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A uniform circular sheet has a radius r. A square sheet whose diagonal length is equal to the radius of the plate is removed. The maximum distance of the centre of mass of the square sheet from the center of original sheet is :

\[\begin {array} {1 1} (a)\;\frac{r}{2(2 \pi-1)} & \quad (b)\;\frac{r}{(\pi-2)} \\ (c)\;\frac{2r}{\pi-3} & \quad  (d)\;\frac{r}{2(\pi-1)}\end {array}\]

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Let $m_1$ be te mass of the square that is chopped off.
$m_1 = \large\frac{M}{\pi r^2}$$ \times \large\frac{r^2}{2}$$ = \large\frac{M}{ 2 \pi}$
$x_{m_1} = \large\frac{M\times0 - m_1\times \large\frac{r}{2} }{M - \large\frac{M}{2 \pi }}$
$x_{m_1} = \large\frac{-\large\frac{M}{2 \pi } \large\frac{r}{2}}{M - \large\frac{M}{2 \pi }}$
$x_{m_1} = \large\frac{-r}{2(2 \pi - 1)}$
Thus, the centre of mass of the remaining part lies at a distance $x_{m_1} = \large\frac{-r}{2(2 \pi - 1)}$ towards right of the origin, or the initial centre of mass of the disc.
Therefore the correct option is (A) $\large\frac{r}{2(2\pi -1)}$
Please note: The new centre of mass does not lie at the origin any more.
answered Nov 21, 2013 by meena.p
edited Mar 25, 2014 by balaji.thirumalai
 

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