Centre of Mass

The center of mass (COM) is a theoretical point in an object or system at which the entire mass is assumed to be concentrated for analysis. It represents the average position of all the mass based on its distribution, and in simple terms, it is the point at which the whole mass of a body appears to act and where the object can be supported or balanced.

Example: A stick can be supported at its middle point so that it does not fall. Similarly, every object has a specific point called the center of mass where its entire mass can be considered to act, and its position is defined relative to the object or system being analyzed.

For uniform and symmetrical shapes, the center of mass usually coincides with the centroid. Example: In a uniform ring or disc, it lies at the geometric center.

The centers of mass of the rectangle, square, and triangular sheets are

Note: As for a ring, its center of mass lies inside the ring, which means the center of mass of a body must lie within the body itself. 

Formula

For uniform and symmetrical bodies, the center of mass lies at the centroid, but for irregular bodies, it depends on mass distribution. The center of mass of a complex object is found by taking the weighted average of the positions of its mass elements. If total mass is M and particles (mi) are at positions (ri), then:

M\vec{r}_{cm} = m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n

Thus, the formula for the location of the center of mass, rcm, is given as:

{r_{cm} =\frac{ m_1r_1 + m_2r_2 + ...+m_nr_n}{M}}

Where,

M = ∑m_i, which is the total mass of the body. 

The above technique uses vector arithmetic. To avoid vector arithmetic, we can find out the center of mass of the body along the x-axis and y-axis, respectively. Formulas for this case are given below: 

{x_{cm} = \frac{ m_1x_1 + m_2x_2 + ...}{M}}

{y_{cm} = \frac{ m_1y_1 + m_2y_2 + ...}{M}}

Centre of Mass of a Continuous Body

For objects with non-uniform structure, the body is divided into infinitely small mass elements (dm), and the center of mass is found by integrating over the entire body. If the total mass is m, then the coordinates of the center of mass are given by:

x_{cm} = \frac{\int x \, dm}{\int dm}

y_{cm} = \frac{\int y \, dm}{\int dm}

z_{cm} = \frac{\int z \, dm}{\int dm}

Centre of Mass and Gravity

Usually, gravity is assumed to be a uniform force acting on the body. In that case, the center of mass and center of gravity are almost the same thing, but as we change the force of gravity to be non-uniform, then the position of both points changes, and we have different centers of mass and centers of gravity for an object in a non-uniform gravitational field.

Rigid Body

While dealing with the center of mass, we came across various types of bodies that are made up of millions of individual particles. These bodies can change their shape or get deformed if force is applied to them. The bodies that do not get deformed by applying force are called rigid bodies. Thus, we can say that rigid bodies are those that, under stress, do not change their shape and whose center of mass remains in the same position.

In the case of any rigid body that has uniform density, its center of mass is generally at its centroid. Some examples of rigid bodies are solid spheres made of steel, solid cylinders, etc.

Centre of Mass for Various Objects

Various formulas for the center of Mass

Some systems occur more frequently in real life than others. While calculating the center of mass for such systems, the traditional method takes time. Certain center of mass formulas should be kept in mind while solving the questions related to the center of mass. These formulas help in simplifying the calculations.

Center of Mass of a System of Two Point Masses  

In such a system, COM lies closer to the heavier mass. 

m₁r₁ = m₂r₂

Distance of COM from mass m₁ = \frac{m_2r}{m_1 + m_2}

Distance of COM from mass m₂ = \frac{m_1r}{m_1 + m_2}

Centre of Mass of a Triangle

The center of mass of any triangle of height h is at a distance h/3 from its base. If we take a uniform triangle of height h, then its center of mass is at height h/3 from the base. The image added below shows the same.

The centre of mass of the triangle lies at the point, which is at a height of h/3

Centre of Mass of Triangle = h/3

where h is the height of the Triangle.

Centre of Mass of Semicircular Disk

The center of mass of any semi-circular disc of radius r is at a distance of 4r/3π from its base. If we take a uniform semicircular disc of radius r, then its center of mass is at a height of 4r/3π from the base. The image added below shows the same.

The center of mass of the semi-circular disc lies at the point, which is at a height of 4r/3π

Centre of Mass of Semi-Circular Disc = 4r/3π

where r is the radius of the semi-circular disc

Centre of Mass of Half Ring

The center of mass of any half ring of radius r is at a distance of 2r/π from its base. If we take a uniform semicircular disc of radius r, then its center of mass is at a height of 2r/π from the base. The image added below shows the same.

The center of mass of the half-ring lies at the point, which is at a height of 2r/π

Centre of Mass of half-ring = \frac{2r} {\pi}

where r is the radius of the half-ring.

