Particles of masses 8 , kg, 2 , kg, 4 , kg,an | vedclass

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(B) Let vertex $A$ be at the origin $(0, 0)$. The side length of the square $s$ is related to the diagonal $d = 80 \, cm$ by $s = d / \sqrt{2} = 80 /

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let vertex $A$ be at the origin $(0, 0)$. The side length of the square $s$ is related to the diagonal $d = 80 \, cm$ by $s = d / \sqrt{2} = 80 / \sqrt{2} = 40\sqrt{2} \, cm$.The coordinates of the vertices are: $A(0, 0)$,$B(40\sqrt{2}, 0)$,$C(40\sqrt{2}, 40\sqrt{2})$,and $D(0, 40\sqrt{2})$.The masses are $m_A = 8 \, kg, m_B = 2 \, kg, m_C = 4 \, kg, m_D = 2 \, kg$. Total mass $M = 8 + 2 + 4 + 2 = 16 \, kg$.The $x$-coordinate of the center of mass is $X_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{M} = \frac{8(0) + 2(40\sqrt{2}) + 4(40\sqrt{2}) + 2(0)}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.The $y$-coordinate of the center of mass is $Y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{M} = \frac{8(0) + 2(0) + 4(40\sqrt{2}) + 2(40\sqrt{2})}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.The distance from $A(0, 0)$ is $R = \sqrt{X_{cm}^2 + Y_{cm}^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30 \, cm$.

Particles of masses $8 \, kg, 2 \, kg, 4 \, kg$,and $2 \, kg$ are placed at the vertices $A, B, C$,and $D$ of a square respectively. The distance of the center of mass of the system from vertex $A$ is $....... \, cm$. The length of the diagonal of the square is $80 \, cm$.

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