The centre of mass is a unique point in a system of particles or an extended body where the entire mass is considered to be concentrated for describing translational motion. It behaves as if all external forces are applied directly at this point.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of centre of mass for discrete particle systems, symmetric rigid bodies, motion of centre of mass, conservation of linear momentum, and internal forces.
PGT extension
Continuous mass distributions, use of integration for irregular objects, variable mass systems, frame of reference transformations, and centre of mass in central force problems.
Prerequisites
Vector algebra and coordinate geometry
Newton's laws of motion
Calculus (integration and differentiation)
Definition and Position Vector
For a discrete system of particles, the position vector R of the centre of mass is given by the weighted average of the position vectors of the individual particles.
$R = (\sum(m_i r_i)) / (\sum(m_i))$, where m_i is the mass of the i-th particle and r_i is its position vector.
The centre of mass depends solely on the mass distribution and is independent of the coordinate system chosen.
Motion of the Centre of Mass
The total external force acting on a system equals the total mass of the system multiplied by the acceleration of its centre of mass: $F_ext = M A_cm$.
Internal forces between particles cancel out in pairs due to Newton's third law and do not affect the acceleration of the centre of mass.
If the net external force is zero, the centre of mass moves with a constant velocity, or remains at rest, regardless of internal motion.
Centre of Mass for Continuous Bodies
For continuous mass distributions, summation is replaced by integration: $R = (1 / M) \int r dm$.
In Cartesian coordinates, the coordinates are $X_cm = (1 / M) \int x dm$, $Y_cm = (1 / M) \int y dm$, and $Z_cm = (1 / M) \int z dm$.
Symmetric objects with uniform density have their centre of mass located at their geometric center.
Centre of Mass Frame (C-Frame)
A reference frame attached to the centre of mass is known as the centre of mass frame or zero-momentum frame.
In this frame, the total linear momentum of the system is always zero.
It significantly simplifies the analysis of collisions, two-body central force problems, and rotational dynamics.
Derivations
Position Vector of Centre of Mass for a Two-Particle System
Conditions: Two point masses m_1 and m_2 located at position vectors r_1 and r_2.
Define the total mass of the system as $M = m_1 + m_2$.
Apply the definition of the centre of mass position vector for discrete masses: $R = (m_1 r_1 + m_2 r_2) / (m_1 + m_2)$.
Differentiate the position vector with respect to time to obtain the velocity of the centre of mass: $V_cm = (m_1 v_1 + m_2 v_2) / M$.
Differentiate again with respect to time to get the acceleration of the centre of mass: $A_cm = (m_1 a_1 + m_2 a_2) / M$.
Result: $R = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$
Centre of Mass of a Uniform Semicircular Ring
Conditions: Uniform semicircular ring of radius R and total mass M lying in the xy-plane.
Consider a differential mass element dm corresponding to an angular element d\theta at angle \theta, where d$m = (M / \pi) d\theta$.
The coordinates of the mass element are x = R \cos\theta and y = R \sin\theta.
By symmetry about the y-axis, the x-coordinate of the centre of mass, $X_cm = 0$.
Evaluate the y-coordinate using integration: $Y_cm = (1 / M) \int y dm = (1 / M) \int_0^\pi (R \sin\theta) (M / \pi) d\theta$.
$m_i = mass of i-th particle$, $r_i = position vector$ Metres (m)
Discrete particle system
Centre of Mass (Continuous)
$$R_{cm} = \frac{1}{M} \int r \, dm$$
M = total mass, dm = mass element, r = position vector Metres (m)
Continuous mass distribution
Velocity of Centre of Mass
$$V_{cm} = \frac{\sum m_i v_i}{\sum m_i}$$
$v_i = velocity of i-th particle$ Metres per second (m/s)
In inertial frame
Acceleration of Centre of Mass
$$A_{cm} = \frac{F_{ext}}{M}$$
$F_ext = net external force$, M = total mass Metres per second squared (m/s^2)
Constant total mass
Total Momentum using CM
$$P_{total} = M V_{cm}$$
M = total mass, $V_cm = velocity of centre of mass$ Kilogram metre per second (kg m/s)
Valid for any system of particles
CM of Uniform Rod
$$X_{cm} = \frac{L}{2}$$
L = length of the rod, origin at one end Metres (m)
Uniform linear mass density
CM of Uniform Semicircular Disc
$$Y_{cm} = \frac{4R}{3\pi}$$
R = radius of the disc Metres (m)
Uniform surface mass density
Shift in CM when a part is removed
$$\Delta X = \frac{A_2 d}{A_1 - A_2}$$
$A_1 = original area$, $A_2 = removed area$, d = distance between their centres Metres (m)
Uniform thickness and density
Graphs and experiments
Explosion of a Projectile in Mid-Air
Draw: A parabolic trajectory of a projectile with a dotted continuation representing the unexploded path, and fragment trajectories diverging symmetrically around the continuing parabolic path of the centre of mass.
