Centre of mass (Point Mass) Questions in English

(D) The centre of mass of each individual ball is located at its geometric centre.
Since the three balls are identical and their centres form an equilateral triangle,the system is symmetric.
The centre of mass of a system of particles is the weighted average of the positions of the individual centres of mass.
For three identical masses placed at the vertices of an equilateral triangle,the centre of mass coincides with the centroid of the triangle.
The centroid of an equilateral triangle is the point of intersection of its medians.

(C) The centre of mass is a unique point defined for a system of particles or a rigid body.
By definition,it is the point where the entire mass of the body is assumed to be concentrated for the purpose of describing the translational motion of the system.
Therefore,the correct option is $C$.

(B) The centre of mass of a body is the point where the entire mass of the body can be considered to be concentrated for translational motion. However,when a non-uniform body like a cricket bat is cut at its centre of mass,the two resulting pieces do not necessarily have equal mass.
In a cricket bat,the mass is concentrated more towards the bottom (the blade) than the handle. The centre of mass lies closer to the heavier end (the blade). When the bat is cut at the centre of mass,the portion containing the blade (the bottom piece) contains more material and thus has a larger mass compared to the handle piece.

(B) For a system of two particles with masses $m_1$ and $m_2$ at distances $r_1$ and $r_2$ from the centre of mass,the condition is $m_1 r_1 = m_2 r_2$.
This implies $r_1 / r_2 = m_2 / m_1$.
Therefore,the distance $r$ of the centre of mass from a particle is inversely proportional to the mass of that particle $(r \propto 1/m)$.

(C) For a uniform triangular lamina,the centre of mass is located at the centroid of the triangle.
If the vertices of the triangle are at $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$,the coordinates of the centroid $(x_{cm}, y_{cm})$ are given by:
$x_{cm} = \frac{x_1 + x_2 + x_3}{3}$ and $y_{cm} = \frac{y_1 + y_2 + y_3}{3}$.
From the figure,the vertices are $(0, 0)$,$(b, 0)$,and $(0, h)$.
Substituting these values:
$x_{cm} = \frac{0 + b + 0}{3} = \frac{b}{3}$
$y_{cm} = \frac{0 + 0 + h}{3} = \frac{h}{3}$
Thus,the coordinates of the centre of mass are $(\frac{b}{3}, \frac{h}{3})$.

(C) For a system of two particles with masses $m_1$ and $m_2$ located at distances $r_1$ and $r_2$ from their centre of mass,the condition for the centre of mass is given by $m_1 r_1 = m_2 r_2$.
Rearranging this equation,we get $\frac{r_1}{r_2} = \frac{m_2}{m_1}$.
This shows that the ratio of the distances of the particles from the centre of mass is inversely proportional to the ratio of their masses.
Therefore,the centre of mass divides the distance between the two particles in the inverse ratio of their masses.

(B) The position of the center of mass is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
For the center of mass to remain unchanged,the change in the position of the center of mass must be zero,i.e.,$\Delta R_{cm} = 0$.
Therefore,$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
Here,mass $m_1$ is moved by a distance $d$ towards the center of mass. Let the displacement be $\Delta x_1 = d$.
Substituting this into the equation: $m_1 d + m_2 \Delta x_2 = 0$.
Solving for $\Delta x_2$,we get $\Delta x_2 = -\frac{m_1}{m_2} d$.
The negative sign indicates that mass $m_2$ must be moved in the opposite direction to $m_1$ (i.e.,towards the center of mass) by a distance of $\frac{m_1}{m_2} d$.

(B) The linear density of the rod varies with distance as $\lambda = 2 + x$.
For a small element of length $dx$ at a distance $x$,the mass is $dm = \lambda \ dx = (2 + x) \ dx$.
The position of the centre of mass $x_{cm}$ is given by:
$x_{cm} = \frac{\int x \ dm}{\int dm} = \frac{\int_{0}^{3} x(2 + x) \ dx}{\int_{0}^{3} (2 + x) \ dx}$
Calculating the numerator:
$\int_{0}^{3} (2x + x^2) \ dx = [x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) - 0 = 9 + 9 = 18$.
Calculating the denominator (total mass $M$):
$\int_{0}^{3} (2 + x) \ dx = [2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) - 0 = 6 + 4.5 = 10.5 = \frac{21}{2}$.
Therefore,$x_{cm} = \frac{18}{21/2} = \frac{18 \times 2}{21} = \frac{36}{21} = \frac{12}{7} \ m$.

(A) For an object to undergo pure translational motion,the applied force must pass through its center of mass $(CM)$.
Let the mass of the horizontal rod $AB$ be $m$. Since the vertical rod $CD$ has length $2l$ and the horizontal rod $AB$ has length $l$,assuming uniform density,the mass of rod $CD$ is $2m$.
The center of mass of rod $AB$ is at its midpoint $D$,which is at a distance $2l$ from $C$. Let this be $y_1 = 2l$.
The center of mass of rod $CD$ is at its midpoint,which is at a distance $l$ from $C$. Let this be $y_2 = l$.
Taking $C$ as the origin,the position of the center of mass of the system along the vertical axis is:
$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$
$y_{cm} = \frac{m(2l) + (2m)(l)}{m + 2m} = \frac{2ml + 2ml}{3m} = \frac{4ml}{3m} = \frac{4l}{3}$
Thus,the force must be applied at a distance of $\frac{4l}{3}$ from $C$ to ensure pure translational motion.