Centre of Mass of a Solid Hemisphere

The center of mass of a solid hemisphere is located at the intersection of its axis of symmetry and the plane of its circular base. To find the coordinates of the center of mass of a solid hemisphere of radius R and uniform density, we can use the following formula:

x_{cm} = \frac{3R}{8\pi}

The y-coordinate and z-coordinate of the center of mass are both zero.

Centre of Mass of Solid Cone

The center of mass of a solid cone is located along its axis of symmetry, at a distance of 3/4 times the height of the cone from its base.

To find the coordinates of the center of mass of a solid cone of radius R, height H, and uniform density, we can use the following formula:

x_{cm} = \frac{3H}{4\pi}

The y-coordinate and z-coordinate of the center of mass are both zero.

Centre of Mass of Hollow Cone

The center of mass of a hollow cone is located along its axis of symmetry, at a distance of 1/4 times the height of the cone from its base.

To find the coordinates of the center of mass of a hollow cone of radius R, height H, and uniform density, we can use the following formula:

x_{cm} = \frac{H}{4} \left( 1 + \left( \frac{R}{r_0} \right)^2 \right)

System of Particles

As we have found the center of mass for rigid bodies, we also need to consider bodies that are not rigid and consist of infinitely many particles free to move individually. These particles interact through internal forces and may move differently from one another. However, there exists a specific point where the entire mass of the system can be assumed to be concentrated for analysis, and this point is called the center of mass of the body.

Derivation

The derivation of the center of Mass of the System of Particles is explained below:

Suppose we have n particles of masses,

m₁, m₂, m₃,..., m_n. And m₁, m₂, m₃, ....., m_n = M

Let their position vector be, \vec{r_1},~\vec{r_2},~\vec{r_3},~.....~\vec{r_n}

Then the centre of mass of this system of particles with respect to the origin is,

\vec{r_{cm}}~=~\frac{m_1\vec{r_1}~+~m_2\vec{r_2}~+~m_3\vec{r_3}~+.....~+m_n\vec{r_n}}{m_1~+~m_2~+~m_n~+....+~m_n}

We know that,

m₁, m₂, m₃, …, m_n = M

\vec{r_{cm}}~=~\frac{m_1\vec{r_1}~+~m_2\vec{r_2}~+~m_3\vec{r_3}~+.....~+m_n\vec{r_n}}{m_1~+~m_2~+~m_n~+....+~m_n}

Two-particle system

For two-particle system we take two particles of mass m₁ and m₂ and their position vectors \vec{r_1}~and~\vec{r_2}.\vec{r_1}~and~\vec{r_2}

Then the centre of mass of this system of particles with respect to the origin is,

\vec{r_{cm}}~=~\frac{m_1\vec{r_1}~+~m_2\vec{r_2}}{m_1~+~m_2}

In the Cartesian coordinate system,

X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}

Centre of Mass of a Body with a Cavity

Suppose we take some part of the body out of the whole body; then its center of mass gets affected. Now the centre of mass of the body is calculated as,

Mass of Body with Cavity = Original Mass of Body(M) + {-Mass of Removed Part(m)}

Suppose the mass of the body is M and the mass of the cavity is m, the centre of mass of the original body is at (x, y, z), and the centre of mass of the cavity is at (x₁, y₁, z1); then the centre of mass of the object with the cavity is,

X_{cm} = \frac{M x - m x_1}{M - m}

Y_{cm} = \frac{M y - m y_1}{M - m}

Z_{cm} = \frac{M z - m z_1}{M - m}

Applications

• Stability: Structures like towers and buildings are designed with their center of mass kept low to ensure stability.
• Motion: In sports and performing arts, shifting the center of mass helps control balance and movement.
• Engineering: Engineers use the concept to design stable machines and structures such as automobiles, cranes, and bridges.
• Astronomy: The center of mass plays an important role in studying the motion of celestial bodies.

Solved Problems

Question 1: Two point masses, m₁ = 5 Kg and m₂ = 2 Kg, are located at x = 2 m and x = 6 m, respectively. Find the center of mass. 

Solution:  Formula for the Centre of mass is given by, 

x_{cm} =\frac{ m_1x_1 + m_2x_2 + ...}{M}

 m₁ = 5Kg, m₂ = 2Kg and x = 2 m and x = 6 m. 

M = m₁ + m₂

⇒ M = 5 + 2 = 7

x_cm = \frac{ m_1x_1 + m_2x_2 + ...}{M}

⇒ x_cm = \frac{ m_1x_1 + m_2x_2}{M}

⇒ x_cm = \frac{ (5)(2) + (2)(6)}{7}

⇒ x_cm = \frac{22}{7}

Question 2: Two point masses, m₁ = 5 Kg and m₂ = 2 Kg, are located at y = 10 m and y = -5 m, respectively. Find the center of mass. 