Interpret: Internal forces causing an explosion do not affect the trajectory of the centre of mass, which continues to follow the original parabola under gravity.
Exam use: Tested conceptually in TGT and numerically in PGT regarding fragment landing positions.
Two-Body System on a Smooth Surface
Draw: Two blocks of unequal masses connected by a spring on a frictionless horizontal floor, showing displacement vectors from a fixed origin to the individual masses and their common centre of mass.
Interpret: Shows that in the absence of external horizontal forces, the centre of mass remains stationary while internal oscillations occur.
Exam use: Used to evaluate understanding of internal vs external forces and conservation of momentum.
Suspension Method for Finding Centre of Gravity
Draw: An irregular lamina suspended from a pinhole with a plumb line hanging vertically downwards, intersecting the lamina along a straight line.
Interpret: The centre of gravity (and hence centre of mass in a uniform gravitational field) always lies on the vertical line passing through the point of suspension.
Exam use: Laboratory-based conceptual question in TGT exams.
Solved numericals
TGT core
Three point masses of 1 kg, 2 kg, and 3 kg are located at the vertices of an equilateral triangle of side 1 m. Find the coordinates of the centre of mass, placing the 1 kg mass at the origin.
Set coordinate system: 1 kg at (0, 0), 2 kg at (1, 0), and 3 kg at (0.5, \sqrt{3}/2).
A uniform circular disc of radius R has a circular hole of radius R/2 cut from it. The centre of the hole is at a distance R/2 from the centre of the original disc. Find the position of the centre of mass of the remaining portion.
Let the mass of the complete disc be M and surface mass density be \sigma.
The area of the complete disc is $A_1 = \pi R^2$, so $M_1 = M$.
The radius of the removed circular hole is R/2, so its area is $A_2 = \pi (R/2)^2 = \pi R^2 / 4 = A_1 / 4$.
The mass of the removed portion is $M_2 = M / 4$.
Let the centre of the original disc be at the origin (0, 0). Place the centre of the hole at (R/2, 0).
Use the formula for the centre of mass of the remaining part: $X_cm = (M_1 X_1 - M_2 X_2) / (M_1 - M_2)$.
Answer: -R/6 from the centre of the original disc along the line joining the centres
TGT core
A man of mass 60 kg is standing at one end of a 2 m long boat of mass 140 kg floating in still water. If the man walks to the other end of the boat, through what distance does the boat move relative to the water?
Total mass $M = m_man + m_boat = 60 + 140 = 200 kg$.
Since no external horizontal force acts on the system (man + boat), the centre of mass remains at rest.
Let the displacement of the boat relative to the water be x. The displacement of the man relative to the water is (L - x), where L = 2 m.
Apply the condition for zero displacement of the centre of mass: m_man (L - x) - m_boat x = 0.
The centre of mass concept assumes macroscopic bodies where classical mechanics is valid, ignoring quantum mechanical wave-packet spreading.
In extended bodies, the gravitational centre of mass (centre of gravity) coincides with the geometric centre of mass only if the gravitational field is uniform across the entire body.
Common misconceptions
Common error: The centre of mass of a body must always lie inside the material of the body.
Correct: The centre of mass can lie outside the physical body, as seen in hollow objects like rings, hollow spheres, and L-shaped brackets.
The definition $R = (1/M) \int r dm relies on mass distribution in space$, not on material continuity.
Common error: Internal forces can change the velocity of the centre of mass of a system.
Correct: Internal forces can only change the relative positions and velocities of individual particles within the system, leaving the centre of mass motion unaffected.
According to Newton's third law, internal forces occur in equal and opposite pairs that cancel out when summing forces for the entire system.
Quick revision
The centre of mass is the point where the entire mass of a system is assumed to be concentrated for translational motion analysis.