(C) Let the positions of the four bodies be $a(-x, -y)$,$b(x, -y)$,$c(x, y)$,and $d(-x, y)$.
The centre of mass $(X_{cm}, Y_{cm})$ of two bodies of equal mass $m$ at positions $(x_1, y_1)$ and $(x_2, y_2)$ is given by $X_{cm} = \frac{x_1 + x_2}{2}$ and $Y_{cm} = \frac{y_1 + y_2}{2}$.
For the centre of mass to remain at the origin $(0, 0)$,we must have $x_1 + x_2 = 0$ and $y_1 + y_2 = 0$.
$1$. For pair $a$ and $c$: $a$ is at $(-x, -y)$ and $c$ is at $(x, y)$. The sum of coordinates is $(-x+x, -y+y) = (0, 0)$. Thus,the centre of mass is at the origin.
$2$. For pair $b$ and $d$: $b$ is at $(x, -y)$ and $d$ is at $(-x, y)$. The sum of coordinates is $(x-x, -y+y) = (0, 0)$. Thus,the centre of mass is at the origin.
Looking at the options,the combination $a$ and $c$ ensures the centre of mass remains at the origin.

(B) Let the masses of the three spheres be $m_1 = m_2 = m_3 = 1 \ kg$.
Let the positions of the centres $P, Q, R$ be $x_P, x_Q, x_R$ along the $x$-axis.
Since $P$ is the origin,$x_P = 0$.
The position of the centre of mass $(x_{cm})$ is given by the formula:
$x_{cm} = \frac{m_1 x_P + m_2 x_Q + m_3 x_R}{m_1 + m_2 + m_3}$
Substituting the values:
$x_{cm} = \frac{1 \times 0 + 1 \times PQ + 1 \times PR}{1 + 1 + 1} = \frac{PQ + PR}{3}$
Since all centres lie on the $x$-axis,the $y$-coordinate of the centre of mass is $y_{cm} = 0$.
Thus,the distance of the centre of mass from $P$ is $\frac{PQ + PR}{3}$.

(A) Let the length of the ladder be $L$. The center of mass is at a distance $L/2$ from the bottom end. Let the angle with the floor be $\theta$. The coordinates of the center of mass $(x, y)$ are given by $x = (L/2) \cos \theta$ and $y = (L/2) \sin \theta$. Squaring and adding these,we get $x^2 + y^2 = (L/2)^2 (\cos^2 \theta + \sin^2 \theta) = (L/2)^2$. This is the equation of a circle with radius $L/2$ centered at the origin (the corner where the wall meets the floor). Thus,the center of mass moves along a circular arc centered at the corner.

(B) Since equal and opposite forces are applied to the two bodies,the net external force on the system is zero.
According to the property of the centre of mass,if the net external force on a system is zero,the centre of mass remains at rest.
Therefore,the two bodies will meet at their common centre of mass.
The position of the centre of mass is determined by the relation $M x_A = m x_B$,where $x_A$ and $x_B$ are the distances of the bodies from the centre of mass.
Since $M > m$,it follows that $x_A < x_B$.
This implies that the centre of mass is closer to the body with the larger mass,which is body $A$.

(A) Before the explosion,the particle was moving along the $x$-axis,which means it had no $y$-component of velocity. Consequently,the center of mass of the system does not move in the $y$-direction,so $y_{CM} = 0$.
Using the formula for the center of mass: $y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$.
Substituting the given values: $0 = \frac{(m/4)(+15) + (3m/4)(y)}{m}$.
This simplifies to: $0 = \frac{15}{4} + \frac{3y}{4}$.
Solving for $y$: $\frac{3y}{4} = -\frac{15}{4}$,which gives $y = -5 \ cm$.

(D) The center of mass of the rod is located at its midpoint,which is at a distance of $l/2$ from the pivot.
When the rod is displaced by an angle $\theta = 60^o$,the center of mass rises by a vertical height $h$.
From the geometry of the figure,the vertical distance of the center of mass from the pivot in the displaced position is $(l/2) \cos \theta$.
The initial vertical distance of the center of mass from the pivot is $l/2$.
Therefore,the rise in height $h$ is given by:
$h = \frac{l}{2} - \frac{l}{2} \cos \theta = \frac{l}{2}(1 - \cos \theta)$
The increase in potential energy $\Delta U$ is given by $\Delta U = mgh$.
Given: $m = 400 \ g = 0.4 \ kg$,$l = 1 \ m$,$\theta = 60^o$,and taking $g = 10 \ m/s^2$.
Substituting the values:
$\Delta U = 0.4 \times 10 \times \frac{1}{2}(1 - \cos 60^o)$
$\Delta U = 4 \times 0.5 \times (1 - 0.5)$
$\Delta U = 2 \times 0.5 = 1 \ J$.

(B) The linear density of the rod varies with distance $x$ as $\lambda = \frac{dm}{dx} = 2 + x$.
The position of the centre of mass $x_{cm}$ is given by the formula:
$x_{cm} = \frac{\int x \, dm}{\int dm} = \frac{\int_{0}^{3} x \lambda \, dx}{\int_{0}^{3} \lambda \, dx}$
Substituting $\lambda = 2 + x$:
$x_{cm} = \frac{\int_{0}^{3} x(2 + x) \, dx}{\int_{0}^{3} (2 + x) \, dx} = \frac{\int_{0}^{3} (2x + x^2) \, dx}{\int_{0}^{3} (2 + x) \, dx}$
Evaluating the integrals:
Numerator: $\int_{0}^{3} (2x + x^2) \, dx = [x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) - 0 = 9 + 9 = 18$
Denominator: $\int_{0}^{3} (2 + x) \, dx = [2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) - 0 = 6 + 4.5 = 10.5 = \frac{21}{2}$
Therefore,$x_{cm} = \frac{18}{21/2} = \frac{18 \times 2}{21} = \frac{36}{21} = \frac{12}{7} \ m$.

