A uniform square wooden sheet of side a has i | vedclass

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Question

(B) The centre of mass of the original square plate of side $a$ is at its geometric centre. Let the origin be at the top-right corner of the original

Vedclass · Accepted Answer

$a / b=(\sqrt{5}+1) / 2$

Vedclass · Answer

(B) The centre of mass of the original square plate of side $a$ is at its geometric centre. Let the origin be at the top-right corner of the original square. The coordinates of the centre of mass of the original square are $(x_1, y_1) = (a/2, a/2)$.Let $k$ be the mass per unit area. The mass of the original square is $m_1 = k a^2$.The removed square portion of side $b$ has its centre of mass at $(x_2, y_2) = (b/2, b/2)$ relative to the same origin.The mass of the removed portion is $m_2 = k b^2$.The remaining $L$-shaped sheet has its centre of mass at point $P$,which is given as $(b, b)$ relative to the chosen origin.Using the formula for the centre of mass of a system with a cavity: $X_{CM} = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2}$.Substituting the values: $b = \frac{(k a^2)(a/2) - (k b^2)(b/2)}{k a^2 - k b^2}$.$b = \frac{a^3 - b^3}{2(a^2 - b^2)}$.$2b(a^2 - b^2) = a^3 - b^3$.$2ba^2 - 2b^3 = a^3 - b^3$.$a^3 - 2ba^2 + b^3 = 0$.Dividing by $b^3$: $(a/b)^3 - 2(a/b)^2 + 1 = 0$.Let $x = a/b$. Then $x^3 - 2x^2 + 1 = 0$.Since $x=1$ is a root,we divide by $(x-1)$: $(x-1)(x^2 - x - 1) = 0$.Since $a > b$,$x > 1$,so $x^2 - x - 1 = 0$.Solving for $x$: $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 + \sqrt{5}}{2}$ (taking the positive root).Thus,$a/b = (\sqrt{5} + 1) / 2$.

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