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 inner corner point $P$ of the cut

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 inner corner point $P$ of the cut-out. The coordinates of the centre of mass of the original square are $(x_1, y_1) = (a/2 - b, a/2 - b)$.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 origin $P$.The mass of the removed portion is $m_2 = k b^2$.The centre of mass of the remaining $L$-shaped sheet is at the origin $(0, 0)$.Using the formula for the centre of mass of a system with a cavity: $X_{CM} = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2} = 0$.$m_1 x_1 = m_2 x_2 \implies (k a^2)(a/2 - b) = (k b^2)(-b/2)$.$a^2(a/2 - b) = -b^3/2$.$a^3/2 - a^2 b = -b^3/2$.$a^3 - 2a^2 b + b^3 = 0$.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 $x$: $x = \frac{1 + \sqrt{5}}{2}$.Thus, $a/b = (\sqrt{5} + 1) / 2$.

$A$ uniform square wooden sheet of side $a$ has its centre of mass located at its geometric centre. $A$ square portion of side $b$ is cut out from one corner to produce an $L$-shaped sheet. The centre of mass of the $L$-shaped sheet lies at the inner corner point $P$ of the cut-out, when:

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