Centre of Mass

The center of mass (COM) is essentially the "balance point" of an object or a system of particles. It's the location where you can think of all the mass of the system being concentrated when analyzing motion. To put it simply, it's like the average position of all the mass in an object, but weighted by how much mass is at each point.This concept is super important for understanding how objects move, especially when you’re dealing with things like forces, torque, or the conservation of momentum. Whether you’re looking at individual particles, complex shapes, or even a continuous mass, knowing the center of mass helps you predict how the entire system will respond to external forces.

1.0Centre of Mass

• The centre of mass is a point where we can concentrate the whole mass of the body and it behaves in a similar manner as a point object behaves under the same circumstances. OR
• The point in a system at which the whole mass of the system may be assumed to be concentrated for all translational effects of the system is called the Centre of Mass (CM).

2.0Properties of Centre of Mass

1. For symmetrical bodies having uniform distribution of mass, it coincides with the centre of symmetry or geometrical centre.

2. For a given shape it depends on the distribution of mass within the body and is closer to the massive part.

3. There may or may not be any mass present physically at the centre of mass and it may be within or outside the body.

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3.0Centre of Mass of Discrete Mass System

Position vector of centre of mass of all n-particles

${\vec{r}}_{\text{cm}}=\frac{\sum m_i{\vec{r}}_i}{\sum m_i}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2+m_3{\vec{r}}_3+\cdots +m_n{\vec{r}}_n}{m_1+m_2+m_3+\cdots +m_n}$

$x_{\text{cm}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+\cdots +m_n{x}_n}{m_1+m_2+m_3+\cdots +m_n}$

$y_{\text{cm}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+\cdots +m_n{y}_n}{m_1+m_2+m_3+\cdots +m_n}$

$z_{\text{cm}}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3+\cdots +m_n{z}_n}{m_1+m_2+m_3+\cdots +m_n}$

Note: If COM is at origin, ${\vec{r}}_{\text{cm}}=0$

$m_1{\vec{r}}_1+m_2{\vec{r}}_2+m_3{\vec{r}}_3+\cdots =0$

Centre of Mass of Two Particle System

${\vec{r}}_{\text{cm}}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2}{m_1+m_2}$ ,    $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ ,  $y_{cm}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

 If COM is at origin, ${\vec{r}}_{cm}=0$

$m_1{\vec{r}}_1+m_2{\vec{r}}_2=0\Rightarrow \frac{r_1}{r_2}=\frac{m_2}{m_1}$

$r=r_1+r_2$

$r_1=\frac{m_{2}r}{m_1+m_2}$, $r_2=\frac{m_{1}r}{m_1+m_2}$

4.0Centre of Mass of Continuous Mass Distribution

${\vec{r}}_{cm}=\frac{\int \vec{r}dm}{m}$          where $m=\int dm$

$x_{cm}=\frac{\int xdm}{\int dm}=\frac{\int xdm}{m}$  ,         $y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int ydm}{m}$,     $z_{cm}=\frac{\int zdm}{\int dm}=\frac{\int zdm}{m}$

5.0Types of Mass Densities

1. Linear Mass Density (For 1-D Objects :$\int dm=\int \lambda dx$

2. Areal Mass Density (For 2-D Objects): $\int dm=\int \sigma dA$

3. Volumetric mass density (For 3-D Objects): $\int dm=\int \rho dV$

6.0Uniform Symmetric Bodies

For symmetrical bodies having homogeneous distribution of mass, CM coincides with the centre of symmetry or the geometrical centre.

7.0Centre of Mass of Composite Bodies

Composite bodies are made up of two or more individual bodies. To find the center of mass, each component is treated as a particle with mass equal to the body’s mass, located at its center of mass. The center of mass of the entire body is then calculated using this approach.

$x_{CM}=\frac{m_d{x}_d+m_s{x}_s}{m_d+m_s}$             $y_{CM}=\frac{m_d{y}_d+m_s{y}_s}{m_d+m_s}$

$m_d$ and $m_s$ are masses of disc and square $x_d$ and $x_s$ are x-coordinates of centre of mass of disc and square respectively $y_d$ and $y_s$ are y-coordinates of centre of mass of disc and square respectively.

8.0Centre of Mass of Truncated Bodies

• To find the centre of mass of truncated bodies or bodies with cavities we can make use of superposition principle
• Restore the removed portion in its original position to get the complete body.
• Add the removed portion back to the truncated body, keeping its position fixed relative to the coordinate frame.
• Treat the remaining portion as if it's a new, intact body after the removal.

[Original mass (M) – mass of the removed part (m)] = {original mass (M)} + {– mass of the removed part (m)}

$x_{cm}=\frac{Mx-m{x}^{\prime }}{M-m}$,          $y_{cm}=\frac{My-m{y}^{\prime }}{M-m}$ ,        $z_{cm}=\frac{Mz-m{z}^{\prime }}{M-m}$

Where x', y' and z' represent the coordinates of the centre of mass of the removed part.

