Center of Mass: Have you ever wondered why does the block whose height is more topples more easily? Have you ever heard of stone balancing or rock balancing? What is the principle behind it? There are numerous examples in our daily life where we knowingly or unknowingly use the concept of the Center of Mass. So what is the centre of mass? The Centre of mass is a very important concept in Physics. This concept lets us analyse rigid bodies that are not point sized easily and more efficiently.

System of Particles

System of particles refers to a collection of two or more particles that may or may not interact with each other. Any interactive forces between the system particles are known as internal forces, and any force which is applied by an agent outside the system of particles is known as external forces. The individual particles of the system may themselves be rigid bodies.

It is to be noted that particles are considered to have zero volume which means they have zero length, zero breath and zero height; therefore, they are point sized.

Rigid Body

In practical scenarios, most of the time, we do not deal with particles. Instead, we deal with continuous bodies which have some volume and occupy space. A Rigid body can be considered to be a collection of infinite numbers of particles. Rigid bodies are non-deformable. Thus, the distance between any two constituent particles always remains the same. That is if we mark any two points on the rigid body, then regardless of the orientation of the rigid body, the separation between the two points will not change.

Example: A block of iron.

What is Center of Mass in Physics?

If we throw a ball in the air at some angle without any spin, then we know that the ball will follow the parabolic path, but the motion of a body that has a different shape will appear different for example, if we throw a hammer, the hammer will also have some rotation and the orientation of the hammer at different instances will be different.

In the above example, if we analyse, we will find a point that follows the parabolic path. This point is the centre of mass of the hammer.

So, in general, the centre of mass it’s a point at which the whole mass of the rigid body or the system of particles can be assumed to be concentrated. Using this concept, we can analyse the system of particle or rigid body more easily and efficiently.

Center of Mass for System of Particles

If we consider a system which two particles then the center of mass will lie on the line joining the two particles.

The centre of mass will be given by,

\(\overrightarrow {{r_{{\rm{cm}}}}} = \frac{{{m_1}{{\vec r}_1} + {m_2}{{\vec r}_2}}}{{{m_1} + {m_2}}}\)

If we consider the first particle to be at origin and the second particle, add some distance

Then the centre of mass will be,

\(\overrightarrow {{r_{{\rm{cm}}}}} = \frac{{{m_2}\left( {{{\vec r}_2} – {{\vec r}_1}} \right)}}{{{m_1} + {m_2}}}\)

If the two particles are off equal mass, then the centre of mass will lie on the midpoint off the line joining the two particles.

\(\overrightarrow {{r_{{\rm{cm}}}}} = \frac{{\left( {\overrightarrow {{r_2}} – {{\vec r}_1}} \right)}}{2}\)

For a system comprising off n number of particles,

the centre of mass is given by,

\(\overrightarrow {{r_{{\text{cm}}}}}  = \frac{{\mathop \sum \nolimits_{i = 1}^n {m_i}\overrightarrow {{r_i}} }}{{\mathop {\mathop {\sum {m_i}}\limits_{i = 1} }\limits^n }}\)

Where,

\(\overrightarrow {{r_{{\rm{cm}}}}} \) is the position of centre of mass.

\({\vec r_i}\) is the position of the particle.

\({m_i}\) is the mass of the particle.

The velocity of the centre of mass is given by,

\(\overrightarrow {{v_{{\rm{cm}}}}} = \frac{{{m_1}\overrightarrow {{v_1}} + {m_2}\overrightarrow {{v_2}} }}{{{m_1} + {m_2}}}\)

where, \(\overrightarrow {{v_1}} \) and \(\overrightarrow {{v_2}} \) are velocities of particles.

Center of Mass for Rigid Bodies

Since the rigid bodies are a continuous volume of mass, we use integration instead of summation to find the centre of mass.

Centre of mass of a rigid body is given by,

\(\overrightarrow {{r_{{\rm{cm}}}}} = \frac{{\int {\vec r} dm}}{{\int d m}}\)

Where,

\(dm\) is the differential mass element.

\(\vec r\) is the position of the differential mass element.

Center of Mass of Some Common Figures

Semi-circular Ring

Semi-circular Disc

Hollow Cone

Solid Cone

Hollow Hemisphere

Solid Hemisphere

Center of Mass and Linear Momentum

The Centre of mass finds an important application in problems related to linear momentum. If there is no external force acting on the system, then the velocity off the centre of mass remains constant even if the velocity of the individual particles of the system changes.

Example: If the body moving in a projectile motion explodes mid-air and breaks into a number of particles, each particle with its own velocity. In this situation, since there is no external force, thus the motion of the centre of mass remains unchanged and follows a parabolic path.

For a rigid body to determine the linear momentum, any external force can be assumed to be acting on the centre of mass.

Center of Mass and Torque

Centre of Mass finds an important role in understanding the rotational motion.

Any general motion of a rigid body can be represented as in the combination of translational and rotational motion.

If the resultant of all the force passes through the centre of mass, then the torque produced will be zero, and the body will not have angular acceleration.

The moment of inertia about the centre of mass is minimum.

Center of Mass and Pseudo Force

When analysing from a non-inertial frame, we need to apply a pseudo force on the body in order to write force equations. The pseudo force is applied to the centre of mass of the body.

Toppling

Suppose the rigid party is subjected to a force that produces some acceleration and torque about the end of the surface in contact with the plane on which it is moving. At a particular instance, it may happen the torque produced by the gravitational force on the centre of mass becomes less than the torque produced by the applied force about the end of the surface in contact. In this situation, the rigid body topples or rolls over in the direction of the force applied. this phenomenon is known as toppling.

