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Question

A circular plate of diameter d is kept in contact with a square plate of edge d as show in figure. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be (a) inside the circular plate (b) inside the square plate (c) at the point of contact (d) outside the system.

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Solution

(b) inside the square plate Let m1 be the mass of circular plate and m2 be the mass of square plate. The thickness of both the plates is t. $\mathrm{mass}=\mathrm{densit}y×\mathrm{volume}\phantom{\rule{0ex}{0ex}}{m}_{1}=\rho \pi {\left(\frac{d}{2}\right)}^{2}t\phantom{\rule{0ex}{0ex}}{m}_{2}=\rho {d}^{2}t$ Centre of mass of the circular plate lies at its centre. Let the centre of circular plate be the origin. ${\stackrel{\to }{r}}_{1}=0$ Centre of mass of the square plate lies at its centre. ${\stackrel{\to }{r}}_{2}=2d\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}R=\frac{{m}_{1}{\stackrel{\to }{r}}_{1}+{m}_{2}{\stackrel{\to }{r}}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}×0+\rho {\mathit{d}}^{2}t×2d}{\rho \pi {\left(\frac{d}{2}\right)}^{2}t+\rho {d}^{2}t}\phantom{\rule{0ex}{0ex}}=\frac{2d}{\frac{\pi }{4}+1}=1.12d\phantom{\rule{0ex}{0ex}}⇒R>d\phantom{\rule{0ex}{0ex}}$ $\therefore$ Centre of mass of the system lies in the square plate.

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