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[Post New]11 Apr 2008 15:39:12 IST
    Subject: Re:D C Pandey Q frm rotational motion 2
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anchitsaini (4352)

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Olaaa!! Perrrfect answer. 796  [982 rates]
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\mbox{The slope of the lines are m }\\ \\ \mbox{Let the equations of lines be - }\\ \\ y = mx + f\\ \\ y = mx + g\\ \\ d = \frac{f - g}{\sqrt{m^2 + 1}}\\ \\ \mbox{Let the point be h,k from which ang momentum is to be calculated }\\ \\ L_1 = mvr_1\\ \\ =mv\frac{mh - k + f}{m^2 + 1} \\ \\ \mbox{similarly }\\ \\ L_2 = mv\frac{mh - k + g}{m^2 + 1}\\ \\ \mbox{Since they are moving in opposite directions,net ang momentum = }\\ \\ L = L_1 - L_2 \\ \\ =\frac{mv}{\sqrt(m^2 + 1} * (mh - k + f - (mh - k + g))\\ \\ =\frac{mv(f-g)}{\sqrt(m^2 + 1}\\ \\ =mvd \mbox{ which is constant }
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