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(A) Let the surface mass density be $\sigma$. The diameter of the circular plate is $\ell$,so its radius is $R = \ell/2$. The area of the circular pla

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Inside the square plate

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(A) Let the surface mass density be $\sigma$. The diameter of the circular plate is $\ell$,so its radius is $R = \ell/2$. The area of the circular plate is $A_1 = \pi R^2 = \pi(\ell/2)^2 = \pi \ell^2/4$. The mass of the circular plate is $m_1 = \sigma A_1 = \sigma(\pi \ell^2/4)$.The side length of the square plate is $\ell$,so its area is $A_2 = \ell^2$. The mass of the square plate is $m_2 = \sigma A_2 = \sigma \ell^2$.Let the center of the circular plate be at the origin $(0,0)$. The center of mass of the circular plate is at $x_1 = 0$. The center of the square plate is at a distance of $R + \ell/2 = \ell/2 + \ell/2 = \ell$ from the origin. So,$x_2 = \ell$.The $x$-coordinate of the center of mass of the composite system is given by:$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{\sigma(\pi \ell^2/4)(0) + \sigma \ell^2(\ell)}{\sigma(\pi \ell^2/4) + \sigma \ell^2} = \frac{\ell^3}{\ell^2(\pi/4 + 1)} = \frac{\ell}{\pi/4 + 1} = \frac{4\ell}{\pi + 4}$.Since $\pi \approx 3.14$,$\pi + 4 \approx 7.14$. Thus,$x_{cm} = \frac{4\ell}{7.14} \approx 0.56\ell$.The square plate extends from $x = \ell/2 = 0.5\ell$ to $x = 3\ell/2 = 1.5\ell$. Since $0.5\ell < 0.56\ell < 1.5\ell$,the center of mass lies inside the square plate.

If the density of the material of a circular plate and a square plate as shown is the same,the center of mass of the composite system will be:

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