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Thread: Relative accelleration Problem

  1. #1
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    Relative accelleration Problem

    Iam having some difficulty with this one iff some one could help.
    What is the accelleration of point D (the centre)
    Attached Thumbnails Attached Thumbnails Relative accelleration Problem-lastscan2.jpg  
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  2. #2
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    Could you please give the problem statement in full?
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    There is a disk,it is ONE piece.It has a large diam and small diam.
    The small dia rolls on a desk to the right.The large dia section does not touch the desk,but has a tangential acceleration as shown.
    That is all that is known .
    Find the accelleration of centre of the disk.
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    I think the thing to do is to find out the angular acceleration $\displaystyle \alpha$. Once you do that, you employ the (assumed) condition that the inner wheel rolls without slipping in order to translate the angular acceleration to linear acceleration. I do not think you have correctly computed the angular acceleration, although you have the correct general equation there:

    $\displaystyle a_{\text{tan}}=r\,\alpha.$

    Try computing that again. What do you get?
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  5. #5
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    Thanks for that ,appreciated,I worked it out
    ans is 15m/s^2 to the left

    5m/s^2=r alpha
    where r =(2-1.5) /2 (because where disk touches desk accell is 0 in the x directions) ie 5m/s^2 wrt desk.
    alpha=20
    accell of ctr = 20*1.5/2
    Last edited by heatly; Oct 13th 2010 at 05:30 AM.
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  6. #6
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    Hmm. Not sure I agree. To start out, you'd have

    5.0 m/s^2 = (1 m)(alpha), which implies alpha = 5.0 1/s^2. Then how would you continue?
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    oops,fixed up now,thanks
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  8. #8
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    r =(2-1.5) /2
    That is incorrect. You need to measure the radius from the center of both concentric circles to the outermost edge. Use the 1m I used.
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    Ok,I'll look into it thansk.
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