Centre of Mass (Discrete): $R_{cm} = \frac{\sum m_i r_i}{\sum m_i}$
Centre of Mass (Continuous): $R_{cm} = \frac{1}{M} \int r \$, dm
Must remember
The position vector of the centre of mass for a discrete system is given by $R_{cm} = \frac{\sum m_i r_i}{\sum m_i}$.
The centre of mass depends solely on the mass distribution and is independent of the coordinate system chosen.
Internal forces cancel out in pairs; only external forces can change the acceleration of the centre of mass ($F_{ext} = M A_{cm}$).
If the net external force is zero, the centre of mass moves with constant velocity or remains at rest.
For continuous mass distributions, the summation is replaced by integration: $R_{cm} = \frac{1}{M} \int r \, dm$.
The centre of mass can lie outside the material body, such as in a uniform circular ring or a hollow sphere.
In the centre of mass frame (C-frame), the total linear momentum of the system is identically zero.
The centre of mass of a uniform semicircular disc of radius $R$ is located at a height of $Y_{cm} = \frac{4R}{3\pi}$ from its straight edge.
Glossary
Centre of MassThe point where the entire mass of a system may be considered concentrated for describing translational motion.
Discrete SystemA system composed of distinct, individual particles with separate masses and position vectors.
Continuous BodyAn object where mass is distributed continuously throughout a region, requiring integration for analysis.
Centre of Mass Frame (C-Frame)A non-inertial or inertial reference frame attached directly to the centre of mass of a system, where total momentum is zero.
Internal ForcesForces exerted by parts of the system upon one another, obeying Newton's third law.
Discrete vs Continuous Mass Distribution
Feature
Discrete System
Continuous System
Mathematical Operator
Summation ($\sum$)
Integration ($\int$)
Position Vector Formula
$R_{cm} = \frac{\sum m_i r_i}{\sum m_i}$
$R_{cm} = \frac{1}{M} \int r \, dm$
Application
Systems of distinct point masses
Rigid bodies and extended objects
Practice Quiz
Choose an option to reveal the explanation.
1. What is the position vector of the centre of mass for a discrete system of particles given by?
Explanation: The position vector of the centre of mass for a discrete system is the weighted average of the position vectors given by $R_{cm} = \frac{\sum m_i r_i}{\sum m_i}$.
2. Which of the following statements is true regarding the location of the centre of mass?
Explanation: The centre of mass is defined by spatial mass distribution and can lie outside the material body, as in hollow objects like rings and spherical shells.
3. If the net external force acting on a system of particles is zero, what can be said about the motion of its centre of mass?
Explanation: From $F_{ext} = M A_{cm}$, if $F_{ext} = 0$, then $A_{cm} = 0$, meaning the centre of mass moves with a constant velocity or remains at rest.
4. Why do internal forces not affect the acceleration of the centre of mass?
Explanation: Internal forces between particles occur in equal and opposite pairs (Newton's third law) and thus cancel out when summing forces for the entire system.
5. For a continuous mass distribution, how is the summation in the discrete formula replaced?
Explanation: For continuous bodies, discrete summation is replaced by integration over the mass elements: $R_{cm} = \frac{1}{M} \int r \, dm$.
6. What is the total linear momentum of a system when viewed from its centre of mass frame (C-frame)?
Explanation: By definition, the centre of mass frame is a zero-momentum frame where the total linear momentum of the system is always zero.
7. Where is the centre of mass of a uniform thin rod of length $L$ located, if the origin is taken at one of its ends?
Explanation: Due to uniform linear mass density, the centre of mass of a uniform rod of length $L$ lies at its geometric midpoint, i.e., $X_{cm} = \frac{L}{2}$ from either end.
8. What is the $Y$-coordinate of the centre of mass of a uniform semicircular disc of radius $R$?
Explanation: For a uniform semicircular disc of radius $R$ with its base on the x-axis, the centre of mass lies on the axis of symmetry at $Y_{cm} = \frac{4R}{3\pi}$.
9. Which of the following equations correctly expresses Newton's second law for the centre of mass of a system of total mass $M$?
Explanation: The motion of the centre of mass is governed solely by external forces: $F_{ext} = M A_{cm}$, as internal forces cancel out.
10. What does the term $dm$ represent in the continuous mass distribution formula for the centre of mass?
Explanation: In continuous systems, $dm$ represents an infinitesimal mass element of the extended body.
