Center of Mass and Linear Momentum of Systems

Centre Of Mass

When dealing with systems of multiple particles or extended bodies, it is often convenient to describe the motion of the entire system using a single point that represents the average position of the system's mass. This point is called the centre of mass (CM).

The centre of mass is a hypothetical point where, for the purpose of describing overall motion, the entire mass of the system can be imagined to be concentrated. When an external force acts on a system, the centre of mass moves as if the entire mass of the system were concentrated at that point and the external force were applied directly to it.

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Definition for a System of Discrete Particles

For a system consisting of $n$ discrete particles with masses $m_1, m_2, ..., m_n$ located at position vectors $\vec{r}_1, \vec{r}_2, ..., \vec{r}_n$ respectively, the position vector of the centre of mass ($\vec{R}$) is given by:

$ \vec{R} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ... + m_n\vec{r}_n}{m_1 + m_2 + ... + m_n} $

This can be written using summation notation:

$ \vec{R} = \frac{\sum_{i=1}^{n} m_i\vec{r}_i}{\sum_{i=1}^{n} m_i} $

Let the total mass of the system be $M = \sum_{i=1}^{n} m_i$. Then,

$ \vec{R} = \frac{1}{M} \sum_{i=1}^{n} m_i\vec{r}_i $

In terms of Cartesian coordinates $(x, y, z)$, the coordinates of the centre of mass $(X, Y, Z)$ are:

$ X = \frac{\sum_{i=1}^{n} m_ix_i}{\sum_{i=1}^{n} m_i} \quad , \quad Y = \frac{\sum_{i=1}^{n} m_iy_i}{\sum_{i=1}^{n} m_i} \quad , \quad Z = \frac{\sum_{i=1}^{n} m_iz_i}{\sum_{i=1}^{n} m_i} $

The position vector of the $i$-th particle is $\vec{r}_i = x_i\hat{i} + y_i\hat{j} + z_i\hat{k}$, and the position vector of the centre of mass is $\vec{R} = X\hat{i} + Y\hat{j} + Z\hat{k}$.

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Definition for a Continuous Mass Distribution (Rigid Body)

For an extended object or a rigid body, which can be considered as a continuous distribution of mass, the summation is replaced by integration. If $dm$ is an infinitesimal mass element located at position vector $\vec{r}$ within the body, and $M$ is the total mass of the body, the position vector of the centre of mass is:

$ \vec{R} = \frac{1}{M} \int \vec{r} \, dm $

The integral is taken over the entire volume or extent of the body. In terms of coordinates:

$ X = \frac{1}{M} \int x \, dm \quad , \quad Y = \frac{1}{M} \int y \, dm \quad , \quad Z = \frac{1}{M} \int z \, dm $

For a body of uniform density ($\rho = \text{constant}$), $dm = \rho \, dV$, where $dV$ is an infinitesimal volume element. The total mass is $M = \int \rho \, dV = \rho \int dV = \rho V$. The position of the CM is then:

$ \vec{R} = \frac{1}{\rho V} \int \vec{r} (\rho \, dV) = \frac{1}{V} \int \vec{r} \, dV $

For bodies with uniform density and high symmetry (like a uniform sphere, cylinder, rectangle, etc.), the centre of mass is located at the geometric centre of the body.

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Location of the Centre of Mass

The centre of mass is a point defined mathematically; it does not necessarily have to lie within the physical boundaries of the object, especially for objects with holes or unusual shapes (e.g., the centre of mass of a ring is at its centre, which is an empty space).

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Motion Of Centre Of Mass

The motion of the centre of mass of a system is particularly significant because it simplifies the analysis of the overall motion of the system, regardless of the complex interactions between its constituent particles or parts.

