Centre of Mass: Of Rigid body, Of Thin Rod & Centre of Gravity with Conditions of Equilibrium

Centre of Mass and Centre of Gravity are hypothetical points or imaginary points. The location where the distribution of mass is constant in all directions called the centre of mass. The location where weight is equally distributed in all directions is known as the centre of gravity. The body’s mass serves as the foundation for the centre of mass. The weight of the body determines the centre of gravity.

These terms are used to solve multiple problems in Physics and Mechanics. We will study the Centre of Mass for a system of particles. We will further study the conditions of equilibrium of a body. Let’s get started.

Centre of Mass

System of particles: A group of particles that undergoes a particular type of motion together is called the system of particles.

Centre of Mass: A particular point where the whole of the mass of the system of particles appears to be concentrated is called the centre of mass.

Check the Application of Thermodynamics article here.

Let us consider two-particle systems having masses m1 and m2 located on X-axis at x1 and x2 respectively.

\(\begin{align}
X= {{m_1x_1 + m_2x_2} \over {m_1+m_2}} \\ \text {If } {m_1 = m_2 = m,}\\
X={ {x_1+x_2} \over {2}}
\end{align}\)

If we have n particles having masses m1, m2, m3, …….…mn on a straight line then the Centre of mass is given by:

The Cartesian coordinate directions can be treated independently, as it is done for vectors. Now suppose we have three particles of masses m1, m2 and m3 in space at (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3). Then the Centre of the mass of the system is located at (X, Y, Z) which is given by:

\(\begin{align}
\text{X-coordinate (X)}= {{m_1x_1+m_2x_2+m_3x_3} \over{m_1+m_2+m_3}} \\ \text{Y-coordinate (Y)}= {{m_1y_1+m_2y_2+m_3y_3} \over{m_1+m_2+m_3}} \\
\text{Z-coordinate (Z)}= {{m_1z_1+m_2z_2+m_3z_3} \over{m_1+m_2+m_3}}
\end{align}\)

Centre of Mass of A Rigid Body

We can take the rigid body as a continuous distribution of mass and take a small element of mass dm at a distance of x, y and z then the Centre of the mass of the rigid body is given by:

\(\begin{align}
\text{X-coordinate (X)}= {{1} \over{M}} \int {x dm} \\ \text{Y-coordinate (Y)}= {{1} \over{M}} \int {y dm} \\ \text{Z-coordinate (Z)}= {{1} \over{M}} \int {z dm}
\end{align}\)

Get the Magnetic Effect of Electric Current in detail.

Centre of Mass of A Thin Rod

Let us take a thin rod of mass M and length L.

We will find the COM of the rod from one end (let’s say left end).

Take a small element of mass dm and length dx at a distance of x from the left end.

Here the rod is a uniform rod.

\(\begin{align}
\text{mass of element (dm)}= {M \over L}dx \\
\text{COM of rod}= {1 \over M} {\int_0^L xdm} = {1 \over M} {\int_0^L x {M \over L}dx} = {1 \over L} {\int_0^L xdx}= { {L^2-0^2} \over 2L}= {L \over 2} \\
\text{COM of uniform rod} = {L \over 2}
\end{align}\)

Centre of Mass of A Semi-Circular Ring

Let us take a semi-circular ring of mass M and radius R.

Here we will take the Centre of the mass of the ring from point O.

\(\begin{align}
\text{X-coordinate (X)}=0 \\
\text{Y-coordinate (Y)}= {2R \over {\pi}}
\end{align}\)

1. There are three particles in a space of masses 2 kg, 3 kg and 5 kg located at (1, 2, 3), (1, 3, 2) and (2, 3, 1). Find the Centre of the mass of the system.

