Centre of Mass

# Centre of Mass | Physics Class 11 - NEET PDF Download

 Table of contents Concept of Centre of Mass Centre of Mass of a System of 'n' Particles Position of COM of two Particles Centre of Mass of a Continuous Mass Distribution Centre of Mass of a Uniform Rod Centre of Mass of a Uniform Semicircular Wire Centre of Mass of Uniform Semicircular Plate Some Important Figures Centre of Mass After Partial Removal Motion of Centre of Mass

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.

## Concept of Centre of Mass

(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.

## Centre of Mass of a System of 'n' Particles

Consider a system of n particles as shown in the figure below, having masses m1, m2, m3, .................. mn  and their position vectors r1, r2, r3, .............. rrespectively. The position vector of the centre of mass is rcm with respect to the origin,

Here,

M is the total mass of the system.

Here the terms m1r1, m2r2 ,..........., mnrn  are called the moment of masses of particles m1, m2,......., mn with respect to the origin.
Further,

and So, the cartesian co-ordinates 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.

## Position of COM of two Particles

For a system comprising of two particles of masses mand m2 , positioned at co-ordinates (x1 ,y1 ,z1) and (x2 ,y2 ,z2) , respectively, we have:

For a two-particle 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.

Question for Centre of Mass
Try yourself:Can the centre of mass of the rigid body be located outside of the body?

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 m1 and m2 , if the distances of the particles from the centre of mass are r1 and r2 respectively then it is seen that m1 r1 = m2 r2

Since both the particles lie on the x-axis, the COM will also lie on the x-axis. Let the COM be located at

x = x, then

r1 = distance of COM from the particle of mass 1 kg = x

and r2 = 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

x2kg = 5 cos 37° = 4 m

y2kg = 5 sin 37° = 3 m

So this system is like a two-particle system of mass 4 kg and 2kg are located (0, 0) and  (4, 3) respectively, then

xcom =  =  =  =

ycom =  =  = 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 r1 & r2 respectively. Then from relation

m1r1 = m2r2

2r1 = 4r2                   ...(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 m1r1 = m2r2'
⇒  m1(r1 + x) = m2 (r2 + y)
2(r1 + x) = 4(r2 + 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

m1x = m2y

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 r1 & r2 distance apart from mass m1 & m2 respectively as shown in the figure

from relation m1 r1 = m2 r2

⇒ (1) r1 = 2r2

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 m1r1' = m2r2'

⇒ 1 (x + r1 + 2) = 2 (10 + r2 - 2)
⇒ x + r1 + 2 = 20 + 2r2 - 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

Co-ordinate of A = (0, 0)

Co-ordinate of B = (1 cos60°,1 sin60°) =

Co-ordinate of C = (1, 0)

Position of centre of mass
Distance of C.O.M from point

## Centre of Mass of a Continuous Mass Distribution

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 vice-versa.

## Centre of Mass of a Uniform Rod

• Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L. Mass per unit length of the rod l =
• Hence, dm, (the mass of the element dx situated at x = x is) = l dx
• The coordinates of the element dx are (x, 0, 0). Therefore, the x-coordinate of COM of the rod will be

xCOM =  =  =
• The y-coordinate of COM is
yCOM =  = 0
• Similarly, zCOM = 0
i.e., the coordinates of COM of the rod are , i.e., it lies at the centre of the rod.

Question for Centre of Mass
Try yourself:Does the explosion of a projectile in mid-air effects the motion of the centre of mass?

## Centre of Mass of a Uniform Semicircular Wire

• To derive the expression for the coordinates of the centre of mass of a semicircular ring with linear density σ and radius r, we can use the method of integration.
• Let's assume that the semicircular ring is in the first quadrant, with the base of the semicircle on the X-axis, and the center of the equivalent circle placed at the origin of the Cartesian plane as shown in the image below:
Semicircular ring
• Consider a small mass element ��dm on the semicircular ring. The mass of this element can be expressed as ��=�⋅��dm=σds, where ��ds is the differential arc length. The position vector of this mass element ��dm with respect to the centre of the semicircular ring can be expressed as �⃗=�cos⁡(�)�^+�sin⁡(�)�r=rcos(θ)i+rsin(θ)j, where θ is the angle made by the position vector with the positive x-axis.
• Then, the total mass of the ring is m=σπr.
• �=���.Let the radius touching this point of the ring make an angle θ with the positive X axis. Consider a small portion of the ring of length ds=rdθ.��=���
• The mass of this portion is dm=σrdθ. The distance of this portion from the base of the semicircle is, y=rsinθ. The centre of mass (�cm,�cm)(xcm,ycm) of the semicircular ring can be calculated using the formula:

• On substituting the required values we get,

• For x- coordinate:

• For y-coordinate:

• The centre of mass of a uniform semicircular ring of radius r is (0, 2R/π)(0,2�where the centre of the equivalent circle is at the origin of the Cartesian plane and the base of the semicircle is on the X-axis.