(A) Let the disc lie in the $xy$-plane with its centre at the origin $O(0,0)$. The disc occupies the first quadrant.
Due to symmetry about the line $y=x$,the centre of mass $(X_{cm}, Y_{cm})$ must lie on this line,so $X_{cm} = Y_{cm}$.
For a semi-circular disc of radius $R$,the centre of mass is at a distance $\frac{4R}{3\pi}$ from the diameter along the axis of symmetry.
For a quarter disc,the centre of mass coordinates are $X_{cm} = \frac{4R}{3\pi}$ and $Y_{cm} = \frac{4R}{3\pi}$.
The distance of the centre of mass from the origin $O$ is given by $d = \sqrt{X_{cm}^2 + Y_{cm}^2}$.
$d = \sqrt{\left(\frac{4R}{3\pi}\right)^2 + \left(\frac{4R}{3\pi}\right)^2} = \sqrt{2 \left(\frac{4R}{3\pi}\right)^2} = \sqrt{2} \frac{4R}{3\pi}$.

(D) Let the mass of the horizontal rod be $m_1$ and the mass of the vertical rod be $m_2$.
The centre of mass of the horizontal rod is at $(L/2, 0)$ and the centre of mass of the vertical rod is at $(0, L/2)$.
The coordinates of the centre of mass $(X_{cm}, Y_{cm})$ of the system are given by:
$X_{cm} = \frac{m_1(L/2) + m_2(0)}{m_1 + m_2} = \frac{m_1 L}{2(m_1 + m_2)}$
$Y_{cm} = \frac{m_1(0) + m_2(L/2)}{m_1 + m_2} = \frac{m_2 L}{2(m_1 + m_2)}$
Since the rods are made of different materials,$m_1 \neq m_2$. Let $\lambda = m_2/m_1$. Then $X_{cm} = \frac{L}{2(1+\lambda)}$ and $Y_{cm} = \frac{\lambda L}{2(1+\lambda)}$.
Note that $X_{cm} + Y_{cm} = \frac{L}{2(1+\lambda)} + \frac{\lambda L}{2(1+\lambda)} = \frac{L}{2}$.
Checking the options,only $(L/3, L/6)$ satisfies $X_{cm} + Y_{cm} = L/2$ (i.e.,$L/3 + L/6 = L/2$).
Thus,the correct option is $(L/3, L/6)$.

(D) The centre of mass $(C.M.)$ of the ring is at the origin $(0, 0)$.
The centre of mass of the chord $AB$ lies at its midpoint,which is $(R/2, R/2)$.
The $C.M.$ of the combined system must lie on the line segment connecting the $C.M.$ of the ring and the $C.M.$ of the chord.
This line segment is the line $y = x$ for $0 \le x \le R/2$.
Therefore,the $C.M.$ of the system must have coordinates $(x, x)$ where $0 \le x \le R/2$.
Option $(A)$ $(R/3, R/3)$ satisfies this condition.
Option $(B)$ $(R/3, R/2)$ does not satisfy $x = y$.
Option $(C)$ $(R/\sqrt{2}, R/\sqrt{2})$ has $x = R/\sqrt{2} \approx 0.707R$,which is greater than $R/2 = 0.5R$,so it lies outside the possible range.
Thus,both $(B)$ and $(C)$ cannot be the centre of mass.

(C) The correct positions of the Centre of Mass $(CM)$ from the geometric centre for objects of radius $R$ are as follows:
$[1]$ For a uniform semicircular disc,the $CM$ is at $\frac{4R}{3\pi}$.
$[2]$ For a uniform semicircular ring,the $CM$ is at $\frac{2R}{\pi}$.
$[3]$ For a solid hemisphere,the $CM$ is at $\frac{3R}{8}$.
$[4]$ For a hemispherical shell,the $CM$ is at $\frac{R}{2}$.
Comparing these with the given statements,only statement $[4]$ is correct.

(D) The moment of inertia of the system about axis $AA$ (passing through $m_A$) is $I_{AA} = m_B l^2$.
The moment of inertia of the system about axis $BB$ (passing through $m_B$) is $I_{BB} = m_A l^2$.
Given that $\frac{I_{BB}}{I_{AA}} = 3$,we have $\frac{m_A l^2}{m_B l^2} = 3$,which implies $\frac{m_A}{m_B} = 3$ or $m_A = 3 m_B$.
Let $x_A$ be the distance of the centre of mass from mass $m_A$. Then the distance from $m_B$ is $(l - x_A)$.
By the definition of the centre of mass,$m_A x_A = m_B (l - x_A)$.
Substituting $m_A = 3 m_B$,we get $3 m_B x_A = m_B (l - x_A)$.
Dividing by $m_B$,we get $3 x_A = l - x_A$,which simplifies to $4 x_A = l$.
Therefore,$x_A = \frac{l}{4}$.

(D) Let the rod have mass $M$ and length $L$. The rod is placed vertically on a horizontal surface with friction.
Since the surface has friction,the point of contact $P$ remains stationary as the rod falls.
There are no external horizontal forces acting on the rod-surface system in the horizontal direction (assuming the friction is sufficient to prevent slipping).
According to the principle of conservation of momentum,the horizontal position of the centre of mass $(CM)$ of the rod remains unchanged.
Initially,the rod is vertical,so its $CM$ is directly above the point of contact $P$.
Therefore,the horizontal coordinate of the $CM$ is at $P$.
When the rod becomes horizontal,its $CM$ must still be at the same horizontal position as it was initially.
Thus,the $CM$ lies at $P$.

