A uniform metal disc of radius \$R\$ is taken and out of it a disc of diameter \$\cfrac{R}{2}\$ is cut off from the end. The centre of mass of the remaining part will be:

Density is uniform, so we can use geometrical method. Area of disc, \${{A}_{1}}=\pi {{R}^{2}}\$ Area of cutoff disc, \${{A}_{2}}=\dfrac{\pi }{4}{{\left( \dfrac{R}{2} \right)}^{2}}=\dfrac{\pi {{R}^{2}}}{16}\$ \${{x}_{com}}=\dfrac{{{A}_{1}}\times 0-{{A}_{2}}{{x}_{2}}}{{{A}_{1}}-{{A}_{2}}}=\dfrac{-\dfrac{\pi {{R}^{2}}}{16}\times \dfrac{3R}{4}}{\pi {{R}^{2}}-\dfrac{\pi {{R}^{2}}}{16}}=\dfrac{-3R}{15\times 4}=\dfrac{-R}{20}\$ Hence, center of mass is at \$\dfrac{R}{20}\$ from the center.

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Centre of Mass for Collection of Particles

Q: Three particles of mass \$1\ kg, 2\ kg\$ and \$3\ kg\$ are placed at the corners \$A, B\$ and \$C\$ respectively of an equilateral triangle \$ABC\$ of edge \$1\ m\$. Find the distance of their centre of mass from \$A\$.

Assume , Point \$A\$ is origin.From the figure, \$CP= 1\times sin(60^o)=\dfrac{\sqrt 3}{2}\$Centre of mass from Point \$A\$ in direction of \$x-axis\$, \$x_{cm}= \dfrac{m_Ax_A+m_Bx_B+ m_Cx_C}{m_A+m_B+m_C}\$\$\Rightarrow x_{cm}= \dfrac{1\times 0+ 2\times 1+ 3\times \dfrac{1}{2}}{1+2+3}=\dfrac{7}{12}\$Centre of mass from Point \$A\$ in direction of \$y-axis\$, \$y_{cm}= \dfrac{m_Ay_A+m_By_B+ m_Cy_C}{m_A+m_B+m_C}\$\$\Rightarrow y_{cm}= \dfrac{1\times 0+ 2\times 0+ 3\times \dfrac{\sqrt 3}{2}}{1+2+3}=\dfrac{\sqrt 3}{4}\$So coordinate of centre of mass \$\left ( \dfrac{7}{12} , \dfrac{\sqrt 3}{4} \right )\$, Distance from \$A\$, \$= \sqrt{\left (\dfrac{7}{12}\right )^2+ \left (\dfrac{\sqrt 3}{4}\right )^2}=\dfrac{\sqrt {19}}{6}\$

Q: Find centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particle are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5 m long.

Given that, \$ {{m}_{1}}=100g \$ \$ {{m}_{2}}=150g \$ \$ {{m}_{3}}=200g \$ a = 0.5 m = 50 cm Now, according to diagram Center of mass coordinates \$ {{X}_{cm}}=\dfrac{0\times 100+150\times 50+200\times 25}{450} \$ \$ {{X}_{cm}}=\dfrac{7500+5000}{450} \$ \$ {{X}_{cm}}=27.8cm \$ \$ {{X}_{cm}}=0.28\,m \$ Now, \$ {{Y}_{cm}}=\dfrac{0\times 100+0\times 150+\sqrt{3}\times 25\times 200}{450} \$ \$ {{Y}_{cm}}=\dfrac{8660.3}{450} \$ \$ {{Y}_{cm}}=19.2\,cm \$ \$ {{Y}_{cm}}=0.2\,m \$ Hence, the center of mass of three particles at the vertices of an equilateral triangle is \$X_{cm}= 0.28 cm\$ and \$Y_{cm}=0.2 cm\$ respectively

Q: Find the center of mass of a uniform L shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 Kg.

