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    I've attempted a part of a vertical circular motion question but for some reason cannot get the same answer as my textbook so I was wondering if anybody could have a go and confirm/show me their method:

    A bead, of mass m, is threaded onto a smooth circular ring, of radius r, which is fixed in a vertical plane. The bead is moving on the wire. Its speed, v, at the highest point of its path is one quarter of its speed at the lowest point.
    (a) Show that v = (4gr/15) square rooted

    (Sorry I don't know how to write that in a better format.) Any help would be appreciated thanks.
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    (Original post by NeeIam)
    I've attempted a part of a vertical circular motion question but for some reason cannot get the same answer as my textbook so I was wondering if anybody could have a go and confirm/show me their method:

    A bead, of mass m, is threaded onto a smooth circular ring, of radius r, which is fixed in a vertical plane. The bead is moving on the wire. Its speed, v, at the highest point of its path is one quarter of its speed at the lowest point.
    (a) Show that v = (4gr/15) square rooted

    (Sorry I don't know how to write that in a better format.) Any help would be appreciated thanks.
    What have you attempted?

    Draw a diagram. Denote the speed at the bottom of the circle as u and as \frac{1}{4}u at the top.

    Now consider what the total energy is at the top and make it equal to the total energy at the bottom by the principle of conservation of energy. In this particular model there's gravitational potential energy and kinetic energy.
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    (Original post by NeeIam)
    x.
    I let the horizontal line through the center of the circle be my base level (ie height=0) and then used GPE + KE at top = GPE + KE at bottom and got it answer. Post working if youre still having trouble and ill try see where you went wrong
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    (Original post by DylanJ42)
    I let the horizontal line through the center of the circle be my base level (ie height=0) and then used GPE + KE at top = GPE + KE at bottom and got it answer. Post working if youre still having trouble and ill try see where you went wrong
    So I did
    1/2 x m x (1/4v)^2 + 2mgr = 1/2 x m x v^2

    Where have I gone wrong? Because when I do that I don't get the correct answer
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    (Original post by NeeIam)
    So I did
    1/2 x m x (1/4v)^2 + 2mgr = 1/2 x m x v^2

    Where have I gone wrong? Because when I do that I don't get the correct answer
    by any chance are you getting  \displaystyle v = \sqrt{\frac{64gr}{15}}
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    (Original post by DylanJ42)
    by any chance are you getting  \displaystyle v = \sqrt{\frac{64gr}{15}}
    Yep that's what I'm getting
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    (Original post by NeeIam)
    Yep that's what I'm getting
    yea youve worked out the speed at the bottom of the circle by mistake. as RDK said, if you let the speed at the bottom be U, then 4V = U, since V is four times smaller than the speed at the bottom.
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    (Original post by DylanJ42)
    yea youve worked out the speed at the bottom of the circle by mistake. as RDK said, if you let the speed at the bottom be U, then 4V = U, since V is four times smaller than the speed at the bottom.
    OHhh that makes sense now! Lol thank you
 
 
 
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