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A small object density rolls up a curved surface with an initial velocity V. It reaches up to a maximum height of $ \large\frac{3V^2}{4g}$ with respect to the initial position. The object is

$\begin{array}{1 1} ring \\ solid \;sphere \\ hollow\;sphere \\ disc \end{array} $

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Energy conservation
$\large\frac{1}{2}$$mv^2 +\large\frac{1}{2}$$Iw^2=mgh$
$\large\frac{v^2}{2g} \bigg[ 1+ \large\frac{k^2}{r^2}\bigg]$$=h$
$h= \large\frac{3v^2}{4g}$


answered Dec 5, 2013 by meena.p

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