Choosing the X and Y axes. we have the coordinates of the vertices of the L. shaped lamina as given in the figure. We can think of the L-shape to constant of 3 square each of length 1m. The mass of each square is 1 Kg. Since the lamina is uniform. The center of mass of each square in X,Y coordinate \${{C}_{1}}\left( 1/2,1/2 \right),\,\,{{C}_{2}}\left( 3/2,1/2 \right).\text{ }\!\!~\!\!\text{ }and\,{{C}_{3}}\left( 1/2,3/2 \right)\$ of the square are by, Formula for center of mass. \$X=\dfrac{{{m}_{a}}{{x}_{a}}+{{m}_{b}}{{x}_{b}}+{{m}_{c}}{{x}_{c}}}{{{m}_{a}}+{{m}_{b}}+{{m}_{c}}}\$ \$Y=\dfrac{{{m}_{a}}{{y}_{a}}+{{m}_{b}}{{y}_{b}}+{{m}_{c}}{{y}_{c}}}{{{m}_{a}}+{{m}_{b}}+{{m}_{c}}}\$ \$\text{ }\!\!~\!\!\text{ }X=\dfrac{\left[ 1\left( 1/2 \right)+1\left( 3/2 \right)+1\left( 1/2 \right)\text{ }\!\!~\!\!\text{ } \right]}{\left( 1+1+1 \right)}=\dfrac{5}{6}m\$ \$\text{ }\!\!~\!\!\text{ }X=\dfrac{\left[ 1\left( 1/2 \right)+1\left( 1/2 \right)+1\left( 3/2 \right)\text{ }\!\!~\!\!\text{ } \right]}{\left( 1+1+1 \right)}=\dfrac{5}{6}m\$ Hence, center of mass of \$L\$ shape is \$\left( \dfrac{5}{6},\dfrac{5}{6} \right)\$

Q: Three identical spheres, each of mass \$1\$kg are placed touching each other with their centres on a straight line. Their centers are marked P, Q and R respectively. The distance of center of mass of the system from P is:

The centre of mass of from P\$\begin{array}{l} x=\dfrac { { { M_{ p } }\left( { { x_{ p } } } \right) +{ M_{ q } }\left( { { x_{ q } } } \right) +{ M_{ R } }\left( { { x_{ R } } } \right) } }{ { { M_{ p } }+{ M_{ q } }+{ M_{ R } } } } \\ =\dfrac { { M\left( 0 \right) +M\left( { PQ } \right) +M\left( { PR } \right) } }{ { M+M+M } } \\ =\dfrac { { M\left( { PQ+PR } \right) } }{ { 3M } } \\ =\dfrac { { PQ+PR } }{ 3 } \end{array}\$Option B.

Q: A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of the discs coincide. The centre of mass of the new disc is \$\alpha R\$ from the centre of the bigger disc. The value of \$\alpha\$ is

\$\textbf{Step 1 - Draw the given situation Refer Figure}\$Let \$\sigma =\$ Surface mass density.So, \$M_{1} = \sigma A_{1} = \sigma \cdot 4\pi R^{2}\$ \$M_{2} = \sigma A_{2} = \sigma \pi R^{2}\$\$\textbf{Step 2 - Calculation of centre of mass}\$\$X_{cm} = \dfrac {M_{1}x_{2} + M_{2}x_{2}}{M_{1} + M_{2}}\$\$X_{cm} = \dfrac {(\sigma \cdot 4\pi R^{2}) \times 0 - (\sigma \cdot \pi R^{2})\cdot R}{(\sigma \cdot 4\pi R^{2}) - (\sigma \cdot \pi R^{2})}\$\$X_{cm} = \dfrac {-\sigma \cdot \pi R^{3}}{3\sigma \cdot \pi R^{2}}\$\$X_{cm} = \dfrac {-R}{3}\$Due to symmetry about x-axisSo, \$Y_{cm} = 0\$Thus, \$\vec {r}_{cm} = X_{cm} \hat i + Y_{cm} \hat {j}\$ \$\vec {r}_{cm} = \dfrac {-R}{3} \hat i + 0\$ \$|\vec {r}_{cm}| = \dfrac {R}{3}\$ \$....(1)\$as a given - \$|\vec {r}_{cm}| = \alpha R\$ \$....(2)\$So, comparing \$(1)\$ and \$(2)\$\$\alpha = \dfrac {1}{3}\$Hence, option (A) is correct.

Q: The center of mass of a system of three particles of masses \$1\ g, 2\ g\$ and \$3\ g\$ is taken as the origin of a coordinate system. The position vector of a fourth particle of mass \$4\ g\$ such that the center of mass of the four particle system lies at the point \$(1, 2, 3,)\$ is \$\alpha \left( {\hat i + 2\hat j + 3\hat k} \right)\$, where \$\alpha\$ is a constant. The value of \$\alpha\$ is: -