## Centre of Mass of Uniform Semicircular Plate

• To derive the expression for the coordinates of the centre of mass of a semicircular plate with radius r, we can use the method of integration.

Semicircular Plate

• For a semicircular plate, we'll use polar coordinates for easier integration. The coordinates of the centre of mass (�ˉ,�ˉ)(x,y) are given by the equations:

• Here A is the area of the semicircle, and ��dA is an infinitesimal area element of the semicircle. In polar coordinates, an element of area ��dA can be represented as � �� ��rdrdθ, where r is the radius from the origin to the area element, and ��dθ is the differential angle in radians. For a semicircle, r ranges from 00 to R, and θ ranges from 00 to π.
• x- coordinate:
The x coordinate of a point in polar coordinates is given by �=�cos⁡(�)x=rcos(θ). However, due to symmetry about the y-axis, we know that for every point with a positive x, there's a corresponding point with a negative x of equal mass, making the net x-coordinate of the centre of mass 00.
• y- coordinate:
The y coordinate of a point in polar coordinates is �=�sin⁡(�)y=rsin(θ). Substituting this into the formula for y, we get:
• Simplifying, we distribute r and move constants outside of the integral:

• The centre of mass of a uniform semicircular plate of radius R is (0, 4R/3π).

## Some Important Figures

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 objective-type questions in NEET :

• L: Total length of the rod or side length of the rectangle/square.
• W: Width of the rectangle.
• R: Radius of the circle or semicircle.
• b: Base length of the triangle.
• h: Height of the triangle from the base to the opposite vertex.
• H: Height of the cylinder or cone.

Question for Centre of Mass
Try yourself:A solid sphere and a hollow cone are placed side by side. The solid sphere has a radius R of 6 cm, and the hollow cone has a base radius R of 6 cm and a height ℎ of 18 cm. What is the distance between the centers of mass of the solid sphere and the hollow cone?

## Centre of Mass After Partial Removal

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 (�original,�originalxoriginalyoriginal) before any mass is removed and the coordinates of the centre of mass of the removed portion (�1,�1x1,y1). The mass of the original body is m, and the mass of the removed portion is �1m1. The task is to find the coordinates of the new centre of mass of the remaining body (�new,�newxnew,ynew).
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 cut-out 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 cut-out square: 1 m x 1 m
Mass of the cut-out square (�1m1): Since the area of the cut-out square is 111/12 of the total area, its mass is 112×24 kg=2 kg(1/12)×24kg=2kg.
The original centre of mass of the rectangular sheet (�original,�originalxoriginal,yoriginal): (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 (0.5 m,0.5 m)(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.

## Motion of Centre of Mass

Consider a system of n particles with masses �1,�2,…,��m1,m2,,mn and velocities �1,�2,…,��v1,v2,,vn, respectively. The total mass M of the system is the sum of the masses of all particles, �=∑�=1���M=∑m(i=1 to i=n). The velocity of the centre of mass (�COMvCOM) is defined as the rate of change of its position

(�COMrCOM) 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 (�COMaCOM) is the rate of change of �COMvCOM 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 x-axis. 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:

The document Centre of Mass | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
All you need of NEET at this link: NEET

## Physics Class 11

118 videos|470 docs|189 tests

### Up next

 Test | 15 ques
 Doc | 29 pages
 Doc | 42 pages

## FAQs on Centre of Mass - Physics Class 11 - NEET

 1. What is the concept of Centre of Mass?
Ans. Centre of Mass is a point in a system or object where the entire mass of the system or object can be assumed to be concentrated. It is a point where the total gravitational force on the body can be considered to act.
 2. How do you find the Centre of Mass of a system of 'n' particles?
Ans. The Centre of Mass of a system of 'n' particles can be found by taking the weighted average of the positions of all the particles, where the weight is the mass of each particle.
 3. How do you determine the Centre of Mass of a uniform rod?
Ans. The Centre of Mass of a uniform rod is located at the midpoint of the rod. This is because the mass distribution is symmetric along the rod, making the Centre of Mass equidistant from both ends.
 4. What is the Centre of Mass of a uniform semicircular plate?
Ans. The Centre of Mass of a uniform semicircular plate lies along the axis of symmetry of the semicircle, at a distance of 4r/(3π) from the diameter of the semicircle, where r is the radius of the semicircle.
 5. How does the motion of the Centre of Mass change after partial removal of a system's mass?
Ans. The motion of the Centre of Mass remains unaffected after the partial removal of a system's mass, as long as no external forces are acting on the system. This is because the Centre of Mass represents the overall motion of the system, regardless of internal changes.

## Physics Class 11

118 videos|470 docs|189 tests

### Up next

 Test | 15 ques
 Doc | 29 pages
 Doc | 42 pages
 Explore Courses for NEET exam

### How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;