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Beads on Semi-Circular wire

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Two beads A and B move along a semicircular wire frame as shown in the figure as shown in figure. The beads are connected by an inelastic string which always remains tight. At an instant the speed of A is u, angle BAC is 45 degrees and angle BOC is 75 degrees, where O is the centre of semicircular arc. The speed of bead B at that instant is.
    23kufc2.jpg


    2. Relevant equations



    3. The attempt at a solution
    I really have no idea on how to begin with this. How should i go on forming equations?
    Please point me in the right direction.

    Thanks!
     
  2. jcsd
  3. Jun 19, 2012 #2
    Hi again :tongue2:

    Since the string is not extensible, the vertical velocity of B will always have to be equal to the velocity of A. Can you frame an equation for this?
     
  4. Jun 19, 2012 #3
    I did that once but i thought i am trying my own foolish things.
    I assumed that at any instant the distance of A from O is x and that of B is y.
    l=(x+y)cos45
    Differentiating with respect to time, i get:
    (dy/dt)=-(dx/dt)

    How can i proceed further?
     
  5. Jun 19, 2012 #4
    Umm noo..how did you get that?

    You can simply equate the vertical velocity component of B to u. The velocity of B will be have its direction tangential to the ring...what does this tell you about its vertical component?
     
  6. Jun 19, 2012 #5

    TSny

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    Hello, Infinitum. I don't see how your statement can be true. If the vertical component of velocity of B equals the velocity of A, then during a time Δt, both particles would move the same distance vertically. But B will also move horizontally during this time. So, it seems to me that this would change the distance between A and B.
     
  7. Jun 19, 2012 #6

    TSny

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    You might give us some context for the problem since there are different methods for solving it. If it's a "related rates problem" from a calculus course then you would be expected to use calculus. Or if it's a problem where you can use any method you want, then you can solve it with just geometry, trig, and some insight.
     
  8. Jun 20, 2012 #7
    Sorry, i think i did not explain it correctly. My assumption can be better shown by this pic:
    108fbkl.jpg
    So what i did was incorrect? and why?

    Well, it isn't specified that i need to do the problem by a specified method, any method will do.

    Would you be so kind to tell me how i can solve by this? :tongue2:
     
  9. Jun 20, 2012 #8

    ehild

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    Hi Pranav,

    The angles change with time so just call them α and β or anything else. Find relation between them and x. You know that u=dx/dt. The speed of the other bead on the circle is v=R|dβ/dt|. You need to differentiate, using that the length of the cord is constant in time, and so is the radius of the circle.

    ehild
     

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  10. Jun 20, 2012 #9
    Hello ehild! :smile:

    As you said i tried finding some relations,
    Ist relation, Lsin(α)=Rsin(β)=d
    IInd Relation, Lcos(α)-Rcos(β)=x
    Differentiating the second relation with respect to time,

    Rsin(β)(dβ/dt)-Lsin(α)(dα/dt)=dx/dt
    I can use the first relation here but i don't seem to reach the answer.
     
  11. Jun 20, 2012 #10
    Instead of using law of sines, why not try using law of cosines.
     
    Last edited: Jun 20, 2012
  12. Jun 20, 2012 #11
    Hi TSny. You are right, I meant to write that the velocity component of A along the length of the rod has to be equal to the velocity component of B along the length of the rod, so that the rod length remains the same.
     
  13. Jun 20, 2012 #12
    Oh, I see. You had mentioned that the distance of B from O is y earlier, that led me into confusion :tongue2:

    I also made a little mistake while saying their vertical components are equal. Instead, their velocities along the rod have to be the same, for no extension.
     
  14. Jun 20, 2012 #13
    Umm..so what i did is incorrect?

    How should i proceed now?
     
  15. Jun 20, 2012 #14
    Yes, you got "l=(x+y)cos45"

    But it should be [itex]lcos45 = x+y[/itex]

    And you cannot simply differentiate this because the angle is changing continuously, as ehild indicated.

    An easier approach is equating the velocities of A and B along the rod. Let the velocity of B be v, and then try to solve the problem.
     
  16. Jun 20, 2012 #15
    Oh, that's a nice hint, thanks for the help, i got the answer. :smile:
     
  17. Jun 20, 2012 #16
    Way to go! :approve:
     
  18. Jun 20, 2012 #17

    TSny

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    That's a very nice way to solve it.

    I had solved it by setting up a velocity triangle based on

    VB = VA + VB/A . (vector addition!)

    where VB/A is the velcity of B relative to A. VB/A must be perpendicular to line AB in order for AB to keep a fixed distance apart. This is similar to your statement that the velocity components of A and B along line AB must be equal. From the velocity triangle you can then use the law of sines to get the result.

    But I like your approach much better. It gets the answer right away.

    Thanks.
     
  19. Jun 20, 2012 #18

    ehild

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    It is ingenious, Infinitum!

    My method is much more complicated.
    The speed of the bead on the vertical line is dx/dt=u.
    The speed of the bead on the circle is v=Rdβ/dt=Rω

    From the cosine law in the tringle AOB,

    [tex]L^2=R^2+x^2+2Rxcos(\beta)[/tex]

    With implicit differentiation,

    [tex]0=x u+R u cos(\beta)-Rx\sin(\beta) ω[/tex]

    isolating v=Rω

    [tex]v=u\frac{(1+\frac{R}{x}cos(\beta))}{\sin(\beta)}[/tex]

    Form the sine law, [tex]\frac{R}{x}=\frac{\sin(\alpha)}{\sin(\beta-\alpha)}[/tex]

    Plugging in and simplifying, we get:

    [tex]v=u\frac{cos(\alpha)}{\sin(\beta-\alpha)}[/tex]

    ehild
     

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  20. Jun 21, 2012 #19
    It's really complicated. :tongue:
    Thanks for an alternative solution. :smile:
     
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