JEE Main Study Notes for Center of Mass: The concept of the center of mass is that of an average of the masses factored by their distances from a reference point. In one plane, that is like the balancing of a seesaw about a pivot point with respect to the torques produced. This section has a weightage of 5-6% in JEE Main. Candidates can expect 1-2 questions on this topic in the examination. 

JEE Main Study Notes for Center of Mass include some important topics such as the Formulas, System of Particles, Rigid Body, Motion of Center of Mass, Equilibrium, and State of Equilibrium. Check JEE Main Physics Syllabus

As this section has a 5-6% weightage, Candidates have to be more focused on this section to bring a good score. Candidates have to attend mock tests and solve the previous years’ question papers to be well prepared. They have to make a proper preparation plan for all the sections of Physics. Check JEE Main Physics Preparation Tips

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Understanding Definition of Center of Mass

The Center of the mass of a body or system of a particle is explained as a point at which the whole of the mass of the body or all the masses of a system of the particle is concentrated. At the time of understanding the dynamics of the motion of the system of a particle as a whole, there is no need to worry about the dynamics of individual particles of the system.

• The motion of this point is similar to the motion of a single particle whose mass is equal to the overall sum of all individual particles of the system. 
• The resultant forces exerted on all the particles of the system by nearby bodies action of a field of force is exerted directly to that particle. 
• This point is being referred to as the center of mass of the system of particles. The whole concept of center of mass (COM) is useful to calculate the challenging motion of the system of objects, especially in case two and more objects collide or an object explodes into fragments.

The formula of the Center of Mass

Center of Mass if a Part of a Body is removed

In case a portion of a body is taken out, the remaining portion might be considered as the original mass (M) minus the mass of the removed part (m)

= {Original mass (M)} + {- mass of the removed part (m)}

The formula in this situation will change to- Xcm = (Mx-mx')/(M-m) and Ycm = (My-my')/(M-m) where the initial ones represent the coordinate of the C.M. of the removed part.

Center of Mass for a Continuous Distribution

For a continuous distribution of mass, an element of mass dm can be treated at any position as a point mass and it can change the summation by integration as discussed beneath:

• Rcm = 1/M ∫ x dm
• So, we get Xcm equals to 1/M ∫ x dm
• Ycm = 1/M ∫ y dm
• Zcm = 1/M ∫ z dm,

For center of mass, the following facts are self-explanatory.

(1) There might or might not be any mass present at the center of mass. The first body has mass present at the center of mass while in the second example of a ring; there is no mass at the center of mass.

(2) The position will be dependent on the actual shape of the body. It will be closer to the region where more mass is concentrated.

(3) Center of mass and center of gravity might not be at the same points.

Must Read:

• JEE Main Study Notes on Kinematics
• JEE Main Study Notes for Current Electricity

Sample Question on Center of Mass

Question: A system has two objects and has a total momentum of (18 kgm/sec)i. The center of mass has the velocity of (3 m/s)i. One of the objects has the mass 4 kg and velocity (1.5 m/s)i. Calculate the mass and velocity of the other objects?

1. 2 kg, (6 m/s)i
2. 2 kg, (-6 m/s)i
3. 2 kg, (3 m/s)i
4. 2 kg, (-3 m/s)i

Solution: (1)

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System of Particles

The term system of particles is known as a collection of a large number of particles that may or may not interact with each other or are connected. They might be the real particle of rigid bodies in translational motion. It is being observed that the particle which interacts with each other also applies force on each other.

The force of mutual interaction between the particles of the system is referred to as the internal force of the system. These forces always exist in the pair of equal magnitude and opposite directions. In addition to the internal forces, these forces appear on all or a few of the particles. Here, the external force is that force that acts on any one particle and is included in the system by some other body outside the system.

What is the meaning of a rigid body?

