swing that was too low to the ground? You know method I'm talking about; you thrust the swing forward with enough velocity to go all the way around. The velocity at which you must make the swing travel when you push it forward (the velocity at the bottom of its ark, where you release it) is exactly the same whether the swing is empty or occupied.
     
    Last edited: Jul 15, 2004
  14. Jul 14, 2004 #13

    Doc Al

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    nit picking :-)

    At the point where I apply Newton's 2nd law (the top of the motion) the tangential acceleration is zero. :smile:
     
  15. Jul 14, 2004 #14
    THANKS DOC AL AND EVERYONE ELSE! Thanks too fo the link given by eJavier!

    I pose a new challenge:
    Can anyone find an explicit function for the velocity as a function of time?
     
  16. Jul 14, 2004 #15

    turin

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    I'm pretty sure that the diff. eq. is nonlinear. Diff. eqs. of this kind are not generally solvable analytically (which basically means it is not generally possible to find the explicit solution, even in principle). The solution is probably some obscure special function defined by the diff. eq. that you can find.
     
  17. Jul 14, 2004 #16

    Janus

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    Well you can neglect the mass increase because even with a near inifinite swing length the velocity of the swing won't exceed escape velocity (11205 meters/sec) and at that velocity the mass would only increase by a factor of 1.0000000007. :wink:

    As for the varience in "g", if you absolutely have to, you could use the formula:

    [tex]v =\sqrt{2GM \left\{\frac{r}{2(R+2r)^{2}}-\frac{1}{R+2r}+\frac{1}{R}\right\}} [/tex]

    Where r is the radius of the swing and R is the distance of the bottom of the swing from the center of the Earth.

    Now for a swing with a raduis of 4 meters, we would get an answer of:

    14.03029073 m/s or 31.56815415 mph, using Doc Al's equation.

    Using the new equation, we get an answer of:

    14.03028017 m/s or 31.56813039 mph.

    Thus we see that Doc Al's equation, though slightly less accurate, fudges a bit to the high side and still gives an answer that would safely get you over the top of the bar. :smile:
     
  18. Jul 15, 2004 #17
    Ah ok could you tell me more about these differential equations? this is the problem I originally had. I was trying to aproach the problem by finding the functions for each variable. Darn.
     
  19. Jul 15, 2004 #18
    janus

    Thank you Janus , I wondered if Doc would pick up on the fact that the guy was still safe -- but he just rolled his eyes instead -- clearly not a physicist of the first water - whereas you have genius quality . Ray
    Now about length contraction -------------------.
     
    Last edited: Jul 15, 2004
  20. Jul 15, 2004 #19
    If I remember correctly large pendulum swings involve integral(1/sqr(cos())) functions at least for 'period' and belong to eliptic functions -- angels fear to tread --
    Ray.
     
  21. Jul 15, 2004 #20
    Would the function for velocity along a pendulum even fit? I mean it seems like the bottom half of the circle would have one equation and the top half would have another, because of the fact that on the bottom hemisphere of motion the deceloration of gravity is on an inclined plane, and on the top half it is straight down
     
  22. Jul 15, 2004 #21

    turin

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    If you require that you should "safely" go all the way around, then your "orbit" in phase space would be well outside a particular orbit that is known as the "separatrix" (and, incidently, the 1-D pendulum, which is basically this problem, has been subjected to a tiresome share of analysis in classical mechanics). In phase space, the orbit is quite simply a sinusoid (I'm pretty sure). You can read the velocity off of the location in phase space directly, as velocity (or, more appropriately, momentum) is essentially one of the coordinates of phase space. Based on this argument, I would wave my hands and say that the velocity is a well defined function in the sense that it obtains a particular single finite value at any given point in time, but the function of time does not correspond directly to the function of position as a sinusoid. As rayjohn has mentioned, I smell elliptic integrals (which are defined with trig functions).
     
    Last edited: Jul 15, 2004
  23. Jul 16, 2004 #22

    Gza

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    I thought this was a high school level problem of a simple swing, I didn't know we would end up at elliptic integrals. :smile:
     
  24. Jul 16, 2004 #23

    turin

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    At the highschool level, they usually make the simplifying approximation of small oscillations. That allows one to approximate the motion as simple harmonic motion. On a slightly deeper level, you can thing of the problem as a 1-D problem in configuration (the angular position of the pendulum bob) which leads to a 2-D phase space (the other D is for the momentum or velocity). Then, you can define a potential energy and expand it in a Taylor series:

    V = a0 + a1x + a2x2 + a3x3 + ...

    The gauge invariance allows you to set a0 = 0 without changing the physical analysis.

    For small oscillations, the pendulum stays close to the straight down position. If you consider the expansion about that point, then the potential increases in both directions, and thus a1 = 0. This gives an expasnsion of the potential as:

    V = a2x2 + a3x3 + ...

    Since the displacement is small, the lowest order term in the expansion dominates, so you can approximate the potential as:

    V ~ a2x2.

    This is the potential of the harmonic oscillator, which has sinusoidal solutions. Notice that this expression for the potential depended on the small oscillation approximation. Going all the way around the swing is not a small displacement, so you must include all of the even order terms which are signigicant:

    V = a2x2 + a4x4 + a6x6 + ...

    It turns out that the potential itself is actually a sinusoid:

    V = Acos(kx).

    When you stick this in the diff. eq., it gets nasty. Also notice that this is somewhat related to the fact that there is no analytical way in general to find the roots of an infinite order polynomial.
     
  25. Jul 18, 2004 #24
    Could you explain more about why you neglected the odd powers of Velocity's Taylor series?

    Second question: would it be worthwhile to try and solve this by using series? would i still have to solve using differential equations?
     
  26. Jul 18, 2004 #25
    Sorry had to amend -went too fast
    V^2 = r.g.(1 + 2.(1 - cos(theta)))
    where theta is angle 0 at top
    r is radius
    g gravity
    and V is the peripheral speed ( not angular)

    Barring errors this is simply derived from energy
    BUTTTTT the period is given by the integral of 1/v which is really nasty
    and thats where I was told by a mathematician one gets into eliptical
    integrals . So the above does not involve complicated maths but the concepts of energy exchange ( kinetic to potential in gravity) may be unfamiliar to some. I'm not going post a proof unless specifically asked because I do not know how to post diagrams. Ray
    The period can be obtained either by numerical integration or by setting up a simulation for those that enjoy programming
    This problem can be extended to long period pendulums by using two minutely different masses at both ends of a beam. have a good day Ray.
     
    Last edited: Jul 18, 2004
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