Centre of mass formulas list with solved examples and problems

This article contains a list of the center of mass formulas along with some centre of mass problems and examples.

Centre of mass formula list

  1. For a system of N particles,the position vector of center of mass is \[\vec{R}_{CM}=\frac{m_1\vec{r}_1+m_2\vec{r}_2+………..+m_N\vec{r}_N}{m_1+m_2+…..+m_N}=\frac{\sum_{i=1}^N{m_i\vec{r}_i}}{M}\]
  2. The position vector of the center of mass of a two-particle system is \[\vec{R}_{CM}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}\]
  3. The Cartesian coordinates of the center of mass are given by \[X_{CM}=\frac{m_1x_1+m_1x_2+………..+m_Nx_N}{m_1+m_2+…..+m_N}=\frac{\sum_{i=1}^N{m_ix_i}}{M}\] \[Y_{CM}=\frac{m_1y_1+m_1y_2+………..+m_Ny_N}{m_1+m_2+…..+m_N}=\frac{\sum_{i=1}^N{m_iy_i}}{M}\] \[Z_{CM}=\frac{m_1z_1+m_1z_2+………..+m_Nz_N}{m_1+m_2+…..+m_N}=\frac{\sum_{i=1}^N{m_iz_i}}{M}\]
  4. For Continuous mass distribution \[\vec{R}_{CM}=\frac{1}{M}\int{\vec{r}dm}\]
    where dm is the mass element located at position \vec{r}. Also \[X_{CM}=\frac{1}{M}\int{xdm}\] \[Y_{CM}=\frac{1}{M}\int{ydm}\] \[Z_{CM}=\frac{1}{M}\int{zdm}\]
  5. Velocity of center of mass of a two particle system is \[\vec{v}_{CM}=\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}\]

The motion of the Center of mass examples

A few examples of motion of the center of mass based on the list of the center of mass formulas.

  • Binary stars:- Binary stars are two stars bound to each other by gravitational force. Binary stars orbit around their common center of mass.
  • The Earth-moon system is one such system center of mass of the Earth-Moon system moves around the Sun in an elliptical path, not the earth and moon themselves.

center of mass formulas

Problems based on Center of mass formulas with solutions

Problem 1 
Determine the center of mass of a hemisphere of radius \(R\) and uniform density \(\rho\).
We will divide the given hemisphere into co-axial discs and then find the relation for the center of mass of this disc using the center of mass formula for a continuous distribution of mass. We will integrate this relation for the limits \(0\) to \(R\) for the x-axis. From symmetry  \(Y_{CM}\) and \(Z_{CM}\) would be 0.
The mass of the strip can be found using the fact that density is mass per unit volume i.e. \(\rho=m/V\)
Let \(\rho\) be the density of the material of the hemisphere. Take \(O\) as the origin. The hemisphere can be assumed to be made up of a large number of co-axial discs as shown below in the figure.

See also  Effects of Friction in our daily life
center of mass formulas

Consider one such elementary disc of radius \(y\) and thickness \(dx\) at a distance \(x\) from the origin.
\(\text{Mass of the elementary disc}= volume \times density\) \[dm=\pi y^2dx\times \rho =\pi \left( R^2-x^2 \right) dx\rho\] The coordinates of the center of mass of the hemisphere can be determined as follows \[X_{CM}=\frac{1}{M}\int{xdm}\] \[=\frac{1}{M}\int_0^R{x\pi \left( R^2-x^2 \right) \rho dx}\] \[=\frac{\pi \rho}{M}\int_0^R{\pi \left( xR^2-x^3 \right) \rho dx}\] \[=\frac{\pi \rho}{M}\left[ R^2\frac{x^2}{2}-\frac{x^4}{4} \right] _{0}^{R}\] Solving and simplifying it we get
\[X_{CM}=\frac{\pi \rho}{M}\left[ \frac{R^4}{4} \right] \] Since, \[M=\frac{2}{3}\pi R^3\times \rho\] for this hemisphere, thus we have \[X_{CM}=\frac{\pi \rho}{\frac{2}{3}\pi R^3\rho}\left[ \frac{R^4}{4} \right] \] Simplifying it we get \[X_{CM}=\frac{3}{8}R\] \(Y_{CM}=0\) and \(Z_{CM}=0\) because of symmetry.
Hence the coordinates of the center of mass of the hemisphere are \[\left( \frac{3}{8}R,\text{0,}0 \right)\]

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