In physics, when we study how things move, we often consider that an object is just a point with mass, meaning it has mass but no size (no length, width, or height). Even when dealing with real things like tennis balls, boxes or even humans, we like to imagine them as point masses. This simplification makes it easier to solve many problems. However, real objects do have size and shape, and these factors can affect how they move. Still, in many situations, we can use the same formulas and methods we use for point masses on more complicated objects. We can do this by pretending that the complex object is just a point mass located at its centre of mass, which is sometimes called its centre of gravity.
In this document, we will be studying the centre of mass and its motion in detail.
(a) When we look at a group of moving particles as a whole, we don't have to worry about each particle's movement. Instead, we can just focus on one special point that represents the entire group.
(b) This point is called the centre of mass (COM). The motion of this point is just like the motion of a single particle, but its mass is the total mass of all the particles in the group.
(c) The forces acting on all the particles can be thought of as acting directly on this centre of mass.
(d) It's a handy concept when we're trying to understand how a bunch of objects move together, especially when things like collisions or explosions happen.
(e) So, instead of dealing with the complexity of each particle's movement, we simplify things by thinking about the motion of the centre of mass.
Consider a system of n particles as shown in the figure below, having masses m_{1}, m_{2}, m_{3}, .................. m_{n }and their position vectors r_{1}, r_{2}, r_{3}, .............. r_{n }respectively. The position vector of the centre of mass is r_{cm} with respect to the origin,
Here,
M is the total mass of the system.
Here the terms m_{1}r_{1}, m_{2}r_{2} ,..........., m_{n}r_{n} are called the moment of masses of particles m_{1}, m_{2},......., m_{n} with respect to the origin.
Further,
and So, the cartesian coordinates of the COM will be,
Note:
 If the origin is taken at the centre of mass then = 0. hence, the COM is the point about which the sum of "mass moments" of the system is zero.
 If we change the origin then changes. So also changes but exact location of center of mass does not change.
For a system comprising of two particles of masses m_{1 }and m_{2} , positioned at coordinates (x_{1} ,y_{1} ,z_{1}) and (x_{2} ,y_{2} ,z_{2}) , respectively, we have:
For a twoparticle system, COM lies closer to the particle having more mass, which is rather obvious. If COM’s coordinates are made zero, we would observe that the distances of individual particles are inversely proportional to their masses.
Q1. Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their centre of mass. (JEE Mains 2018)
Solution: For the system of particles of masses m_{1} and m_{2} , if the distances of the particles from the centre of mass are r_{1} and r_{2} respectively then it is seen that m_{1} r_{1} = m_{2} r_{2}
Since both the particles lie on the xaxis, the COM will also lie on the xaxis. Let the COM be located at
x = x, then
r_{1} = distance of COM from the particle of mass 1 kg = x
and r_{2} = distance of COM from the particle of mass 2 kg = (3 x)
Using
or
or x = 2 m
thus, the COM of the two particles is located at x = 2m.
Q2. Two particles of mass 4 kg & 2 kg are located as shown in the figure then find out the position of centre of mass.
Solution: First, find out the position of the 2 kg mass
x_{2kg} = 5 cos 37° = 4 m
y_{2kg} = 5 sin 37° = 3 m
So this system is like a twoparticle system of mass 4 kg and 2kg are located (0, 0) and (4, 3) respectively, then
x_{com} = = = =
y_{com} = = = 1 m
So the position of C.O.M is
Q3. Two particles of mass 2 kg and 4 kg lie on the same line. If 4 kg is displaced rightwards by 5 m then by what distance 2 kg should be moved for which centre of mass will remain at the same position?
Solution: Let us assume that C.O.M. lie at point C and the distance of C from 2 kg and 4 kg particles are r_{1} & r_{2} respectively. Then from relation
m_{1}r_{1} = m_{2}r_{2}
2r_{1} = 4r_{2} ...(i)
Now 4 kg is displaced rightwards by 5 m then assume 2 kg is displaced leftwards by x distance to keep the C.O.M. at rest.
from relation m_{1}r_{1} = m_{2}r_{2}'
⇒ m_{1}(r_{1} + x) = m_{2} (r_{2} + y)
2(r_{1} + x) = 4(r_{2} + 5) ...(ii)
2x = 20
x = 10 m
To keep the C.O.M at rest 2 kg displaced 10 m leftwards
Alter: If the centre of mass is at rest then we can write
m_{1}x = m_{2}y
2 × x = 4 × 5
x = 10 m.
Q4. Two particles of mass 1 kg and 2 kg lie on the same line. If 2 kg is displaced 10 m rightwards then by what distance 1 kg be displaced so that the centre of mass will displaced 2m rightwards?
