The position of the centre of mass of the uni | vedclass

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(D) Let the uniform plate be considered as a large rectangle of sides $2a$ and $2b$ with a smaller rectangle of sides $a$ and $b$ removed from it.Let

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$\left(-\frac{a}{6},-\frac{b}{6}\right)$

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(D) Let the uniform plate be considered as a large rectangle of sides $2a$ and $2b$ with a smaller rectangle of sides $a$ and $b$ removed from it.Let the mass of the removed rectangle be $m_1$ and the mass of the remaining shaded plate be $m_2$.The area of the removed rectangle is $A_1 = a 	imes b$. Thus,$m_1 = \sigma ab$.The area of the full rectangle is $A = 2a 	imes 2b = 4ab$. The area of the remaining plate is $A_2 = 4ab - ab = 3ab$. Thus,$m_2 = 3\sigma ab = 3m_1$.The centre of mass of the full rectangle is at $(-a/2, -b/2)$ relative to the origin $(0,0)$.The centre of mass of the removed rectangle is at $(a/2, b/2)$.Using the principle of superposition for the centre of mass: $X_{cm} = \frac{m_2 x_2 + m_1 x_1}{m_2 + m_1} = 0$ (since the full rectangle's $CM$ is at the origin if we shift coordinates,but here we use the formula for a cavity).For a cavity,the position of the centre of mass of the remaining part is given by $\vec{R} = \frac{M\vec{R}_{full} - m_1\vec{r}_1}{M - m_1}$.Here,the full rectangle has its centre at $(-a/2, -b/2)$. Its mass is $M = 4m_1$.The removed part has its centre at $(a/2, b/2)$.$X_{cm} = \frac{4m_1(-a/2) - m_1(a/2)}{3m_1} = \frac{-2m_1a - 0.5m_1a}{3m_1} = \frac{-2.5a}{3} = -\frac{5a}{6}$.Wait,re-evaluating based on the coordinate system provided: The origin $(0,0)$ is at the corner of the cavity.The full rectangle (if it were complete) would span from $x = -a$ to $x = a$ and $y = -b$ to $y = b$. Its $CM$ is at $(0,0)$.The removed rectangle spans $x = 0$ to $a$ and $y = 0$ to $b$. Its $CM$ is at $(a/2, b/2)$.$X_{cm} = \frac{M(0) - m_1(a/2)}{3m_1} = -a/6$.$Y_{cm} = \frac{M(0) - m_1(b/2)}{3m_1} = -b/6$.Thus,the position is $(-\frac{a}{6}, -\frac{b}{6})$.

The position of the centre of mass of the uniform plate as shown in the figure is

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