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Every single physical system has associated itself with a certain point whose motion characterizes the motion of the whole system. When the system moves under an external force, then this point moves as if the entire mass of the system were concentrated at it, and also the external force was applied at it. This point is therefore called the â€˜center of massâ€™ of the system. The motion of a system can be described in terms of the motion of its center of mass.Â Â Â

While studying the dynamics of the movement of the system of particles as a whole, do not be bothered about the dynamics of an individual particle of the system. Rather focus only on the dynamics of a unique point corresponding to that system. The movement of this unique point is analogous to the movement of a single particle whose mass is equivalent to the sum of all independent particles of the system and the outcome of all the forces exerted on all the particles of the system by surrounding bodies (or) action of a field of force is exerted directly to that particle. This point is known as the center of mass of the system of particles. The idea of the center of mass (COM) is functional in analyzing the complicated movement of the system of objects, especially when two and more objects collide or an object explodes into fragments.

Center of mass formula for point objects: Z_{com}=\[\frac{\sum_{i=0}^{n}m_{i}z_{i}}{M}\]

Center of mass formula for any geometric shape: X_{com}= \[\frac{l}{2}\]

Let us consider a system of two particles of masses m_{1} and m_{2 }located at points A and B respectively.Â

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Let r_{1} and r_{2 }be the position vectors of the particles relative to a fixed origin O. Then, the position vector r_{cm} of the center of mass C of the system is defined byÂ

(m_{1}+m_{2}) r_{cm} =m_{1} r_{1}+m_{2} r_{2}.

The product of the total mass of the system and the position vector of the center of mass is equal to the sum of the products of the masses of the two particles and their respective position vectors.Â Â Â Â

Example 1: In the HCL molecule, the separation between the nuclei of the two atoms is 1.27 A. Find the location of the center of mass of the molecule. A chlorine atom is 35.5 times heavier than a hydrogen atom.Â Â Â

Solution: The center of mass of the HCL molecule will be on the line joining H and CL atoms. Let the HCL molecule be along the X-axis, the H atom being at the origin (x=0). The center of mass relative to the H atom is given byÂ

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x_{cm}= \[\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}\]

Where m1 and m2are the respective masses of H and CL atoms, and x1and x2their distances relative to the H atom.Â

Here, m_{2}=35.5 m_{1}, x_{1}=0 and x_{2}=1.27 A

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x_{cm}= \[\frac{(m_{1}\times 0 ) + (35.5 m_{1} \times 1.27 \text A)}{m_{1} + 35.5 \text m_{1}}\] Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = \[\frac{35.5\times1.27\text A}{36.5}\] = 1.235 A

Example 2: Two bodies of masses 0.5 kg and 1 kg are lying in the X-Y plane at points (-1, 2) and (3,4) respectively. Locate the center of mass of the system.Â Â

Solution : Here, we haveÂ

m_{1}=0.5 kg, m_{2}=1 kg, x_{1}= -1, y_{1}=2, x_{2}=3, y_{2}=4.

By definition, the coordinates of the center of mass of the two bodies are given by:Â

x_{cm}= \[\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}\] =\[\frac{(0.5)(-1)+(1)(4)}{(0.5)+(1)}\]= Â \[\frac{5}{3}\]

And,Â y_{cm}=
\[\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}+m_{2}}\]Â = \[\frac{(0.5)(2)+(1)(4)}{(0.5)+(1)}\] = \[\frac{10}{3}\]Â

The coordinates of C>M> are (5/3, 10/3).

Follow and practice more center of mass examples to get a better understanding.

FAQ (Frequently Asked Questions)

Question 1: Give a Real-Life Center of Mass Example.Â Â

Answer: We know that the trajectory of a projectile under gravity is a parabola. Therefore, when a firecracker is projected from the earth, it goes up along a parabolic path and explodes in mid-air at some point. Let us say P explodes into many fiery fragments which fly off along their own parabolic paths, depending on the position of P. Since the explosion is caused by internal forces alone, the â€˜center of massâ€™ of all the fragments continues to move along the same initial parabolic path that the unexploded cracker would have followed. (In fact, after the explosion, there is nothing at the location of the center of mass which is just a mathematical point.)

Question 2: What is a Rigid Body?

Answer: In practice, we come across extended bodies that may either be deformable or non-deformable, that is, rigid. An extended body is also a system of an infinitely large number of particles that have infinite small separations between them. When a body deforms, the disconnection between the distance of its particles and their relative locations also changes. Whereas, a rigid body is an extended object in which the separations and relative location of all of its constituent particles will always remain the same under all circumstances.

It is the average position of all the parts of the system, weighted based on their masses. While dealing with a simple rigid object that has a uniform density, the center of mass will be located at the centroid.