(D) Let the center of mass be at the origin. Initially,the positions of the particles are $-x_1$ and $x_2$ such that $m_1(-x_1) + m_2(x_2) = 0$,which implies $m_1 x_1 = m_2 x_2$ (Equation $1$).
When the first particle is moved by distance $d$ towards the center of mass,its new position becomes $-(x_1 - d)$. Let the second particle be moved by distance $d'$ towards the center of mass,so its new position becomes $(x_2 - d')$.
To keep the center of mass at the origin,the new condition is:
$m_1(-(x_1 - d)) + m_2(x_2 - d') = 0$
$-m_1 x_1 + m_1 d + m_2 x_2 - m_2 d' = 0$
Since $m_1 x_1 = m_2 x_2$ from Equation $1$,the terms $m_1 x_1$ and $m_2 x_2$ cancel out:
$m_1 d - m_2 d' = 0$
$m_2 d' = m_1 d$
$d' = \frac{m_1}{m_2} d$

(A) Consider a solid cone of height $h$ and base radius $R$. Let the density of the material be $\rho$.
We place the vertex at the origin $(0,0)$ and the axis of the cone along the $y$-axis.
At a distance $y$ from the vertex,consider a thin elemental disk of thickness $dy$ and radius $r$.
By similar triangles,$\frac{r}{y} = \frac{R}{h}$,which implies $r = \frac{R}{h}y$.
The mass of this elemental disk is $dm = \rho \cdot \pi r^2 dy = \rho \pi \left(\frac{R}{h}y\right)^2 dy$.
The centre of mass $z_0$ (or $y_{cm}$) is given by:
$z_0 = \frac{\int y dm}{\int dm} = \frac{\int_0^h y \cdot \rho \pi \frac{R^2}{h^2} y^2 dy}{\int_0^h \rho \pi \frac{R^2}{h^2} y^2 dy}$
$z_0 = \frac{\int_0^h y^3 dy}{\int_0^h y^2 dy} = \frac{[y^4/4]_0^h}{[y^3/3]_0^h} = \frac{h^4/4}{h^3/3} = \frac{3h}{4}$.

(A) The mass of a small element of length $dx$ at a distance $x$ from the end is $dm = \lambda dx = \frac{k x^3}{L^3} dx$.
The position of the centre of mass $X_{cm}$ is given by the formula:
$X_{cm} = \frac{\int x dm}{\int dm}$
Substituting the expression for $dm$:
$X_{cm} = \frac{\int_0^L x (\frac{k x^3}{L^3} dx)}{\int_0^L (\frac{k x^3}{L^3} dx)}$
$X_{cm} = \frac{\frac{k}{L^3} \int_0^L x^4 dx}{\frac{k}{L^3} \int_0^L x^3 dx}$
$X_{cm} = \frac{[\frac{x^5}{5}]_0^L}{[\frac{x^4}{4}]_0^L} = \frac{L^5 / 5}{L^4 / 4} = \frac{4}{5} L$

(D) Let the positions of the three masses be $m_1 = M$ at $(L, 0)$, $m_2 = M$ at $(-L, 0)$, and $m_3 = M$ at $(L \cos \theta, L \sin \theta)$.
The coordinates of the center of mass $(X, Y)$ are given by:
$X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{M(L) + M(-L) + M(L \cos \theta)}{3M} = \frac{L \cos \theta}{3}$
$Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{M(0) + M(0) + M(L \sin \theta)}{3M} = \frac{L \sin \theta}{3}$
Squaring and adding the equations for $X$ and $Y$:
$X^2 + Y^2 = (\frac{L \cos \theta}{3})^2 + (\frac{L \sin \theta}{3})^2 = \frac{L^2}{9} (\cos^2 \theta + \sin^2 \theta) = \frac{L^2}{9}$
Thus, the path traced by the $COM$ is $x^2 + y^2 = L^2/9$.

(C) First,we identify the center of mass of each individual cube. Since each cube has a side length of $1$ unit and they are placed along the axes,their centers are located at:
Cube $1$ (at origin): $(0.5, 0.5, 0.5)$
Cube $2$ (along $x$-axis): $(1.5, 0.5, 0.5)$
Cube $3$ (along $y$-axis): $(0.5, 1.5, 0.5)$
Cube $4$ (along $z$-axis): $(0.5, 0.5, 1.5)$
Since all four cubes are identical,we can treat the center of mass of each cube as a point particle of equal mass $m$ located at its respective center.
The coordinates of the center of mass of the system are given by:
$x_{\text{COM}} = \frac{m(0.5) + m(1.5) + m(0.5) + m(0.5)}{4m} = \frac{3.0}{4} = 0.75$
$y_{\text{COM}} = \frac{m(0.5) + m(0.5) + m(1.5) + m(0.5)}{4m} = \frac{3.0}{4} = 0.75$
$z_{\text{COM}} = \frac{m(0.5) + m(0.5) + m(0.5) + m(1.5)}{4m} = \frac{3.0}{4} = 0.75$
Thus,the center of mass is $(0.75, 0.75, 0.75)$ or $(3/4, 3/4, 3/4)$.

(C) Let the mass of the $1^{st}$ block be $m_1 = m$ and its length be $x_1 = x$. The mass of the $n^{th}$ block is $m_n = m/4^{n-1}$ and its length is $x_n = x/2^{n-1}$.
To find the center of mass of the system relative to the left edge (axis $AA'$),we calculate the weighted average of the center of mass of each block.
The center of mass of the $n^{th}$ block from the axis $AA'$ is $x_{cm,n} = \frac{1}{2} \cdot \frac{x}{2^{n-1}} = \frac{x}{2^n}$.
The total mass of the system is $M = \sum_{n=1}^{\infty} \frac{m}{4^{n-1}} = m \left( 1 + \frac{1}{4} + \frac{1}{16} + \dots \right) = m \left( \frac{1}{1 - 1/4} \right) = \frac{4m}{3}$.
The position of the center of mass of the entire system is $X_{CM} = \frac{1}{M} \sum_{n=1}^{\infty} m_n x_{cm,n} = \frac{3}{4m} \sum_{n=1}^{\infty} \left( \frac{m}{4^{n-1}} \cdot \frac{x}{2^n} \right) = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^n \cdot 4^{n-1}} = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^n \cdot 2^{2n-2}} = \frac{3x}{4} \sum_{n=1}^{\infty} \frac{1}{2^{3n-2}} = \frac{3x}{4} \left( \frac{1/2}{1 - 1/8} \right) = \frac{3x}{4} \left( \frac{1/2}{7/8} \right) = \frac{3x}{4} \cdot \frac{4}{7} = \frac{3x}{7}$.
For the system to be on the verge of toppling,the center of mass must lie exactly at the edge of the table,so $L = X_{CM} = 3x/7$.