Center of Gravity

For a rigid body, each differential element of mass experiences gravitational force, the centre of gravity is the point at which the total weight of the rigid body can be considered to be concentrated.

The Centre of gravity for a body is a point around which the resultant torque due to gravitational force is zero.

The bodies with smaller dimensions, especially height, will have a centre of gravity add centre of mass at the same point as the value of gravitational acceleration remains the same for every element of the body, but if the body is of significant height, the variation of the gravitational acceleration becomes significant then the centre of gravity and centre of mass will not coincide.

Application

The Centre of mass is an important consideration while designing buildings, vehicles. The turbines of the generators should be perfectly aligned with the centre of mass.

Solved Examples- Center of Mass

Q.1. Three masses are placed on the \(x\)-axis, \(300\;{\rm{g}}\) at origin, \(500\;{\rm{g}}\) at \(x = 40\;{\rm{cm}}\) and \(400\;{\rm{g}}\) at \(x = 70\;{\rm{cm}}\). The distance of the centre of mass from the origin is
Ans: Position of centre of mass of discrete particle system is given as,

\({X_{{\rm{CM}}}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}}}{{{m_1} + {m_2} + {m_3}}}\)

where \(m\) is the mass of the particle.

\(x\) is its position.

Putting in the values, we get,

\({X_{{\rm{CM}}}} = \frac{{(300 \times 0) + (500 \times 40) + (400 \times 70)}}{{300 + 500 + 400}}\)

On solving we get,

\({X_{{\rm{CM}}}} = \frac{{(500 \times 40) + (400 \times 70)}}{{1200}}\)

\( \Rightarrow {X_{{\rm{CM}}}} = 40\;{\rm{cm}}\)

Q.2. Consider a two particles system with particles having masses \({m_1}\) and \({m_2}\). If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved so as to keep the centre of mass at the same position?
Ans: To keep the centre of mass at the same position, the velocity of the centre of mass is zero. So,

\(\overrightarrow {{v_{{\rm{cm}}}}} = \frac{{{m_1}{{\overrightarrow {{v_1}}}} + {m_2}\overrightarrow {{v_2}}}}{{{m_1} + {m_2}}} = 0\)

where, \({v_1}\) and \({v_2}\) are velocities of particles \(1\) and \(2\), respectively.

\( \Rightarrow {m_1}\frac{{{\rm{d}}\overrightarrow {{r_1}} }}{{\;{\rm{d}}t}} + {m_2}\frac{{{\rm{d}}\overrightarrow {{r_1}} }}{{\;{\rm{d}}t}} = 0\)

As

\(\left[\because {\vec v = \frac{{{\rm{d}}\overrightarrow r}}{{\;{\rm{d}}t}}} \right]\)

\( \Rightarrow {m_1}\;{\rm{d}}{\vec r_1} + {m_2}\;{\rm{d}}{\vec r_1} = 0\)

\({\rm{d}}{r_1}\) and \({\rm{d}}{r_2}\) represent the change in displacement of particles.

Let the second particle be displaced by distance \(x\),

\( \Rightarrow {m_1}(d) + {m_2}(x) = 0\)

\( \Rightarrow x = – \frac{{{m_1}d}}{{{m_2}}}\)

Here, the negative sign donates that displacement of the second particle is in the opposite direction to the displacement of the first particle.

Summary

Particles are dimensionless or point size. For a rigid body, the separation between any two points on the body always remains the same. The whole mass of the system can be considered to be concentrated at the centre of mass. For a system of two particles of equal mass, the centre of mass lies on the mid-point of the line joining the two particles.

The centre of mass of a system of particles is given by,

\(\overrightarrow {{r_{{\text{cm}}}}}  = \frac{{\mathop \sum \nolimits_{i = 1}^n {m_i}\overrightarrow {{r_i}} }}{{\mathop {\mathop {\sum {m_i}}\limits_{i = 1} }\limits^n }}\)

Centre of mass of a rigid body is given by,

\(\overrightarrow {{r_{{\rm{cm}}}}} = \frac{{\int {\vec r} dm}}{{\int d m}}\)

For uniform rigid bodies, the centre of mass will lie on the line of symmetry, if any.

The moment of inertia is least about the centre of mass.

The Centre of gravity is a point about which the torque due to the gravitation force is zero.

FAQs on Center of Mass

Q.1. Is the centre of gravity the same as the centre of mass?
Ans: No, the centre of gravity and centre of mass have different significance, but when we consider gravity to be constant, the centre of gravity and centre of mass coincides.

Q.2. What is the expression for the centre of mass for particles?
Ans: The centre of mass of the system of particles is given by,
\(\overrightarrow {{r_{{\text{cm}}}}}  = \frac{{\mathop \sum \nolimits_{i = 1}^n {m_i}\overrightarrow {{r_i}} }}{{\mathop {\mathop {\sum {m_i}}\limits_{i = 1} }\limits^n }}\)

Q.3. Can the centre of mass of the rigid body be located outside of the body?
Ans: Yes, the centre of mass of the body can be located outside of the body if the shape of the body is concave or the body is hollow.

Q.4. Does the explosion of a projectile in mid-air effects the motion of the centre of mass?
Ans: No, the motion of the projectile does not affect the motion of the centre of mass as the explosion is an internal force.

Q.5. Can the particles of the system be in motion while the centre of mass is at rest?
Ans: Yes, the centre of mass can still be at rest while the individual particles are in motion.
Example: Planets moving in a circular path under the influence of their mutual force of attraction.