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M2 Centres of mass-Please can u explain why this is what it is watch

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    Hi,

    When a question regarding centres of mass with a hanging lamina asks: find the angle between OA and the horizontal, I never know what to do. Is it the angle between OA and the new horizontal or old horizontal?

    Here are two questions I'm stuck on, please can you explain why this is the required angle:

    http://img402.imageshack.us/img402/1511/89244940.png
    http://img34.imageshack.us/img34/9748/65272731.png

    Thanks in advance.

    JT80 :-)
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    (Original post by JTeighty)
    ...
    There is only one horizontal.

    Since they are free to rotate, the Centre of Gravity will be vertically below the axis when they are allowed to hang in equilibrium.

    So for the first one, AG will be vertical, and the angle between OA and the vertical will be OAG. Since you want the angle with the horizontal you want 90 minus that, which is the angle AGO, as stated.

    A similar argument for the second one.

    It would be clearer, if they had drawn the diagrams showing the shapes hanging in equilibrium, but....
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    (Original post by ghostwalker)
    There is only one horizontal.

    Since they are free to rotate, the Centre of Gravity will be vertically below the axis when they are allowed to hang in equilibrium.

    So for the first one, AG will be vertical, and the angle between OA and the vertical will be OAG. Since you want the angle with the horizontal you want 90 minus that, which is the angle AGO, as stated.

    A similar argument for the second one.

    It would be clearer, if they had drawn the diagrams showing the shapes hanging in equilibrium, but....
    So when it says horizontal, it refers to the original horizontal (AB in first question and BD in the second question).

    My problem is, that when it becomes free to rotate and AG is vertical, AG is the new verticle and then the horizontal should be perpendicular to that.

    According to what your saying, the original axes(horizontal and verticle) remain.

    Is this what you mean?

    Thanks

    JT80
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    (Original post by JTeighty)
    So when it says horizontal, it refers to the original horizontal (AB in first question and BD in the second question).
    You are not told that AB, and DB in the second question, are horizontal; that's just a convenience of the diagram that they appear so (and even if they were, it is irrelevant). The shape could have been drawn at any angle.

    If you imagine holding the shape by the point about which it rotates, the centre of gravity will be directly below the point you are holding it.

    The one and only vertical is the line from the point you are holding it to the centre of gravity.

    The one and only horizontal is a line perpendicular to that.

    Don't know if that helps; I suspect I'm not getting to the source of your confusion.

    I am currently in the middle of moving house (and will be for a week or two yet), so won't be able to post again until tomorrow.
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    (Original post by ghostwalker)
    You are not told that AB, and DB in the second question, are horizontal; that's just a convenience of the diagram that they appear so (and even if they were, it is irrelevant). The shape could have been drawn at any angle.

    If you imagine holding the shape by the point about which it rotates, the centre of gravity will be directly below the point you are holding it.

    The one and only vertical is the line from the point you are holding it to the centre of gravity.

    The one and only horizontal is a line perpendicular to that.

    Don't know if that helps; I suspect I'm not getting to the source of your confusion.

    I am currently in the middle of moving house (and will be for a week or two yet), so won't be able to post again until tomorrow.
    Ok...So how does that explain my original question/s? It doesn't seem to make sense that way?
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    (Original post by JTeighty)
    Ok...So how does that explain my original question/s? It doesn't seem to make sense that way?
    Sorry for the delay.

    As I said, I didn't think I was getting to the source of your confusion.

    What they are doing is finding the angle between "where the vertical would be when it is hanging freely" and "the line in question". Then subtracting that from 90 to get the angle with "where the horizontal would be", since the angle with the horizontal + the angle with the vertical = 90 (considering the acute angles each time).

    Does that clarify? If not, can you try rephrasing your problem, and I might be clearer what the issue is and get a better angle on the problem (groan for the pun).
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    (Original post by ghostwalker)
    Sorry for the delay.

    As I said, I didn't think I was getting to the source of your confusion.

    What they are doing is finding the angle between "where the vertical would be when it is hanging freely" and "the line in question". Then subtracting that from 90 to get the angle with "where the horizontal would be", since the angle with the horizontal + the angle with the vertical = 90 (considering the acute angles each time).

    Does that clarify? If not, can you try rephrasing your problem, and I might be clearer what the issue is and get a better angle on the problem (groan for the pun).
    Haha. Thanks for the response and sorry about the delay (had to revise for chem and bio).

    I thought they did that, but why are they taking the angle of the verticle and OG? It should be with OA? (in 1st q)

    Thanks

    JT80
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    (Original post by JTeighty)

    I thought they did that, but why are they taking the angle of the verticle and OG? It should be with OA? (in 1st q)

    Thanks

    JT80
    In the first question AG will be the vertical, and the angle with OA will be OAG. BUT they want the angle with the horizontal which is 90 - OAG.

    Look at the triangle OAG.

    90 minus the angle OAG is the angle AGO (it's a right angled triangle).

    Make sense?
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    (Original post by ghostwalker)
    In the first question AG will be the vertical, and the angle with OA will be OAG. BUT they want the angle with the horizontal which is 90 - OAG.

    Look at the triangle OAG.

    90 minus the angle OAG is the angle AGO (it's a right angled triangle).

    Make sense?
    Ah now I see. Thanks a lot.
 
 
 

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