(D) Let the mass of each brick be $m$. The $x$-coordinate of the centre of mass of each individual brick is at its geometric centre.
Taking the origin $O$ at the left edge of the bottom brick (brick $1$):
$x_1 = \frac{L}{2}$
$x_2 = \frac{L}{2} + \frac{L}{10}$
$x_3 = \frac{L}{2} + \frac{2L}{10}$
$x_4 = \frac{L}{2} + \frac{3L}{10}$
$x_5 = \frac{L}{2} + \frac{2L}{10}$
$x_6 = \frac{L}{2} + \frac{L}{10}$
$x_7 = \frac{L}{2}$
Since there are $7$ identical bricks,the total mass $M = 7m$.
The $x$-coordinate of the centre of mass of the system is given by:
$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{m(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)}{7m}$
$x_{cm} = \frac{1}{7} [(\frac{L}{2}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{3L}{10}) + (\frac{L}{2} + \frac{2L}{10}) + (\frac{L}{2} + \frac{L}{10}) + (\frac{L}{2})]$
$x_{cm} = \frac{1}{7} [7(\frac{L}{2}) + 2(\frac{L}{10} + \frac{2L}{10} + \frac{3L}{10})]$
$x_{cm} = \frac{1}{7} [\frac{7L}{2} + 2(\frac{6L}{10})] = \frac{1}{7} [\frac{7L}{2} + \frac{6L}{5}]$
$x_{cm} = \frac{1}{7} [\frac{35L + 12L}{10}] = \frac{47L}{70}$
Wait,re-evaluating the displacement pattern from the image: The bricks are stacked symmetrically. The middle brick (brick $4$) is shifted by $3L/10$. The calculation above is correct based on the symmetry shown.

(C) The center of mass of a system of two particles (or two bodies) always lies on the line segment joining their individual centers of mass.
In this case,the centers of mass of the two rods are at points $A$ and $B$.
Therefore,the center of mass of the combined system must lie on the line segment $AB$.
Since the line $AB$ is the boundary between region $1$ and region $2$,and the center of mass must lie on this line,option $C$ is the correct choice.

(A) The center of mass $\vec{R}_{cm}$ is defined as $\frac{\int \vec{r} \, dm}{\int dm}$.
For a body of uniform density $\rho$,the mass of a small volume element $dV$ is $dm = \rho \, dV$.
Substituting this into the center of mass formula:
$\vec{R}_{cm} = \frac{\int \vec{r} \, \rho \, dV}{\int \rho \, dV}$.
Since $\rho$ is constant for a uniform body,it can be taken out of the integral:
$\vec{R}_{cm} = \frac{\rho \int \vec{r} \, dV}{\rho \int dV} = \frac{\int \vec{r} \, dV}{\int dV}$.
This expression is exactly the definition of the center of volume provided in the question.
Therefore,for a body of uniform density,the center of volume is the same as the center of mass.

(D) The position of the centre of mass of a system of particles is defined by the formula $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
From this formula,it is clear that the centre of mass depends on the masses $(m_i)$ and the positions $(\vec{r}_i)$ of the particles.
The relative distance between particles is a function of their positions.
However,the centre of mass is a geometric property of the distribution of mass and does not depend on the external or internal forces acting on the particles.

(D) The position of the centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.
Given the linear mass density $\lambda = kx^2$,the mass element is $dm = \lambda dx = kx^2 dx$.
Substituting these into the formula:
$x_{cm} = \frac{\int_{0}^{L} x (kx^2 dx)}{\int_{0}^{L} kx^2 dx} = \frac{k \int_{0}^{L} x^3 dx}{k \int_{0}^{L} x^2 dx}$.
Evaluating the integrals:
$x_{cm} = \frac{[x^4/4]_{0}^{L}}{[x^3/3]_{0}^{L}} = \frac{L^4/4}{L^3/3} = \frac{L^4}{4} \times \frac{3}{L^3} = \frac{3L}{4}$.

(B) Let $\rho$ be the surface mass density of the discs. Since they have the same thickness and material,$\rho$ is constant for both.
Mass of the large disc $(M_1)$ = $\pi(2R)^2 \rho = 4\pi R^2 \rho$.
The center of mass of the large disc is at the origin,so $x_1 = 0$.
Mass of the small disc $(M_2)$ = $\pi R^2 \rho$.
The center of the small disc is at a distance $R$ from the origin,so $x_2 = R$.
The center of mass of the system $(X_{cm})$ is given by:
$X_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$
$X_{cm} = \frac{(4\pi R^2 \rho)(0) + (\pi R^2 \rho)(R)}{4\pi R^2 \rho + \pi R^2 \rho}$
$X_{cm} = \frac{\pi R^3 \rho}{5\pi R^2 \rho} = \frac{R}{5}$.
Thus,the center of mass is at a distance $\frac{R}{5}$ from the origin towards the smaller disc.