In daily practice, we deal with extended bodies that might be deformable or non-deformable (or) rigid. An extended body is the system of an infinitely large number of particles with smaller separation between them. In case a body deforms, the separation between the distance between its particles and their relative locations also changes. A rigid body is an extended object where the separations and relative location of all of its constituent particles will be the same no matter whatever the circumstances are. Actually, it is the average position of all the parts of the system that’s weighted is according to the masses. The simple rigid object with uniform density has the center of mass located at the centroid.

Sample Question

Question: When a moving bullet hits a solid target resting on a frictionless surface gets embedded in it. What is conserved in it?

1. Momentum Alone
2. KE alone
3. Both Momentum and KE
4. Neither KE nor momentum

Solution: (1)

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Motion of Center of Mass

Since we have the position now, we can now go beyond the concept of the center of mass to velocity and acceleration and can use different tools to understand the motion of a system of particles. We can conclude that:

Vcm = m1v1+m2v2/m1+m2

Thus we have almost the same expression for the velocity of the center of mass. By differentiating again, we can present an expression for acceleration:

acm = m1a1+m2a2/m1+m2

With this set of three equations, we have generated the much-needed elements of the kinematics of a system of particles.

So, (m1+m2) acm = m1a1+m2a2

Fcm = F1+ F2 + ….

Here, Fcm will be equal to M (acm). Therefore M is being referred to as the total mass of the system.

The total mass of the system times the acceleration of its center of mass remains equal to vector sum of all the forces acting on the group of particles.

Sample Question

Question: A body is stationary and has a mass of 3 kg explodes into three equal parts. Here two of the pieces fly off at right angles to each other and they have the velocities 2i m/s and 3j m/s. If the explosion takes place in 10-3 sec. calculate the average force acting on the third piece in N

1. (-2i-3j)10(power3)
2. (2i+3j)10(power3)
3. 2i-3j)10(power-3)
4. none of these

Solution: (1)

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What is Equilibrium?

The rigid body is explained to be in equilibrium in case the forces acting on it do not change its state of rest or of uniform motion. In simpler words, it means that if a body is at rest, it should remain at rest. In case the body is in motion, it should keep on moving with uniform velocity (may be linear or angular).

Based on this, the equilibrium is segregated into two categories mentioned below:

(a) Translatory equilibrium

A body is in translatory equilibrium if the center of mass has no linear acceleration in an inertial frame of reference. For a body to remain in translatory equilibrium, the main criterion is that the vector sum of all the external forces acting on the body must be zero. In that situation only, the body will remain at rest. This equilibrium is static. If a body is moving at a uniform velocity, along a straight line will continue doing that. This equilibrium is being referred to as the dynamic translatory equilibrium.

(b) Rotatory equilibrium

A body will be in rotatory equilibrium if it has no angular acceleration about an axis in an inertial frame. The body will be in rotatory equilibrium if the vector sum of all the external torques acting on the body is zero.

For a body to be in equilibrium, it needs to fulfill both the conditions mentioned above at the same time.

• The vector sum of all the external forces acting on the body needs to be zero.
• The vector sum of all the external torque acting on the body should also vanish.

Check: JEE Main Study Notes for Current Electricity

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What is the State of Equilibrium?

Equilibrium is classified into three categories:

• Stable equilibrium- The equilibrium of a body is said to be stable if on getting slightly disturbed as well, it will come back to its actual position.
• Unstable equilibrium- The body is referred to be in unstable equilibrium in the case of being slightly changed, it shows zero tendencies to come back to its original position and moves away from it.
• Neutral equilibrium- A body will be in neutral equilibrium in the case of being slightly displaced, it remains in a new position.

What are the conditions for Stable Equilibrium?

The degree of stability is based on the height of the center of gravity of the body from the surface of the support. If the height of the center of gravity is smaller, it has greater stability.

