A circular plate of diameter 'a' is kept in c | vedclass

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(B) Let the point of contact be the origin $(0,0)$.The area of the circular plate is $A_1 = \pi (a/2)^2 = \frac{\pi a^2}{4}$. Its centre of mass is at

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Inside the square plate

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(B) Let the point of contact be the origin $(0,0)$.The area of the circular plate is $A_1 = \pi (a/2)^2 = \frac{\pi a^2}{4}$. Its centre of mass is at $x_1 = -a/2$.The area of the square plate is $A_2 = a^2$. Its centre of mass is at $x_2 = a/2$.The centre of mass of the composite system along the x-axis is given by:$x_{cm} = \frac{A_1 x_1 + A_2 x_2}{A_1 + A_2}$$x_{cm} = \frac{(\frac{\pi a^2}{4})(-a/2) + (a^2)(a/2)}{\frac{\pi a^2}{4} + a^2}$$x_{cm} = \frac{-\frac{\pi a^3}{8} + \frac{a^3}{2}}{\frac{\pi a^2}{4} + a^2} = \frac{a^3(\frac{1}{2} - \frac{\pi}{8})}{a^2(1 + \frac{\pi}{4})} = \frac{a(4 - \pi)}{8} \cdot \frac{4}{4 + \pi} = \frac{a(4 - \pi)}{2(4 + \pi)}$Since $\pi \approx 3.14$,$4 - \pi > 0$,therefore $x_{cm} > 0$.Since the positive x-axis lies within the square plate,the centre of mass lies inside the square plate.

$A$ circular plate of diameter '$a$' is kept in contact with a square plate of side '$a$' as shown. The density of the material and the thickness are same everywhere. The centre of mass of the composite system will be ...........

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