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Circular Motion: Mechanics Help!!! Urgent watch

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    So this is the question:

    Name:  Screen Shot 2013-11-17 at 2.18.50 PM.png
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    And this is the solution:
    Name:  Screen Shot 2013-11-17 at 2.20.30 PM.png
Views: 123
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    Isn't the the GPE for the bead when it's on the top  mga(1+cosx)?

    Why is the KE=GPE?
    All in all can someone please explain the solution.
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    (Original post by MAyman12)


    Isn't the the GPE for the bead when it's on the top  mga(1+cosx)?
    yes
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    (Original post by TenOfThem)
    yes
    Ok, then why is it the opposite in the markshceme?
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    (Original post by MAyman12)
    Ok, then why is it the opposite in the markshceme?
    where
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    (Original post by TenOfThem)
    where
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    (Original post by MAyman12)
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    Oh, I looked at the second line

    I have not done this recently so will hope for someone more current to answer
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    (Original post by MAyman12)
    So this is the question:

    Name:  Screen Shot 2013-11-17 at 2.18.50 PM.png
Views: 113
Size:  28.9 KB

    And this is the solution:
    Name:  Screen Shot 2013-11-17 at 2.20.30 PM.png
Views: 123
Size:  30.3 KB

    Isn't the the GPE for the bead when it's on the top  mga(1+cosx)?

    Why is the KE=GPE?
    All in all can someone please explain the solution.
    GPE at A is whatever, it doesn't matter.

    They're using "gain in KE" = "loss in GPE", not KE = GPE.

    At the top angle theta (I'll call it x) we have loss in GPE = mg ( a -acosx)
    Hence mga(1-cosx)
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    (Original post by MAyman12)
    So this is the question:

    Name:  Screen Shot 2013-11-17 at 2.18.50 PM.png
Views: 113
Size:  28.9 KB

    And this is the solution:
    Name:  Screen Shot 2013-11-17 at 2.20.30 PM.png
Views: 123
Size:  30.3 KB

    Isn't the the GPE for the bead when it's on the top  mga(1+cosx)?

    Why is the KE=GPE?
    All in all can someone please explain the solution.
    Gain in KE =Loss of GPE
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    (Original post by ghostwalker)
    GPE at A is whatever, it doesn't matter.

    They're using "gain in KE" = "loss in GPE", not KE = GPE.

    At the top angle theta (I'll call it x) we have loss in GPE = mg ( a -acosx)
    Hence mga(1-cosx)
    Ok. What about the GPE when it's at the bottom?

    Isn't the GPE at the bottom mga(1-cosx) when x<pi/2?
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    (Original post by brianeverit)
    Gain in KE =Loss of GPE
    But what about the GPE of the bead when it's at the bottom?
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    (Original post by MAyman12)
    Ok. What about the GPE when it's at the bottom?

    Isn't the GPE at the bottom mga(1-cosx) when x<pi/2?
    No, the loss in GPE is mg ( a +a cosx) at the bottom position.

    Edit: Position B2 in the diagram should be on the left, but I'm not redoing the whole thing.
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    (Original post by ghostwalker)
    No, the loss in GPE is mg ( a +a cosx) at the bottom position.
    You said that the loss in GPE energy was mga(1-cosx) in the previous post.
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    (Original post by MAyman12)
    You said that the loss in GPE energy was mga(1-cosx) in the previous post.
    That's at the top position, B1 in my diagram.
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    (Original post by ghostwalker)
    That's at the top position, B1 in my diagram.
    So with respect to the position A the GPE lost at B1= mga(1-cosx) and at B2 the GPE lost= mga(1+cosx)?

    So gain in KE1=mga(1-cosx)

    gain in KE2=mga(1+cosx)
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    (Original post by MAyman12)
    So with respect to the position A the GPE lost at B1= mga(1-cosx) and at B2 the GPE lost= mga(1+cosx)?

    So gain in KE1=mga(1-cosx)

    gain in KE2=mga(1+cosx)
    Yes, and those are the first two lines of the given solution.

    Note: Just realised that my B2 position should be on the left rather than the right, but it's a mirror image, and all the values for KE, GPE are the same, and I'm not doing the diagram again.
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    (Original post by ghostwalker)
    Yes, and those are the first two lines of the given solution.

    Note: Just realised that my B2 position should be on the left rather than the right, but it's a mirror image, and all the values for KE, GPE are the same, and I'm not doing the diagram again.
    So for the rest of the question the Kinetic energy at B1 is equal to the GPE at B2 since the conversation of energy right?
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    (Original post by MAyman12)
    So for the rest of the question the Kinetic energy at B1 is equal to the GPE at B2 since the conversation of energy right?
    Huh! Where did that come from?
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    (Original post by ghostwalker)
    Huh! Where did that come from?
    The loss of GPE at B1+Loss in GPE at B2= Total energy, But since Loss of GPE is equal to gain in KE, then GPE at B1 must be equal to KE at B2. The difference between B1 and B2 is pi.
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    (Original post by MAyman12)
    The loss of GPE at B1+Loss in GPE at B2= Total energy.
    This bit doesn't make any sense to me.

    In going from A to B1 we have a loss in GPE, which was worked out previously.

    In going from A to B2 we have a different loss in GPE, again previously worked out.
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    (Original post by ghostwalker)
    This bit doesn't make any sense to me.

    In going from A to B1 we have a loss in GPE, which was worked out previously.

    In going from A to B2 we have a different loss in GPE, again previously worked out.
    So how are we going to use this in finding R1,R2 by radial resolution?

    I'm really sorry. I'm sleep deprived and not thinking straight.
 
 
 
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