Given that: m1 = 2 kg, m2 = 3 kg and m3 = 5 kg

(x1, y1, z1) = (1, 2, 3), (x2, y2, z2) = (1, 3, 2) and (x3, y3, z3) = (2, 3, 1)

\(\begin{align}
\text{X-coordinate (X)}= {{m_1x_1+m_2x_2+m_3x_3} \over {m_1+m_2+m_3}} = {{2 \times 1 + 3\times 1 +5 \times 2} \over {2+3+5}} = {15 \over 10} =1.5 \\
\text{Y-coordinate (Y)}= {{m_1y_1+m_2y_2+m_3y_3} \over {m_1+m_2+m_3}} = {{2 \times 2 + 3\times 3 +5 \times 3} \over {2+3+5}} = {28 \over 10} =2.8 \\
\text{Z-coordinate (Z)}= {{m_1z_1+m_2z_2+m_3z_3} \over {m_1+m_2+m_3}} = {{2 \times 3 + 3\times 2 +5 \times 1} \over {2+3+5}} = {17 \over 10} =1.7 \\
\end{align}\)

Thus, the Centre of the mass of the system will be at (1.5, 2.8, 1.7)

2. Find the Centre of the mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100g, 150g, and 200g respectively. Each side of the equilateral triangle is 0.5m long.

Ans. Given that: m1 = 100 g, m2 = 150 g and m3 = 200 g

(x1, y1) = (0, 0), (x2, y2) = (0.5, 0) and (x3, y3) = (0.25, 0.25√3)

\(\begin{align}
X= {{m_1x_1+m_2x_2+m_3x_3} \over {m_1+m_2+m_3}} = {{100 \times 0 + 150\times 0.5 +200\times 0.25} \over {2+3+5}} = {125 \over 450} ={5 \over 18} \\
Y= {{m_1y_1+m_2y_2+m_3y_3} \over {m_1+m_2+m_3}} = {{100 \times 0 + 150\times 0 +200\times 0.25 \sqrt3} \over {2+3+5}} = {50 \sqrt3 \over 450} = {1 \over {3 \sqrt3}}
\end{align}\)

Motion of Centre of Mass

Now let’s understand how the centre of mass affects the motion of the particles.

Net force on a System of Particles

The total mass of a system of particles times the acceleration of its Centre of mass is the vector sum of all the forces acting on the system of particles.

Total forces on the system (Fnet) = total mass of the system (Mtotal) × acceleration of Centre of mass (aCOM)

Linear Momentum of System of Particles

Linear momentum: The product of mass and velocity is called linear momentum.

The total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its Centre of mass.

Total momentum (P) = total mass of the system (M) × velocity of Centre of mass (VCOM)

Centre of Gravity

 

If a body of mass M is made up a large number of tiny particles of mass m1,m2,m3…mnthen each particle will have its separate gravitation force acting on it which will add up to the weight of the entire body. All these particles are at different distances from the earth’s centre when they experience gravity.The resultant of these parallel forces gives us the weight of the body.

W= m1g1+m2g2+m3g3…..+mngn= Mg

The point at which the resultant weight of the body lies is called the Centre of Gravity.

Since, the body is small as compared to the size of the earth, the acceleration due to gravity can be taken as constant i.e. g

The centre of gravity of a body lies at its geometric centre. It is not necessarily inside the body. For example, the centre of gravity of a circular hoop will lie at its centre.

For bodies of small size, C.G. and C.M coincide. However, in case of big objects like mountains, C.G. lies lower than C.M.

Conditions of Equilibrium of a Rigid Body

A rigid body is said to be in equilibrium when it is in:

1. Translational Equilibrium
2. Rotational Equilibrium

A body is said to be in translational equilibrium if the net force acting in all the directions X, Y and Z is zero. i.e. Fx=0, Fy=0, Fz=0

A body is said to be in rotational equilibrium if the net force acting in all the directions X, Y and Z is zero. i.e. R=0

Note: Equilibrium doesn’t mean absence of force. It means that the resultant force is zero i.e. existing forces nullify each other.

Hope this article helped you understand the important concept of Centre of Mass and Centre of Gravity. Now practice it through problems on our free Testbook App.

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