9.0Motion of Centre of Mass

Velocity of COM

For a system of particles, position of centre of mass is given by,

${\vec{R}}_{CM}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2+m_3{\vec{r}}_3+\cdots +m_n{\vec{r}}_n}{m_1+m_2+m_3+\cdots +m_n}$

$\frac{d({\vec{R}}_{CM})}{dt}=\frac{m_1\left(\frac{d{\vec{r}}_1}{dt}\right)+m_2\left(\frac{d{\vec{r}}_2}{dt}\right)+\dots }{m_1+m_2+m_3+\dots }$

${\vec{v}}_{CM}=\frac{d({\vec{R}}_{CM})}{dt}=\frac{m_1{\vec{v}}_1+m_2{\vec{v}}_2+m_3{\vec{v}}_3+\dots }{m_1+m_2+m_3+\dots }$

Acceleration of COM

For a system of particles,

Acceleration of centre of mass =${\vec{a}}_{CM}=\frac{d({\vec{v}}_{CM})}{dt}=\frac{m_1{\vec{a}}_1+m_2{\vec{a}}_2+m_3{\vec{a}}_3+\dots }{m_1+m_2+m_3+\dots }$

Linear Momentum

$M{\vec{v}}_{CM}=m_1{\vec{v}}_1+m_2{\vec{v}}_2+m_3{\vec{v}}_3+\cdots ={\vec{p}}_1+{\vec{p}}_2+{\vec{p}}_3+\dots [\because \vec{p}=m\vec{v}]$

$M{\vec{v}}_{CM}=\vec{p}\left[\sum {\vec{p}}_i={\vec{p}}_{CM}\right]$

Linear momentum of a system of particles is equal to the product of mass of the system with velocity of its centre of mass. From Newton's second law,${\vec{F}}_{net}=\frac{d(M{\vec{v}}_{CM})}{dt}$

10.0Effect of External Force on Centre of Mass

From Newton's second law, ${\vec{F}}_{ext}=\frac{d(M{\vec{v}}_{CM})}{dt}$

$M{\vec{a}}_{CM}={\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3+\cdots \Rightarrow {\vec{F}}_{net}=M{\vec{a}}_{CM}$

11.0Conservation of Linear Momentum of System

${\vec{F}}_{ext}=\frac{d(M{\vec{v}}_{CM})}{dt}=\frac{d({\vec{p}}_{system})}{dt}$

$If{\vec{F}}_{ext}=0,then{\vec{p}}_{system}=\text{constant}$

• This means that the total linear momentum of a system of particles remains constant over time if no external forces act on the system (i.e., the impulse of external forces is zero). The momentum of the system is unaffected by internal forces, and if the net external impulse during a given time interval is zero, the total momentum of the system is conserved during that period. ${\vec{p}}_{initial}={\vec{p}}_{final}$
• This is the principle of conservation of momentum. Since force, impulse, and momentum are vectors, the momentum component of a system in a specific direction is conserved if the net impulse of external forces in that direction is zero.

Example : 

1. Block-Bullet System

Conserving momentum of bullet and block, $mv+0=(M+m)V$

Velocity of block $V=\frac{mv}{M+m}$

By conservation of mechanical energy

$\frac{1}{2}(M+m)V^2=(M+m)gh\Rightarrow V=\sqrt{2gh}$

Maximum height gained by block $h=\frac{V^2}{2g}=\frac{m^2{v}^2}{2g(M+m)}$

2. If bullet emerges out of the block

Conserving momentum

$mv+0=m{v}_1+M{v}_2$

$m(v-v_1)=M{v}_2$

Conserving energy

$\frac{1}{2}M{v}_2^2=Mgh\Rightarrow v_2=\sqrt{2gh}$

From above equations we get

$m(v-v_1)=M\sqrt{2gh}\Rightarrow h=\frac{m^2(v-v_1{)}^2}{2g{M}^2}$

12.0Sample Questions on Centre of Mass

Q-1. A body of mass 3m at rest explodes into three identical pieces. Two of the pieces move with speed v in perpendicular directions. What is the total kinetic energy released in the explosion?

Solution:  

$m{v}^{\prime }=\sqrt{2}mv\Rightarrow v^{\prime }=\sqrt{2}v$

$K.E.=(\frac{1}{2}m{v}^2)\times 2+\frac{1}{2}m\times (\sqrt{2}v{)}^2=2m{v}^2$

Q-2. A sphere of mass m is rolling down on a wedge of mass M. When sphere reaches at lowest point then find distance moved by wedge in horizontal direction.

Solution: Let distance moved by wedge is 'x'

$m(R-x)-Mx=0$

$\Rightarrow mR-mx-Mx=0\Rightarrow mR-x(M+m)=0\Rightarrow x=\frac{mR}{M+m}$

Q-3. For a given system of particles find out velocity of COM.

Solution:

${\vec{v}}_{cm}=\frac{1(10\hat{i})+2(10\hat{i}+10\hat{j})}{3}=\frac{10\hat{i}+20\hat{i}+20\hat{j}}{3}=\frac{30\hat{i}+20\hat{j}}{3}=10\hat{i}+\frac{20}{3}\hat{j}$

${\vec{v}}_{cm}=\left(10\hat{i}+\frac{20}{3}\hat{j}\right)\text{m/s}$

Q-4. A semi-circular disc of radius r is removed from a rectangular plate of side 'r' and as '2r' as shown in figure. Then find out the position of COM of the remaining part.

Solution:

$x_{cm}=\frac{\sigma \left(2r^2\left(\frac{\pi }{2}\right)-\sigma \left(\frac{\pi r^2}{2}\right)\left(\frac{4r}{3\pi }\right)\right)}{\sigma \left(2r^2\left(\frac{\pi }{2}\right)-\sigma \left(\frac{\pi r^2}{2}\right)\right)}=\frac{r-\frac{2r}{3}}{3-\frac{\pi }{2}}=\frac{2r}{3(4-\pi )}$

Q-5. Find position of COM?

Solution:

$x_{cm}=\frac{1(1)+2(2)+3(3)+\cdots +n(n)}{1+2+3+\cdots +n}$

$x_{cm}=\frac{1^2+2^2+3^2+\cdots +n^2}{1+2+3+\cdots +n}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$