(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)

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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)

The distance between the carbon and oxygen atoms in a carbon monoxide molecule is $1.1 \, \mathring A$. Given that the mass of the carbon atom is $12 \, a.m.u.$ and the mass of the oxygen atom is $16 \, a.m.u.$,find the position of the center of mass of the carbon monoxide molecule.

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$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

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Obtain an expression for the position vector of the centre of mass of a system of $n$ particles in two dimensions.

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Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

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The position of the center of mass of three masses $2 \text{ kg}$, $3 \text{ kg}$, and $15 \text{ kg}$ placed with respect to the midpoint $(p)$ of the base, as shown in the figure, is . . . . . . .

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The distance between the carbon and oxygen atoms in a carbon monoxide molecule is $1.1 \, \mathring A$. Given that the mass of the carbon atom is $12 \, a.m.u.$ and the mass of the oxygen atom is $16 \, a.m.u.$,find the position of the center of mass of the carbon monoxide molecule.

$A$ uniform rod of mass $400 \ g$ and length $1 \ m$ is pivoted at one end. It is displaced by an angle of $60^o$. The increase in its potential energy is ....... $J$.

Obtain an expression for the position vector of the centre of mass of a system of $n$ particles in two dimensions.

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

The position of the center of mass of three masses $2 \text{ kg}$, $3 \text{ kg}$, and $15 \text{ kg}$ placed with respect to the midpoint $(p)$ of the base, as shown in the figure, is . . . . . . .

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