Centre of mass formula list with solved examples and problems

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

Center 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 \[\vec{v}_{CM}=\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}\]

Motion of Center of mass examples


Few of the examples of motion of center of mass are
  • Binary stars :- Binary stars are two stars bound to each other by the gravitational force. Binary stars orbit around their common center of mass.
  • 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 Problems with solutions


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