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    Hey, I've just done the OCR MEI June 2012 paper and I couldn't work out how to do the last part of question 4

    I've looked at the mark scheme and I have no idea why the situation ends up as it is - why the reaction to the surface passes through O.

    The COM is \dfrac{49}{16} from O, ie \dfrac{17}{16} from the center of the surface.

    Question 4bii) from http://www.mei.org.uk/files/papers/m3_june_2012.pdf
    Mark scheme is at the bottom of the document.

    Thanks!
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    (Original post by Duskstar)
    Hey, I've just done the OCR MEI June 2012 paper and I couldn't work out how to do the last part of question 4

    I've looked at the mark scheme and I have no idea why the situation ends up as it is - why the reaction to the surface passes through O.

    The COM is \dfrac{49}{16} from O, ie \dfrac{17}{16} from the center of the surface.

    Question 4bii) from http://www.mei.org.uk/files/papers/m3_june_2012.pdf
    Mark scheme is at the bottom of the document.

    Thanks!
    Alhough I do edexcel I will be home in 15 minutes. I will do this question. Main thing to note is that the slope is tangent to the circle and all tangents are perpendicular to the radius. That sorts the question out easily. Note that G lies on this line aswell.
    Reaction is perpendicular to friction(the slope) hence why it passes through O. Not that Fr and R act at piint of contact and G acts at the COM. Take moments about O.

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    (Original post by physicsmaths)
    Alhough I do edexcel I will be home in 15 minutes. I will do this question. Main thing to note is that the slope is tangent to the circle and all tangents are perpendicular to the radius. That sorts the question out easily. Note that G lies on this line aswell.


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    Thanks, long time no speak too :P

    And actually that sorts it out - the circle has to sit tangent to the slope so the reaction will go through the centre, and I can do the question from there

    Idk why I missed over that but I just couldn't get an image of how it would sit on a slope haha
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    (Original post by Duskstar)
    Thanks, long time no speak too :P

    And actually that sorts it out - the circle has to sit tangent to the slope so the reaction will go through the centre, and I can do the question from there

    Idk why I missed over that but I just couldn't get an image of how it would sit on a slope haha
    Yep
    I did a very similar question on a Delphis paper.


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