(A) Since no external force acts on the system,the center of mass remains at rest.
Let the mass $M$ be at the origin $(x_1 = 0)$ and the mass $4M$ be at $x_2 = 10 \ m$.
The position of the center of mass $X_{cm}$ is given by:
$X_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$
$X_{cm} = \frac{M(0) + 4M(10)}{M + 4M} = \frac{40M}{5M} = 8 \ m$.
The particles will collide at the center of mass.
Therefore,the distance from the smaller particle (mass $M$) is $8 \ m$.

(B) Let the linear mass density of the rods be $\lambda$.
Each rod has its centre of mass at its midpoint. For a rod of length $l_i$ starting from the $y-$axis,the $x-$coordinate of its centre of mass is $x_i = l_i / 2$.
The mass of each rod is $m_i = \lambda l_i$.
The lengths are $L, L/2, L/4, \dots$,so the masses are $M_1 = \lambda L, M_2 = \lambda L/2, M_3 = \lambda L/4, \dots$.
The $x-$coordinates of the centres of mass are $x_1 = L/2, x_2 = L/4, x_3 = L/8, \dots$.
The $x-$coordinate of the centre of mass of the system is given by:
$X_{CM} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(\lambda L)(L/2) + (\lambda L/2)(L/4) + (\lambda L/4)(L/8) + \dots}{\lambda L + \lambda L/2 + \lambda L/4 + \dots}$
$X_{CM} = \frac{\lambda L^2 (1/2 + 1/8 + 1/32 + \dots)}{\lambda L (1 + 1/2 + 1/4 + \dots)}$
$X_{CM} = \frac{L}{2} \cdot \frac{(1 + 1/4 + 1/16 + \dots)}{(1 + 1/2 + 1/4 + \dots)}$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$:
Numerator sum $= \frac{1}{1 - 1/4} = \frac{1}{3/4} = 4/3$.
Denominator sum $= \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
$X_{CM} = \frac{L}{2} \cdot \frac{4/3}{2} = \frac{L}{2} \cdot \frac{2}{3} = L/3$.

(D) The center of mass $x_{cm}$ for a continuous body is given by $x_{cm} = \frac{\int x dm}{\int dm}$.
Given the density $\rho(x) = \rho_0 \frac{x^2}{L^2}$,the mass element $dm$ for a cross-sectional area $A$ is $dm = \rho(x) A dx = \rho_0 \frac{x^2}{L^2} A dx$.
The total mass $M = \int_0^L dm = \int_0^L \rho_0 \frac{x^2}{L^2} A dx = \frac{\rho_0 A}{L^2} \left[ \frac{x^3}{3} \right]_0^L = \frac{\rho_0 A L}{3}$.
The moment of mass about the origin is $\int_0^L x dm = \int_0^L x \left( \rho_0 \frac{x^2}{L^2} A \right) dx = \frac{\rho_0 A}{L^2} \int_0^L x^3 dx = \frac{\rho_0 A}{L^2} \left[ \frac{x^4}{4} \right]_0^L = \frac{\rho_0 A L^2}{4}$.
Therefore,$x_{cm} = \frac{\frac{\rho_0 A L^2}{4}}{\frac{\rho_0 A L}{3}} = \frac{3L}{4}$.

(A) The coordinates of the corners of the square based on the figure are:
$(0,0)$ for $2\,kg$,$(2,0)$ for $3\,kg$,$(2,2)$ for $5\,kg$,and $(0,2)$ for $8\,kg$.
The $x$-coordinate of the centre of mass is given by:
$X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4}{m_1 + m_2 + m_3 + m_4}$
$X_{CM} = \frac{2 \times 0 + 3 \times 2 + 5 \times 2 + 8 \times 0}{2 + 3 + 5 + 8} = \frac{0 + 6 + 10 + 0}{18} = \frac{16}{18} = \frac{8}{9}\,m$
The $y$-coordinate of the centre of mass is given by:
$Y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{m_1 + m_2 + m_3 + m_4}$
$Y_{CM} = \frac{2 \times 0 + 3 \times 0 + 5 \times 2 + 8 \times 2}{2 + 3 + 5 + 8} = \frac{0 + 0 + 10 + 16}{18} = \frac{26}{18} = \frac{13}{9}\,m$
Therefore,the coordinates of the centre of mass are $\left( \frac{8}{9}, \frac{13}{9} \right)$.

(D) The formula for the position vector of the centre of mass is given by $\overrightarrow{r}_{cm} = \frac{m_{1} \overrightarrow{r}_{1} + m_{2} \overrightarrow{r}_{2}}{m_{1} + m_{2}}$.
Given $m_{1} = 1\,kg$,$\overrightarrow{r}_{1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $m_{2} = 3\,kg$,$\overrightarrow{r}_{2} = -2\hat{i} + 3\hat{j} - 4\hat{k}$.
Substituting the values:
$\overrightarrow{r}_{cm} = \frac{1(2\hat{i} + 3\hat{j} + 4\hat{k}) + 3(-2\hat{i} + 3\hat{j} - 4\hat{k})}{1 + 3}$
$\overrightarrow{r}_{cm} = \frac{2\hat{i} + 3\hat{j} + 4\hat{k} - 6\hat{i} + 9\hat{j} - 12\hat{k}}{4}$
$\overrightarrow{r}_{cm} = \frac{-4\hat{i} + 12\hat{j} - 8\hat{k}}{4}$
$\overrightarrow{r}_{cm} = -\hat{i} + 3\hat{j} - 2\hat{k}$.