\$\textbf{Step 1 - Taking 1g, 2g, 3g particles as one system}\$We have \$m_{system} = 1 + 2 + 3\$ \$= 6\ g\$ \$....(1)\$Given \$\vec {r}_{com\ system} = 0\hat {i} + 0\hat {j} +0\hat {k}\$We can now write above as\$m_{1} = 6\ g\$ \$\vec {r}_{1} = 0\hat {i} + 0\hat {j} + 0\hat {k}\$\$\textbf{Step 2 - Now, it is also given}\$ \$m_{2} = 4g\$ \$\vec {r}_{2} = ?\$\$\vec {r}_{com} = (1\hat {i} + 2\hat {j} + 3\hat {k})\$We know that \$\vec {r}_{cm} = \dfrac {m_{1}\vec {r}_{1} + m_{2}\vec {r}_{2}}{m_{1} + m_{2}}\$\$(\hat {i} + 2\hat {j} + 3\hat {k}) = \dfrac {6\times 0 + 4\vec {r}_{2}}{6 + 4} = \dfrac {4\vec {r}_{2}}{10}\$ \$\vec {r}_{2} = \dfrac {10}{4} (\hat {i} + 2\hat {j} + 3\hat {k})\$Comparing above with \$\alpha (\hat {i} + 2\hat {j} + 3\hat {k})\$We get \$\alpha = \dfrac {10}{4} = 2.5\$option B

Q: In the HCl molecule, the separation between the nuclei of the two atoms is about \$1.27\ \mathring{A} (1 \mathring{A} = 10^{-10}\ m)\$. Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Let Mass of H atom \$=m\$Mass of Cl atom \$=35.5m\$Let the centre of mass of the system lie at a distance \$x\ \mathring A\$ from the Cl atom.Distance of the centre of mass from the H atom =\$(1.27-x)\ \mathring{A}\$Let us assume that the centre of mass of the given molecule lies at the origin. Let Hydrogen lie to the left of the origin and chlorine to the right.Position of Hydrogen atom \$=-(1.27-x)\ \mathring{A}\$Position of chlorine atom \$=x\ \mathring{A}\$ Therefore, we can have:\$\dfrac{-m(1.27-x)+35.5mx}{m+35.5m}=0\$\$\implies -m(1.27-x)+35.5mx=0\$\$\implies x=\dfrac{1.27}{35.5+1}=0.035\ \mathring{A}\$Hence, the centre of mass of the HCl molecule lies \$0.035\ \mathring{A}\$ from the Cl atom.

Q: Four particles of masses \$1\$kg,2kg, \$3\$kg and \$4\$kg are placed at the four vertices A, B, C and D, respectively of a square of side \$1 m\$. Fine the position of centre of mass of the particles.

Assuming \$D\$ as the origin. \$DC\$ as x-axis and \$DA\$ as y-axis we have\$m_1 = 1 kg, (x_1, y_1) = (0, 1 m)\$\$m_2 = 2 kg, (x_2, y_2) = (1 m, 1 m)\$\$m_3 = 3 kg, (x_3, y_3) = (1 m, 0)\$\$m_4 = 4 kg, (x_4, y_4) = (0, 0)\$Coordination of their \$CM\$ are calculated as follows.\$x_{CM} = \dfrac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}\$\$= \dfrac{(1)(0) + 2(1) + 3(1) + 4(0)}{1 +2 +3 +4}\$\$\dfrac{5}{10} = \dfrac{1}{2}m= 0.5 m\$Similarly, \$y_{CM} = \dfrac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}\$ \$= \dfrac{(1)(1) + 2(1) + 3(0) + 4(0)}{1 + 2 +3 + 4}\$\$= \dfrac{3}{10}m = 0.3 m\$\$\therefore (x_{CM}, y{CM}) = (0.5 m, 0.3 m)\$

Q: A circular plate of uniform thickness has diameter 56 cm.A circular part of diameter 42 cm is removed from one edge.What is the position of the center of mass of the remaining part

New center of mass is given by:\$x_{com}=\dfrac{m_1x_1-m_2x_2}{m_1-m_2}\$ where, \$m_1\$ is mass of original disc\$m_2\$ is mass of removed disc\$x_1\$ is center of mass position of original disc\$x_2\$ is center of mass position of removed discIf mass per unit area=\$m\$ then, mass of original disc=\$m \times \pi \times (28 \times 10^{-2})^2\$mass of removed disc=\$m \times \pi \times (21 \times 10^{-2})^2\$\$x_{com}==\dfrac { m\times \pi \times (28\times 10^{ -2 })^{ 2 }\times 0-m\times \pi \times (21\times 10^{ -2 })^{ 2 }\times \left( 7\times { 10 }^{ -2 } \right) }{ m\times \pi \times (28\times 10^{ -2 })^{ 2 }\quad -m\times \pi \times (21\times 10^{ -2 })^{ 2 } } \\ =\dfrac { -m\pi { \left( 28\times 10^{ -2 } \right) }^{ 2 }\left( 7\times { 10 }^{ -2 } \right) }{ m\pi \left[ \left( 28\times 10^{ -2 })^{ 2 }-(21\times 10^{ -2 })^{ 2 } \right) \right] } \\ =\dfrac { -0.003087 }{ 0.0343 } \\ =-0.09\quad m\\ =-9\quad cm\$Center of mass shifts by 9 cm.

