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Path integral applied to circular path

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider path given by equation [itex] ( x - 1 )^2 + ( y - 1 ) ^2 = 1 [/itex] that connect the points
    A = ( 0 , 1 ) and B = ( 1 , 0 ) in xy plane ( shown in image attached ).
    A bead falling under influence of gravity from a point A to point B along a curve is given by -

    [tex] T = \int \frac{dt}{v}[/tex]

    where the bead's velocity is [itex] v = \sqrt{2gy}[/itex] , where g is the gravitational constant.

    2. Relevant equations

    [itex] ( x - 1 )^2 + ( y - 1 ) ^2 = 1 [/itex]

    3. The attempt at a solution

    Given curve [itex] ( x - 1 )^2 + ( y - 1 ) ^2 = 1 [/itex] can be parametrized as -

    [itex] x - 1 = \cos\Theta [/itex]
    [itex] y - 1 = \sin\Theta [/itex]

    curve can be written as

    [itex] c( \Theta) = ( 1 + \cos\Theta , 1 + \sin\Theta )[/itex]
    [itex] c'( \Theta) = ( - \sin\Theta , \cos\Theta ) [/itex]
    [itex] \left| c'( \Theta) \right| = 1 [/itex]

    From given relation we have -

    [tex]T = \int \frac{dt}{v}[/tex]

    [tex]T = \int_{c} \frac{ \left| c'( \Theta) \right| d\Theta}{ \sqrt{2gy}} [/tex]

    [tex]T = \int_{\pi}^{ \frac{3\pi}{2}} \frac{d\Theta}{ \sqrt{2g( 1 + \sin\Theta)}} [/tex]


    [tex] As, 1 + \sin\Theta = \left( \cos\frac{\Theta}{2} + \sin\frac{\Theta}{2} \right) ^2 [/tex]

    [tex]T = \frac{1}{ \sqrt{2g}} \int_{\pi}^{\frac{3\pi}{2}} \frac{d\Theta}{ \cos\frac{\Theta}{2} + \sin\frac{\Theta}{2}} [/tex]

    Changing variables ,

    [tex] \frac{\theta}{2} = x , d\Theta = 2 dx [/tex]

    [tex]T = \frac{1}{ \sqrt{2g}} \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \frac{ 2 dx }{ \cos x + \sin x } [/tex]


    [tex] As, \sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} [/tex]

    multiplying Nr and Dr by [itex]\frac{1}{\sqrt{2}} [/itex]

    [tex]T = \frac{1}{\sqrt{g}} \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \frac{dx}{ \sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4} \sin x }[/tex]

    [tex]T = \frac{1}{\sqrt{g}} \int_{ \frac{\pi}{2} } ^ {\frac{3\pi}{4}} \frac{dx} { \sin\left( x + \frac{\pi}{4} \right) } [/tex]

    [tex]T = \frac{1}{\sqrt{g}} \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \csc \left( x + \frac{\pi}{4} \right) dx[/tex]

    [tex]T = \frac{1}{\sqrt{g}} \left| \ln \left[ \tan \left( \frac{ x + \frac{\pi}{4} } {2} } \right) \right] \right |_{\frac{\pi}{2}}^{\frac{3\pi}{4}} [/tex]


    The value of first term after limit substitution is infinity. Here the value of integral i.e. ' time ' should be strictly finite. What is wrong with my solution?
     

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  2. jcsd
  3. May 30, 2010 #2

    vela

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    "A bead is given by T=..." doesn't really make sense.

    The problem is the integral you started with is wrong. You want a time, but if you check the units of the integral, you get time/(distance/time) = time2/distance.
     
  4. May 30, 2010 #3

    Hurkyl

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    What does t have to do with the curve?
     
  5. May 30, 2010 #4

    tiny-tim

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    Welcome to PF!

    Hi symmetric! Welcome to PF! :smile:

    (have a theta: θ and a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    Nooo … your own notation is bad, and you've misunderstood the question's notation. :redface:

    i] never use t for a distance parameter, use s (or x etc)

    ii] you don't need to use the usual θ (starting with θ = 0 along the positive x-axis) … you can choose any distance parameter, so you may as well start with θ = 0 at A, going to θ = π/2 at B

    iii] the y in the given formula, √2gy, is not the y-coordinate, it's the difference in height (so at A it's 0).

    Try again. :smile:
     
  6. May 30, 2010 #5

    vela

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    Re: Welcome to PF!

    D'oh! I just assumed t stood for time. Just ignore my earlier post. :blushing:
     
  7. May 31, 2010 #6
    @vela @Hurkyl @tiny-tim

    Thanks for your reply.

    To clear confusion of variable 't' attaching original problem snapshot.

    From my understanding of given question it's distance travelled along circular path ( specifically for above problem ) .



    i] i generally don't use 't' for distance but given problem follows that notation .

    ii] and iii] But we have to calculate transit time T for bead travelling along path given ( part of circle ). The given formula [itex] \sqrt{2gy}[/itex] is valid for any path ( circular, straight line etc ) with same start and end point. Hence, given curve must be parametrized in order to get transit time T along specific path. If we chose the parametrization mentioned by me, then [itex] \Theta = 0 [/itex] and [itex]\Theta = \frac{\pi}{2}[/itex] will not leads to co-ordinates A(0,1) and B(1,0).
     

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