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Velocity of the Centre of Mass

The velocity of the centre of mass ($\vec{V}$) can be found by taking the time derivative of the position vector of the centre of mass $\vec{R}$ with respect to time:

$ \vec{V} = \frac{d\vec{R}}{dt} = \frac{d}{dt} \left( \frac{\sum_{i=1}^{n} m_i\vec{r}_i}{M} \right) $

Assuming the masses $m_i$ are constant, and $M$ is constant:

$ \vec{V} = \frac{1}{M} \sum_{i=1}^{n} m_i \frac{d\vec{r}_i}{dt} $

Since $\vec{v}_i = \frac{d\vec{r}_i}{dt}$ is the velocity of the $i$-th particle:

$ \vec{V} = \frac{\sum_{i=1}^{n} m_i\vec{v}_i}{M} $

This shows that the velocity of the centre of mass is the weighted average of the velocities of the individual particles, weighted by their masses.

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Acceleration of the Centre of Mass

Similarly, the acceleration of the centre of mass ($\vec{A}$) is the time derivative of the velocity of the centre of mass:

$ \vec{A} = \frac{d\vec{V}}{dt} = \frac{d}{dt} \left( \frac{\sum_{i=1}^{n} m_i\vec{v}_i}{M} \right) $

$ \vec{A} = \frac{1}{M} \sum_{i=1}^{n} m_i \frac{d\vec{v}_i}{dt} $

Since $\vec{a}_i = \frac{d\vec{v}_i}{dt}$ is the acceleration of the $i$-th particle:

$ \vec{A} = \frac{\sum_{i=1}^{n} m_i\vec{a}_i}{M} $

Now, consider the forces acting on each particle. The acceleration $\vec{a}_i$ of the $i$-th particle is caused by the net force acting on it, $\vec{f}_i$. This force $\vec{f}_i$ can be divided into two types:

• Internal forces ($ \vec{f}_{ij} $): Forces exerted on particle $i$ by other particles $j$ within the system. By Newton's Third Law, $\vec{f}_{ij} = -\vec{f}_{ji}$, and the sum of all internal forces in the system is zero: $ \sum_{i \ne j} \vec{f}_{ij} = 0 $.
• External forces ($ \vec{F}_{i}^{ext} $): Forces exerted on particle $i$ by agents outside the system. The total external force on the system is $\vec{F}_{ext} = \sum_i \vec{F}_{i}^{ext}$.

The net force on particle $i$ is $\vec{f}_i = \vec{F}_{i}^{ext} + \sum_{j \ne i} \vec{f}_{ij}$. By Newton's Second Law for particle $i$, $\vec{f}_i = m_i \vec{a}_i$.

$ m_i \vec{a}_i = \vec{F}_{i}^{ext} + \sum_{j \ne i} \vec{f}_{ij} $

Summing over all particles in the system:

$ \sum_{i=1}^{n} m_i \vec{a}_i = \sum_{i=1}^{n} \vec{F}_{i}^{ext} + \sum_{i=1}^{n} \sum_{j \ne i} \vec{f}_{ij} $

The first term on the right side is the total external force on the system, $\vec{F}_{ext} = \sum_i \vec{F}_{i}^{ext}$. The second term is the sum of all internal forces, which is zero according to Newton's Third Law.

$ \sum_{i=1}^{n} m_i \vec{a}_i = \vec{F}_{ext} + 0 = \vec{F}_{ext} $

Substitute $\sum_{i=1}^{n} m_i\vec{a}_i = M\vec{A}$ from the definition of the acceleration of the centre of mass:

$ M \vec{A} = \vec{F}_{ext} $

This is a profound result. It states that the total external force acting on a system is equal to the total mass of the system multiplied by the acceleration of its centre of mass. This is equivalent to Newton's Second Law applied to a single particle of mass $M$ located at the position of the centre of mass and acted upon by the total external force $\vec{F}_{ext}$.

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Significance of the Motion of CM

• The centre of mass moves as if the entire mass of the system were concentrated at that point and the total external force were applied there.
• Internal forces within the system do not affect the motion of the centre of mass. For example, the explosion of a projectile in mid-air might send its fragments in different directions, but if air resistance is negligible (so the only external force is gravity), the centre of mass of the fragments will continue to follow the same parabolic trajectory that the original projectile would have followed.
• If the total external force on a system is zero, the acceleration of the centre of mass is zero ($\vec{A} = 0$). This means the velocity of the centre of mass is constant ($\vec{V} = \text{constant}$). This is the principle of conservation of linear momentum for a system, discussed in the next section.