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Centre of Mass of a System of N Particles

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Important Questions

A uniform metal disc of radius $R$ is taken and out of it a disc of diameter $\frac{R}{2}$ is cut off from the end. The centre of mass of the remaining part will be:

Medium

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Three particles of mass $1 k g, 2 k g$ and $3 k g$ are placed at the corners $A, B$ and $C$ respectively of an equilateral triangle $A B C$ of edge $1 m$ . Find the distance of their centre of mass from $A$ .

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Find centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particle are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5 m long.

Medium

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Find the center of mass of a uniform L shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 Kg.

Medium

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Three identical spheres, each of mass $1$ kg are placed touching each other with their centres on a straight line. Their centers are marked P, Q and R respectively. The distance of center of mass of the system from P is:

Medium

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A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of the discs coincide. The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

Medium

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The center of mass of a system of three particles of masses $1 g, 2 g$ and $3 g$ is taken as the origin of a coordinate system. The position vector of a fourth particle of mass $4 g$ such that the center of mass of the four particle system lies at the point $(1, 2, 3,)$ is $α (\hat{i} + 2 \hat{j} + 3 \hat{k})$ , where $α$ is a constant. The value of $α$ is: -

Hard

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In the HCl molecule, the separation between the nuclei of the two atoms is about $1.27 \overset{˚}{A} (1 \overset{˚}{A} = 1 0^{- 10} m)$ . Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

Medium

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Four particles of masses $1$ kg,2kg, $3$ kg and $4$ kg are placed at the four vertices A, B, C and D, respectively of a square of side $1 m$ . Fine the position of centre of mass of the particles.

Medium

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A circular plate of uniform thickness has diameter 56 cm.A circular part of diameter 42 cm is removed from one edge.What is the position of the center of mass of the remaining part

Medium

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Centre of Mass for Collection of Particles

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REVISE WITH CONCEPTS

Centre of Mass of a System of N Particles

A uniform metal disc of radius R is taken and out of it a disc of diameter 2R​ is cut off from the end. The centre of mass of the remaining part will be:

Three particles of mass 1 kg,2 kg and 3 kg are placed at the corners A,B and C respectively of an equilateral triangle ABC of edge 1 m. Find the distance of their centre of mass from A.

Find centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particle are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5 m long.

Find the center of mass of a uniform L shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is 3 Kg.

Three identical spheres, each of mass 1kg are placed touching each other with their centres on a straight line. Their centers are marked P, Q and R respectively. The distance of center of mass of the system from P is:

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of the discs coincide. The centre of mass of the new disc is αR from the centre of the bigger disc. The value of α is

Four particles of masses 1kg,2kg, 3kg and 4kg are placed at the four vertices A, B, C and D, respectively of a square of side 1m. Fine the position of centre of mass of the particles.

A circular plate of uniform thickness has diameter 56 cm.A circular part of diameter 42 cm is removed from one edge.What is the position of the center of mass of the remaining part

A uniform metal disc of radius $R$ is taken and out of it a disc of diameter $\frac{R}{2}$ is cut off from the end. The centre of mass of the remaining part will be:

Three particles of mass $1 k g, 2 k g$ and $3 k g$ are placed at the corners $A, B$ and $C$ respectively of an equilateral triangle $A B C$ of edge $1 m$ . Find the distance of their centre of mass from $A$ .

Three identical spheres, each of mass $1$ kg are placed touching each other with their centres on a straight line. Their centers are marked P, Q and R respectively. The distance of center of mass of the system from P is:

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of the discs coincide. The centre of mass of the new disc is $α R$ from the centre of the bigger disc. The value of $α$ is

Four particles of masses $1$ kg,2kg, $3$ kg and $4$ kg are placed at the four vertices A, B, C and D, respectively of a square of side $1 m$ . Fine the position of centre of mass of the particles.