Solution: Initially let us assume that C.O.M is at point C which is r_{1} & r_{2} distance apart from mass m_{1} & m_{2} respectively as shown in the figure
from relation m_{1} r_{1} = m_{2} r_{2}
⇒ (1) r_{1} = 2r_{2}
Now 2kg is displaced 10 m rightwards then we assume that 1 kg is displaced x m leftward to move the C.O.M 2m rightwards.
So from relation m_{1}r_{1}' = m_{2}r_{2}'
⇒ 1 (x + r_{1} + 2) = 2 (10 + r_{2}  2)
⇒ x + r_{1} + 2 = 20 + 2r_{2}  4 ...(ii)
from eq. (i) & (ii) x = 14 m (leftwards).
Q5. Three particles of mass 1 kg, 2 kg, and 3 kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1 m. Find the distance of their centre of mass from A.
Solution: Assume that 1 kg mass is placed at the origin as shown in the figure
Coordinate of A = (0, 0)
Coordinate of B = (1 cos60°,1 sin60°) =
Coordinate of C = (1, 0)
Position of centre of mass
Distance of C.O.M from point
For continuous mass distribution, the centre of mass can be located by replacing the summation sign with an integral sign. Proper limits for the integral are chosen according to the situation
= M (mass of the body)
here x, y, z in the numerator of the eq. (i) is the coordinate of the centre of mass of the dm mass.
=
Note :
 If an object has symmetric mass distribution about x axis then y coordinate of COM is zero and viceversa.
Semicircular Plate
Here's a comprehensive table that compiles the values of centre of mass (COM) coordinates for all the important figures, in Cartesian coordinates, suitable for objectivetype questions in NEET :
To find the coordinates of the centre of mass of a body after a portion of it is removed, we can consider the body and the removed part as two separate masses and then use the principle of superposition for the centre of mass. Let's assume we know the coordinates of the centre of mass of the original body (x_{original}, y_{original}) before any mass is removed and the coordinates of the centre of mass of the removed portion (x_{1},y_{1}). The mass of the original body is m, and the mass of the removed portion is m_{1}. The task is to find the coordinates of the new centre of mass of the remaining body (x_{new},y_{new}).
The formula for the new centre of mass, considering the original body and the removed portion as a system, is:Note that this approach assumes you can treat the mass distribution of both the original body and the removed portion as being concentrated at their respective centres of mass.
Q6. A rectangular sheet of metal has dimensions of 4 m by 3 m and a uniform mass of 24 kg. A square piece of side 1 m is cut out from one of the corners of the sheet. Find the coordinates of the centre of mass of the remaining sheet if the original centre of mass of the whole sheet was at the geometric centre. Assume the sheet and the cutout piece have uniform thickness and density.
Solution:
Dimensions of the rectangular sheet: 4 m x 3 m
Mass of the rectangular sheet (m): 24 kg
Dimensions of the cutout square: 1 m x 1 m
Mass of the cutout square (m_{1}): Since the area of the cutout square is 1/12 of the total area, its mass is (1/12)×24kg=2kg.
The original centre of mass of the rectangular sheet (x_{original},y_{original}): (2 m, 1.5 m), the geometric centre of the sheet.
Since the square is cut from one of the corners, its centre of mass will be at the midpoint of the square. Assuming the bottom left corner of the rectangle as the origin (0,0), the centre of mass of the square is at (0.5m,0.5m).
The new centre of mass of the remaining metal sheet, after a square piece is cut out from one of the corners, is approximately
(2.14 m, 1.59 m) from the original bottom left corner.
(r_{COM}) with respect to time. The position of the centre of mass is given by:
Differentiating both sides with respect to time gives:
Following the velocity of the centre of mass, the acceleration of the centre of mass (a_{COM}) is the rate of change of v_{COM} with respect to time. Thus:
Q7. A car of mass 1500 kg is moving east with a speed of 20 m/s, and a truck of mass 3000 kg is moving west with a speed of 10 m/s. Find the velocity of the centre of mass of the system.
Solution:
Q8. Two particles, one of mass 3 kg and the other of mass 2 kg, are moving along the xaxis. If the 3 kg mass has a velocity of 4 m/s to the right and the 2 kg mass has a velocity of 3 m/s to the left, what is the velocity of the centre of mass?
Solution:
118 videos470 docs189 tests

1. What is the concept of Centre of Mass? 
2. How do you find the Centre of Mass of a system of 'n' particles? 
3. How do you determine the Centre of Mass of a uniform rod? 
4. What is the Centre of Mass of a uniform semicircular plate? 
5. How does the motion of the Centre of Mass change after partial removal of a system's mass? 
118 videos470 docs189 tests


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