(A) Let the linear density of wires $AB$ and $AC$ be $\lambda$. Then the linear density of wire $BC$ is $2\lambda$.
Mass of $AB = 5\lambda$,Mass of $AC = 5\lambda$,Mass of $BC = 6 \times 2\lambda = 12\lambda$.
The height of the triangle $ABC$ from $BC$ to $A$ is $h = \sqrt{5^2 - 3^2} = 4\,cm$.
Let $BC$ lie on the $x$-axis with $B$ at $(0,0)$ and $C$ at $(6,0)$. Then $A$ is at $(3,4)$.
The centre of mass of $AB$ is at $(1.5, 2)$,$AC$ is at $(4.5, 2)$,and $BC$ is at $(3, 0)$.
The $y$-coordinate of the centre of mass is $Y_{cm} = \frac{m_{AB}y_{AB} + m_{AC}y_{AC} + m_{BC}y_{BC}}{m_{AB} + m_{AC} + m_{BC}} = \frac{5\lambda(2) + 5\lambda(2) + 12\lambda(0)}{5\lambda + 5\lambda + 12\lambda} = \frac{20\lambda}{22\lambda} = \frac{10}{11}\,cm$.
This is the distance from the base $BC$. The distance from $A$ is $h - Y_{cm} = 4 - \frac{10}{11} = \frac{44-10}{11} = \frac{34}{11}\,cm$.

(A) Let the surface mass density be $\sigma$. The diameter of the circular plate is $\ell$,so its radius is $R = \ell/2$. The area of the circular plate is $A_1 = \pi R^2 = \pi(\ell/2)^2 = \pi \ell^2/4$. The mass of the circular plate is $m_1 = \sigma A_1 = \sigma(\pi \ell^2/4)$.
The side length of the square plate is $\ell$,so its area is $A_2 = \ell^2$. The mass of the square plate is $m_2 = \sigma A_2 = \sigma \ell^2$.
Let the center of the circular plate be at the origin $(0,0)$. The center of mass of the circular plate is at $x_1 = 0$. The center of the square plate is at a distance of $R + \ell/2 = \ell/2 + \ell/2 = \ell$ from the origin. So,$x_2 = \ell$.
The $x$-coordinate of the center of mass of the composite system is given by:
$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{\sigma(\pi \ell^2/4)(0) + \sigma \ell^2(\ell)}{\sigma(\pi \ell^2/4) + \sigma \ell^2} = \frac{\ell^3}{\ell^2(\pi/4 + 1)} = \frac{\ell}{\pi/4 + 1} = \frac{4\ell}{\pi + 4}$.
Since $\pi \approx 3.14$,$\pi + 4 \approx 7.14$. Thus,$x_{cm} = \frac{4\ell}{7.14} \approx 0.56\ell$.
The square plate extends from $x = \ell/2 = 0.5\ell$ to $x = 3\ell/2 = 1.5\ell$. Since $0.5\ell < 0.56\ell < 1.5\ell$,the center of mass lies inside the square plate.

(A) Let the masses $m_1 = 1\,kg$,$m_2 = \frac{3}{2}\,kg$,and $m_3 = 2\,kg$ be located at the vertices $A(0, 0)$,$B(a, 0)$,and $C\left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right)$ respectively.
The $x$-coordinate of the centre of mass is given by:
$X_{CM} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{1(0) + \frac{3}{2}(a) + 2(\frac{a}{2})}{1 + \frac{3}{2} + 2} = \frac{\frac{3a}{2} + a}{\frac{2+3+4}{2}} = \frac{\frac{5a}{2}}{\frac{9}{2}} = \frac{5a}{9}$.
The $y$-coordinate of the centre of mass is given by:
$Y_{CM} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{1(0) + \frac{3}{2}(0) + 2(\frac{\sqrt{3}a}{2})}{1 + \frac{3}{2} + 2} = \frac{\sqrt{3}a}{\frac{9}{2}} = \frac{2\sqrt{3}a}{9} = \frac{2a}{3\sqrt{3}}$.
Thus,the coordinates of the centre of mass are $\left( \frac{5a}{9}, \frac{2a}{3\sqrt{3}} \right)$.

(A) Since the particle is isolated and moving along the $x-$axis,there is no external force acting on it in the $y-$direction. Therefore,the center of mass of the system remains at $y = 0$ (the $x-$axis).
Let $y_1$ and $y_2$ be the $y-$coordinates of the fragments of masses $m_1 = m/4$ and $m_2 = 3m/4$ respectively.
The position of the center of mass in the $y-$direction is given by $Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = 0$.
Substituting the given values: $\frac{(m/4)(15) + (3m/4)(y_2)}{m} = 0$.
This simplifies to: $\frac{15}{4} + \frac{3}{4}y_2 = 0$.
Multiplying by $4$: $15 + 3y_2 = 0$.
Solving for $y_2$: $3y_2 = -15$,which gives $y_2 = -5 \ cm$.

(C) For any object with uniform mass density and high degree of symmetry,the centre of mass coincides with its geometric centre.
Since a cube is a highly symmetric three-dimensional object with uniform density,its centre of mass is located at its geometric centre,which is the point where its body diagonals intersect.

(A) The position vector of the centre of mass for a system of particles is given by the formula:
$\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$
Given:
$m_1 = 1\,kg, \vec{r}_1 = (\hat{i} + 4\hat{j} + \hat{k})\,m$
$m_2 = 2\,kg, \vec{r}_2 = (\hat{i} + \hat{j} + \hat{k})\,m$
$m_3 = 3\,kg, \vec{r}_3 = (2\hat{i} - \hat{j} - 2\hat{k})\,m$
Substituting the values:
$\vec{R}_{cm} = \frac{1(\hat{i} + 4\hat{j} + \hat{k}) + 2(\hat{i} + \hat{j} + \hat{k}) + 3(2\hat{i} - \hat{j} - 2\hat{k})}{1 + 2 + 3}$
$\vec{R}_{cm} = \frac{(\hat{i} + 4\hat{j} + \hat{k}) + (2\hat{i} + 2\hat{j} + 2\hat{k}) + (6\hat{i} - 3\hat{j} - 6\hat{k})}{6}$
$\vec{R}_{cm} = \frac{(1 + 2 + 6)\hat{i} + (4 + 2 - 3)\hat{j} + (1 + 2 - 6)\hat{k}}{6}$
$\vec{R}_{cm} = \frac{9\hat{i} + 3\hat{j} - 3\hat{k}}{6}$
$\vec{R}_{cm} = \frac{3(3\hat{i} + \hat{j} - \hat{k})}{6} = \frac{1}{2}(3\hat{i} + \hat{j} - \hat{k})\,m$

