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Velocity - String(Pulley) constraints

  1. Mar 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A bead C can move freely on a horizontal rod. The bead is connected by blocks B and D by a string as shown in the figure. If the velocity of B is v. Find the velocity of block D.
    14cwojo.png


    2. Relevant equations
    As the string is inextensible the velocity of the string along the length is const.


    3. The attempt at a solution
    The doubt I have is that, the following should be true
    [tex]v_c\ =\ v_b.cos53°\ \ \ \ ...(1)[/tex]

    But in the solution from the book I get that
    [tex]v_b\ =\ v_c.cos53°\ \ \ \ ...(2)[/tex]

    I used relation 1 and got wrong results, I'm just confused how do we get relation 2.
    Kindly help me out.
    Thanks for your time :smile:
     
  2. jcsd
  3. Mar 27, 2014 #2
    How is the tip of the string connected to bead C moving ? What is its velocity ?
     
  4. Mar 28, 2014 #3
    Its velocity is [itex]v_b[/itex],along the length of the string, the same as that of block B.
     
  5. Mar 28, 2014 #4
    That means the tip of the string and bead C have different velocities . Is that so :rolleyes: ?
     
  6. Mar 28, 2014 #5
    I think that so (may be my misconception). The bead is confined to the horizontal bar, and can only have velocity along the horizontal bar.

    Furthermore, is the velocity of TIP and the whole of the string, not the same. :confused:
     
  7. Mar 28, 2014 #6
    The red spot in the picture depicts the tip of the string .Forget about all the pulleys and blocks for a moment.Just focus on the red spot .How does it move ?
     

    Attached Files:

  8. Mar 28, 2014 #7
    It should have the velocity same as that of the bead [itex]v_c[/itex].
     
  9. Mar 28, 2014 #8

    haruspex

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    It hasn't been specified anywhere that I can see, but I presume the bead is tied to the string, so the string cannot slide through it.

    phoenixXL, if the bead moves to the right at velocity vb, what is the component of that towards the top-right pulley?
     
  10. Mar 28, 2014 #9
    haruspex,
    As you are pointing, the following would be the diagram
    ( assuming [itex]v_b[/itex] to be the velocity of the block( or string) and [itex]v_c[/itex] the velocity of the bead. )
    izxawn.png

    But, I got confused and tried to solve the problem using the following diagram,
    2viks5x.png

    So, I will be grateful to know what misconception I do have, and how can I further prevent from getting errors.
     
  11. Mar 28, 2014 #10

    haruspex

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    The question of which vector equals a component of the other comes up in a few guises. In the present case, you just have to remember that it's the string length that's constant, so the rate at which the string passes over the pulley equals the rate at which the bead gets closer to that pulley.
    I find it can also help to imagine the process happening. If you pull down steadily on the string hanging from the pulley, do you expect the bead to get faster or slower?
     
  12. Mar 29, 2014 #11
    How can we conclude this?

    Faster, of course.
     
  13. Mar 29, 2014 #12
    Your misconception is the belief that the opposite ends of the string must have the same velocities. They don't. Only the component of the motion parallel to the string is constrained. The ends of the string are completely free to move in the direction perpendicular to the string.
     
  14. Mar 29, 2014 #13

    haruspex

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    The distance from bead to pulley is the length of string joining them. The rate at which the string passes over the pulley is the rate at which that length decreases.
     
  15. Mar 30, 2014 #14
    Got it.

    For anyone with similar problem
    2hdrxph.png

    Let the distance of the bead from the pulley be r. Then [tex]-\frac{dr}{dt}\ =\ v_b\\
    \implies\ -\frac{d\sqrt{x^2\ +\ y^2}}{dt}\ =\ v_b\\
    \implies\ -\frac{1}{2\sqrt{x^2\ +\ y^2}}.\frac{dx^2}{dt}\ =\ v_b\\
    \implies\ -\frac{1}{2\sqrt{x^2\ +\ y^2}}.2x.v_c\ =\ v_b\\
    \implies\ -v_c.cosθ\ =\ v_b\\
    [/tex]

    Thank you so much when I derive it myself, meager confidence builds up.
    Thanks you so much
     
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