• The vertical drawn from the center of gravity must not pass through the base.
• The center of gravity needs to be at the lowest
• The base needs to be much wider

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Previous Years Solved Questions on Center of Mass 

1. Question: A particle of mass 1 kg is projected upwards having a velocity 60 m/s. Another particle of mass 2 kg is dropped from some height. After two seconds when neither of the particles has collided with the ground. What will be the acceleration and velocity of COM?

Solution: Since we know that, uCOM is equal to [m1a1 + m2a2]/[m1+m2] and uCOM = [m1u1 + m2u2]/[m1+m2]

So, aCOM = [m1a1 + m2a2]/[m1+m2]

= [(1) (-10) + 2 (-10)] / 3

= -10 m/s2

uCOM = m1u1 + m2u2/m1+m2

= [(1) (60) + (2) (0)]/3 = + 20 m/s

From the above calculation, we can conclude that, acceleration of the center of mass will be -10 m/s2 nad velocity of the center of mass will be 20 m/s.

2. Question: A system of two objects has total momentum of (18 kgm/sec)i and its center of mass has the velocity of (3 m/s)i. Out of the, one object has a mass of 4 kg and velocity (1.5 m/s)i. What will be the mass and velocity of the remaining object

(a) 2 kg, (6 m per second)i 

(b) 2 kg, (-6 m per second)i

(c) 2 kg, (3 m per second)i 

(d) 2 kg, (-3 m/s)i

Solution: Here the total momentum is 18 kgm/sec)i, velocity of Center of mass=(3 m/s)i, Mass of one object=4 kg, Velocity of this object=(1.5 m/s)i. Here, let’s assume M as the mass of other object while V means the velocity.

Total momentum is equal to the total massX velocity of center of mass

(18 kgm/sec)i=(m+4)(3 m/s)i.

or m=2 kg

Now Vcm=(m1v1+m2v2)/(m1+m2)

Or, 3i=(41.5i + 2v)/6

So, 18i=6i+2v

Or, v=6i m/sec

From the above observation, it is clear that option (a) is right.

3. Question: The particle of masses 2 kg, 2 kg, 1 kg and 1 kg are kept at the corners A, B, C, D of a square of side L. What will be center of mass of the system?

Solution: If A is taken as origin, then,

xcm = (m1 x1 + m2 x2 + m3 x3 + m4 x4) / (m1 + m2 + m3 + m4 )

= (2.0+2.L+1.L+1.0)/6 = 1/2

ycm = (m1 x1 + m2 x2 + m3 x3 + m4 x4) / (m1 + m2 + m3 + m4)

= (2.0+2.0+1.L+1.L)/6=L/3

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Tips to Solve Questions on Center of Mass

• A rigid body has a fixed shape that can’t be changed at all. The distance between the particles is fixed and doesn’t change at all.
• Any rigid body has an infinite number of particles, and therefore is referred to as the system of particles.
• The whole mass of a body is concentrated at a point that is why it is being referred to as the center of mass (COM)
• For a single rigid body, the center of mass mostly coincides with its centroid. But the COM can be present inside or outside the body.
• The velocity of the rigid body is also referred to as the velocity of its COM.
• In the case of the translation motion, the velocity of all the particles of the rigid body will remain the same. However, this is not true in the case of rotational motion.
• The line along which the object is fixed and the motion of any object is limited is being referred to as the Axis of rotation.
• The particle in the circular motion has angular velocity, ω, which is related to the linear velocity as v = rω.
• The angular acceleration, α, is the rate of change of the angular velocity with time.
• The rotational analog of force, in the rotational motion, is called the moment of force, also called as Torque.
• The torque is defined as the cross product of the perpendicular distance from the axis of rotation and force. i.e. Τ=r×F.
• The analog of mass in rotational motion is known as the Moment of Inertia, it is also known as the Angular mass or the Rotational Inertia. It is the quantity that helps to calculate the torque for angular acceleration along the axis.
• The unit of M.I. is kg-m2 and the dimensional formula is ML2.

JEE Main Center of Mass FAQs