(B) Based on the coordinate system shown in the figure:
- Particle at $D$ (origin) has mass $m_D = 4\,kg$ at $(0, 0)$.
- Particle at $C$ has mass $m_C = 3\,kg$ at $(1, 0)$.
- Particle at $B$ has mass $m_B = 2\,kg$ at $(1, 1)$.
- Particle at $A$ has mass $m_A = 1\,kg$ at $(0, 1)$.
The $x$-coordinate of the centre of mass is:
$X_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{1(0) + 2(1) + 3(1) + 4(0)}{1 + 2 + 3 + 4} = \frac{5}{10} = 0.5\,m$.
The $y$-coordinate of the centre of mass is:
$Y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{1(1) + 2(1) + 3(0) + 4(0)}{1 + 2 + 3 + 4} = \frac{3}{10} = 0.3\,m$.
Thus,the coordinates of the centre of mass are $(0.5\,m, 0.3\,m)$.

(C) The center of mass $X_{cm}$ of a continuous body is given by the formula: $X_{cm} = \frac{\int x dm}{\int dm}$.
Given $\lambda = \frac{M_0 x}{l}$,the mass element $dm$ is $\lambda dx = \frac{M_0 x}{l} dx$.
Substituting this into the formula:
$X_{cm} = \frac{\int_{0}^{l} x (\frac{M_0 x}{l} dx)}{\int_{0}^{l} (\frac{M_0 x}{l} dx)}$
$X_{cm} = \frac{\frac{M_0}{l} \int_{0}^{l} x^2 dx}{\frac{M_0}{l} \int_{0}^{l} x dx}$
$X_{cm} = \frac{[x^3/3]_{0}^{l}}{[x^2/2]_{0}^{l}}$
$X_{cm} = \frac{l^3/3}{l^2/2} = \frac{l^3}{3} \times \frac{2}{l^2} = \frac{2l}{3}$.

(C) Let the length of wire $AB = x$ and $BC = y$. Let the linear mass density be $\lambda$.
The center of mass of $BC$ is at $(y/2, 0)$ and its mass is $m_1 = \lambda y$.
The center of mass of $AB$ is at $(x/2 \cos 60^{\circ}, x/2 \sin 60^{\circ}) = (x/4, x\sqrt{3}/4)$ and its mass is $m_2 = \lambda x$.
The $x$-coordinate of the center of mass of the system is given by:
$X_{cm} = \frac{m_1(y/2) + m_2(x/4)}{m_1 + m_2} = \frac{\lambda y(y/2) + \lambda x(x/4)}{\lambda(x + y)} = \frac{y^2/2 + x^2/4}{x + y}$.
Since the center of mass lies vertically below $A$,its $x$-coordinate must be equal to the $x$-coordinate of point $A$,which is $x \cos 60^{\circ} = x/2$.
Equating the two:
$\frac{y^2/2 + x^2/4}{x + y} = \frac{x}{2} \Rightarrow y^2/2 + x^2/4 = x^2/2 + xy/2$.
Multiplying by $4$:
$2y^2 + x^2 = 2x^2 + 2xy \Rightarrow 2y^2 - 2xy - x^2 = 0$.
Dividing by $x^2$ and letting $r = y/x$:
$2r^2 - 2r - 1 = 0$.
Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$r = \frac{2 \pm \sqrt{4 - 4(2)(-1)}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2\sqrt{3}}{4} = \frac{1 \pm \sqrt{3}}{2}$.
Since $r$ must be positive,$r = \frac{1 + \sqrt{3}}{2} \approx \frac{1 + 1.732}{2} = 1.366 \approx 1.37$.

(B) The center of mass $x_{cm}$ of a rod with variable linear mass density $\mu(x)$ is given by the formula:
$x_{cm} = \frac{\int_{0}^{L} x \mu(x) dx}{\int_{0}^{L} \mu(x) dx}$
Substituting $\mu(x) = a + \frac{bx}{L}$:
$x_{cm} = \frac{\int_{0}^{L} x(a + \frac{bx}{L}) dx}{\int_{0}^{L} (a + \frac{bx}{L}) dx} = \frac{\int_{0}^{L} (ax + \frac{bx^2}{L}) dx}{\int_{0}^{L} (a + \frac{bx}{L}) dx}$
Evaluating the integrals:
Numerator: $[\frac{ax^2}{2} + \frac{bx^3}{3L}]_{0}^{L} = \frac{aL^2}{2} + \frac{bL^2}{3} = L^2(\frac{a}{2} + \frac{b}{3})$
Denominator: $[ax + \frac{bx^2}{2L}]_{0}^{L} = aL + \frac{bL}{2} = L(a + \frac{b}{2})$
Thus,$x_{cm} = \frac{L^2(\frac{a}{2} + \frac{b}{3})}{L(a + \frac{b}{2})} = L \frac{(\frac{3a + 2b}{6})}{(\frac{2a + b}{2})} = L \frac{3a + 2b}{3(2a + b)}$
Given $x_{cm} = \frac{7}{12}L$,we have:
$\frac{3a + 2b}{3(2a + b)} = \frac{7}{12}$
$\frac{3a + 2b}{2a + b} = \frac{7}{4}$
$4(3a + 2b) = 7(2a + b)$
$12a + 8b = 14a + 7b$
$b = 2a$.