(A) The total mass of the system is $M = 6m + m + m + m + m = 10m$.The $y$-coordinates of the centers of mass of the individual parts are:Outer circle

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(A) The total mass of the system is $M = 6m + m + m + m + m = 10m$.The $y$-coordinates of the centers of mass of the individual parts are:Outer circle: $y_1 = 0$,mass $m_1 = 6m$Left inner circle: $y_2 = a$,mass $m_2 = m$Right inner circle: $y_3 = a$,mass $m_3 = m$Horizontal line segment: $y_4 = -a$,mass $m_4 = m$The $y$-coordinate of the center of mass is given by:$Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{M}$$Y_{cm} = \frac{(6m)(0) + (m)(a) + (m)(a) + (m)(-a)}{10m}$$Y_{cm} = \frac{0 + ma + ma - ma}{10m} = \frac{ma}{10m} = \frac{a}{10}$

Class 11 Physics System of Particles and Rotational Motion Centre of mass (Point Mass)

Medium

Look at the figure drawn with ink of uniform linear density. $A$ mass $m$ of ink is used to draw each of the two inner circles and each of the two line segments. $A$ mass $6m$ of ink is used to draw the outer circle. The coordinates of the centers of the different parts are: outer circle $(0, 0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,and the horizontal line segment $(0, -a)$. Find the $y$-coordinate of the center of mass of the ink in the figure.

A
$\frac{a}{10}$
B
$\frac{a}{8}$
C
$\frac{a}{12}$
D
$\frac{a}{3}$

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System of Particles and Rotational Motion

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Centre of mass (Point Mass)

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