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Centre of mass

The centre of mass of a body or system is that point, where entire mass is assumed to be contracted for the purpose of explanation of motion the body or system.

Let a body or system of mass \(M\) is placed in a certain reference frame and it is made up of \(n\) number of particles as shown in Fig. (2.1). Let the mass of \(i^{\text{th}}\) particle is \(m_i\) and its position is \(\vec{r}_i\).

Now let us consider a point \(C\) in the system whose position vector is \(\vec{R}_C\) and the position vector of \(i^{\text{th}}\) particle is \(\vec{r}_{Ci}\) with respect to point \(C\).

In \(\triangle OCP\),

\[ \vec{r}_i = \vec{R}_C + \vec{r}_{Ci} \tag{1} \]

Multiplying the mass \(m_i\) of \(i^{\text{th}}\) particle on both side of equation (1), then summing for all particles. We get

\[ \sum_{i=1}^{n} m_i \vec{r}_i = \sum_{i=1}^{n} m_i \vec{R}_C + \sum_{i=1}^{n} m_i \vec{r}_{Ci} \tag{2} \]

If for a body

\[ \sum_{i=1}^{n} m_i \vec{r}_{Ci}=0, \]

then point \(C\) is called centre of mass i.e. the point in the body or system about which the sum of the product of the mass of the particle and the position vector of that particle is zero, is called centre of mass. Thus

\[ \sum_{i=1}^{n} m_i \vec{r}_i = \sum_{i=1}^{n} m_i \vec{R}_C \] \[ \vec{R}_C = \frac{\sum_{i=1}^{n} m_i \vec{r}_i}{ \sum_{i=1}^{n} m_i} \]

Hence, the position vector of the centre of mass of a system of \(n\) particles is given by

\[ \vec{R}_C = \frac{ m_1\vec{r}_1+m_2\vec{r}_2+\cdots+m_n\vec{r}_n }{ m_1+m_2+\cdots+m_n } \tag{3} \]

where \(m_1, m_2, \ldots, m_n\) are the masses of the \(1^{\text{st}}\), \(2^{\text{nd}}\), ..., \(n^{\text{th}}\) particles, respectively, and \(\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n\) are their respective position vectors with respect to the origin \(O\).

if Total mass of system is M, then \[{ M=m_1+m_2+\cdots+m_n} \] thus \[ \vec{R}_C = \frac{ m_1\vec{r}_1+m_2\vec{r}_2+\cdots+m_n\vec{r}_n }{M} \] Diff. w. r. to t \[ \frac{d}{dt}\vec{R}_c = \frac{1}{M}\frac{d}{dt}(m_1\vec{r}_1+m_2\vec{r}_2+\cdots+m_n\vec{r}_n)\] hence, \[ \vec{V}_C = \frac{ m_1\vec{v}_1+m_2\vec{v}_2+\cdots+m_n\vec{v}_n }{M} \tag{4} \]

where \(v_1, v_2, \ldots, v_n\) are the velocities of the \(1^{\text{st}}\), \(2^{\text{nd}}\), ..., \(n^{\text{th}}\) particles, respectively, and \(\vec{{V}_c}\) is velocity of CM point.

as per defination of momentum \[ M\vec{V}_C =m_1\vec{v}_1+m_2\vec{v}_2+\cdots+m_n\vec{v}_n\] \[ \vec{P}_C =\vec{p}_1+\vec{p}_2+\cdots+\vec{p}_n \tag{5}\] Total momentum of the system is equal to the product of the total mass of the system and velocity of CM point.

again, Diff. w. to. t of equation (4)

\[ \frac{d}{dt}\vec{V}_c = \frac{1}{M}\frac{d}{dt}(m_1\vec{v}_1+m_2\vec{v}_2+\cdots+m_n\vec{v}_n)\] hence, \[ \vec{a}_C = \frac{ m_1\vec{a}_1+m_2\vec{a}_2+\cdots+m_n\vec{a}_n }{M} \] \[M\vec{a}_C =m_1\vec{a}_1+m_2\vec{a}_2+\cdots+m_n\vec{a}_n \] hence, \[\vec{F}_C =\vec{F}_1+\vec{F}_2+\cdots+\vec{F}_n \] Total Force acting on the system as total internal force acting among the particles.

Conservation of linear momentum

law of conservation of linear momentum states that the total linear momentum of a closed, isolated system remains constant if no net external force acts on it. in other words, "The total linear momentum of a closed system before a collision is always equal to the total linear momentum after the collision."