Analysing the motion of the centre of mass allows us to separate the overall translational motion of a system from its internal motions (like rotation, vibration, or the relative motion of its parts). For example, the motion of a complex object like a boomerang or a human diver can be understood by tracking the parabolic path of their centre of mass, even while the object or body is rotating or changing shape relative to the CM.

Linear Momentum Of A System Of Particles

The concept of linear momentum, which for a single particle is defined as the product of its mass and velocity ($\vec{p} = m\vec{v}$), can be extended to a system of multiple particles or an extended body.

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Definition of Total Linear Momentum

The total linear momentum ($\vec{P}$) of a system of $n$ particles is defined as the vector sum of the linear momenta of all individual particles in the system.

If the $i$-th particle has mass $m_i$ and velocity $\vec{v}_i$, its linear momentum is $\vec{p}_i = m_i\vec{v}_i$. The total linear momentum of the system is:

$ \vec{P} = \vec{p}_1 + \vec{p}_2 + ... + \vec{p}_n = \sum_{i=1}^{n} \vec{p}_i = \sum_{i=1}^{n} m_i\vec{v}_i $

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Relation to the Velocity of the Centre of Mass

We know that the velocity of the centre of mass is $\vec{V} = \frac{\sum_{i=1}^{n} m_i\vec{v}_i}{M}$, where $M = \sum_{i=1}^{n} m_i$ is the total mass.

From this definition, we can write:

$ M \vec{V} = \sum_{i=1}^{n} m_i\vec{v}_i $

Comparing this with the definition of total linear momentum $\vec{P} = \sum_{i=1}^{n} m_i\vec{v}_i$, we see that:

$ \vec{P} = M \vec{V} $

This is another important result: the total linear momentum of a system of particles is equal to the total mass of the system multiplied by the velocity of its centre of mass.

This means that, for the purpose of analysing the total linear momentum, the entire system can be treated as a single particle of mass $M$ located at the centre of mass and moving with the velocity of the centre of mass.

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Newton's Second Law for a System of Particles

We previously derived that $M \vec{A} = \vec{F}_{ext}$, where $\vec{A}$ is the acceleration of the centre of mass and $\vec{F}_{ext}$ is the total external force on the system.

Since $\vec{A} = \frac{d\vec{V}}{dt}$, we can write:

$ M \frac{d\vec{V}}{dt} = \vec{F}_{ext} $

We know that $\vec{P} = M\vec{V}$. Assuming the total mass $M$ is constant (no mass is added or removed from the system), we have:

$ \frac{d\vec{P}}{dt} = \frac{d(M\vec{V})}{dt} = M \frac{d\vec{V}}{dt} = M\vec{A} $

Therefore, Newton's Second Law for a system of particles can be stated as:

$ \frac{d\vec{P}}{dt} = \vec{F}_{ext} $

The rate of change of the total linear momentum of a system is equal to the total external force acting on the system.

Internal forces do not affect the total linear momentum of the system, only the external forces do.

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Law of Conservation of Linear Momentum

A particularly significant consequence of this is the Law of Conservation of Linear Momentum.

If the total external force acting on a system is zero ($\vec{F}_{ext} = 0$), then the rate of change of the total linear momentum is zero ($d\vec{P}/dt = 0$). This implies that the total linear momentum of the system remains constant.

Statement of Conservation of Linear Momentum: If the net external force acting on a system is zero, the total linear momentum of the system remains constant (conserved).

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

This means $\vec{P}_{initial} = \vec{P}_{final}$ for an isolated system (where $\vec{F}_{ext} = 0$).

$ \sum_{i=1}^{n} m_i\vec{v}_{i, initial} = \sum_{i=1}^{n} m_i\vec{v}_{i, final} $

The conservation of linear momentum is a powerful principle that applies to collisions, explosions, and other interactions within isolated systems. It is a fundamental conservation law in physics, arising from the translational symmetry of space (i.e., the laws of physics are the same everywhere in space).

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