Solution:  Formula for the Centre of mass is given by, 

y_{cm} = \frac{ m_1y_1 + m_2y_2 + ...}{M}

 m₁ = 5 kg, m₂ = 2 kg and y = 10 m and y = -5 m. 

M = m_{1 +} m₂

⇒ M = 5 + 2 = 7

y_cm = \frac{ m_1y_1 + m_2y_2 + ...}{M}

⇒ y_cm = \frac{ m_1y_1 + m_2y_2}{M}

⇒ y_cm = \frac{ (5)(10) + (2)(-5)}{7}

⇒ y_cm = \frac{40}{7}

Question 3: Two point masses, m₁ = 1 Kg and m₂ = 2 Kg, are located at vector a = 6i + 4j and vector b = -5i + 2j, respectively. Find the center of mass. 

Solution:  Formula for the Centre of mass in the vector notation is given by, 

r_{cm} = \frac{ m_1\vec{r_1} + m_2\vec{r_2} + ...}{M}

 m₁ = 1Kg, m₂ = 2Kg and a = 6i + 4j, b = -5i + 2j

M = m_{1 +} m₂

⇒ M = 1 + 2 = 3

r_cm = \frac{ m_1\vec{r_1} + m_2\vec{r_2} + ...}{M}

⇒r_cm = \frac{ m_1\vec{a} + m_2\vec{b}}{M}

⇒ r_cm = \frac{ (1)(6\hat{i} + 4\hat{j} ) + (2)(-5\hat{i} + 2\hat{j})}{3}

⇒ r_cm = \frac{ 6\hat{i} + 4\hat{j} + -10\hat{i} + 4\hat{j})}{3}

⇒ r_cm = \frac{ -4\hat{i} + 8\hat{j} }{3}

Problem 4: Two point masses, m₁ = 4 Kg and m₂ = 2 Kg, are located at vector a = i + j and vector b = -i + j, respectively. Find the center of mass. 

Solution:  Formula for the Centre of mass in the vector notation is given by, 

 r_{cm} = \frac{ m_1\vec{r_1} + m_2\vec{r_2} + ...}{M}

 m₁ = 4Kg, m₂ = 2Kg and a = i + j, b = -i + j

M = m_{1 +} m₂

⇒ M = 4 + 2 = 6

r_cm = \frac{ m_1\vec{r_1} + m_2\vec{r_2} + ...}{M}

⇒ r_cm = \frac{ m_1\vec{a} + m_2\vec{b}}{M}

⇒ r_cm = \frac{ (4)(\hat{i} + \hat{j} ) + (2)(\hat{i} -\hat{j})}{6}

⇒ r_cm = \frac{ 4\hat{i} + 4\hat{j} + 2\hat{i} -2\hat{j})}{6}

⇒ r_cm = \frac{ 6\hat{i} + 2\hat{j} }{6}

Problem 5: A disk of radius R/2 is removed from a bigger disk of mass M of radius R. Find the center of mass. 

Solution:  Since the density of the disk is uniform, the weight is uniformly distributed over the area. 

Mass "m" of the removed disk = \frac{M (\pi (\frac{R}{2})^2)}{\pi R^2} = \frac{M}{4}

The figure shows the center of mass of the remaining portion and the removed portion. Notice that if both of these are taken together, the center of mass should lie at the Centre. Let the distance of the center of mass of the remaining portion be “x”. 

0 = \frac{x\frac{3M}{4} + \frac{R}{2}\frac{M}{4}}{\frac{3M}{4} + \frac{M}{4}} \\ \Rightarrow 0 = x\frac{3M}{4} + \frac{RM}{8} \\ \Rightarrow -\frac{RM}{8} = x\frac{3M}{4} \\ \Rightarrow x = \frac{-R}{6}

Unsolved Problems

Question 1: Three point masses, 2 kg, 3 kg, and 5 kg, are placed at x = 0 m, 4 m, and 10 m respectively; find the x-coordinate of the center of mass.

Question 2: Two particles of masses 4 kg and 6 kg are located at position vectors (2\hat i + 3\hat j) and (-1\hat i + 5\hat j). Find the center of mass.

Question 3: A uniform rod of length 8 m and mass 4 kg has a 2 kg mass attached at one end; find the position of the center of mass from the left end.

Question 4: A circular disc of radius R and mass M has a smaller disc of radius R/3 removed, whose center is at a distance R/2 from the big disc’s center. Find the shift in the center of mass.

Question 5: Four equal masses m are placed at the corners of a square of side a, and one mass is removed. Find the new center of mass.