11. If two particles of masses $m_1$ and $m_2$ are separated by a distance $r$, how far is the centre of mass from mass $m_1$?
Explanation: Taking $m_1$ at the origin, the distance of the centre of mass from $m_1$ is given by $X_{cm} = \frac{0 \cdot m_1 + r \cdot m_2}{m_1 + m_2} = \frac{m_2 r}{m_1 + m_2}$.
12. Which reference frame is particularly useful for simplifying collisions and two-body central force problems?
Explanation: The centre of mass frame (zero-momentum frame) significantly simplifies analysis in collisions and two-body problems.
13. Does the position of the centre of mass depend upon the choice of the coordinate system?
Explanation: The physical location of the centre of mass relative to the body is invariant and independent of the arbitrary coordinate system chosen.
14. What is the shift in the centre of mass when a circular portion of area $A_2$ is removed from a uniform circular plate of original area $A_1$, with distance between their centers being $d$?
Explanation: The formula for the shift in the centre of mass when a part of area $A_2$ is removed from area $A_1$ is $\Delta X = \frac{A_2 d}{A_1 - A_2}$.
15. Which of the following physical quantities is conserved for an isolated system of particles where net external force is zero?
Explanation: When the net external force is zero, the total linear momentum of the system remains conserved.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. Two particles of masses $1\text{ kg}$ and $3\text{ kg}$ have position vectors $i + j + k$ and $-3i - 2j + k$ respectively. What is the position vector of their centre of mass?
2. A bomb initially at rest explodes into two pieces of masses in the ratio $1:3$. If the heavier piece moves with a velocity of $3i + 4j\text{ m/s}$, what is the velocity of the lighter piece?
Explanation: Since initial momentum is zero and no external force acts, total momentum remains zero. $m_1 v_1 + m_2 v_2 = 0 \implies 1 \cdot v_1 + 3(3i + 4j) = 0 \implies v_1 = -9i - 12j\text{ m/s}$.
3. The centre of mass of a uniform semicircular ring of radius $R$ from its geometric center is given by:
Explanation: For a uniform semicircular wire (ring), the centre of mass is located at $Y_{cm} = \frac{2R}{\pi}$ from the center along the axis of symmetry.
4. A uniform disc of radius $R$ has a circular hole of radius $R/2$ cut from it, such that the edge of the hole passes through the center of the disc. What is the shift in the centre of mass?
Explanation: Area of original disc $A_1 = \pi R^2$, area of hole $A_2 = \pi (R/2)^2 = \frac{\pi R^2}{4}$, distance between centers $d = R/2$. Shift $\Delta X = \frac{A_2 d}{A_1 - A_2} = \frac{(\pi R^2 / 4)(R / 2)}{\pi R^2 - \pi R^2 / 4} = \frac{R/8}{3/4} = \frac{R}{6}$.
5. A system consists of two masses $m$ and $2m$ moving towards each other with speeds $2v$ and $v$ respectively. The velocity of the centre of mass of the system is:
Explanation: Let one move in $+x$ direction ($2v$) and other in $-x$ direction ($v$). $V_{cm} = \frac{m(2v) + 2m(-v)}{m + 2m} = \frac{2mv - 2mv}{3m} = 0$.
6. If a man of mass $M$ stands at one end of a boat of mass $M$ and length $L$ floating at rest in still water, and walks to the other end, how much does the boat shift?
Explanation: Since no external horizontal force acts on the boat-man system, the centre of mass does not move. The shift of the boat is given by $\Delta x = \frac{m \cdot L}{m + M} = \frac{M \cdot L}{M + M} = \frac{L}{2}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following coordinates represents the $X$-centre of mass for a continuous body?
Explanation: In Cartesian coordinates, the $X$-coordinate of the centre of mass for a continuous body is $X_{cm} = \frac{1}{M} \int x dm$.
2. What is the total momentum of a system expressed in terms of total mass $M$ and velocity of centre of mass $V_{cm}$?
Explanation: The total linear momentum of any system of particles is equal to the total mass multiplied by the velocity of its centre of mass: $P_{total} = M V_{cm}$.
3. In a two-body system, if the centre of mass remains stationary and no external forces act, what can be said about their individual trajectories?
Explanation: By definition of the centre of mass, $\sum m_i r_i' = 0$ in the centre of mass frame, meaning the particles balance each other about the CM.
4. The acceleration of the centre of mass of a system depends upon:
Explanation: Internal forces cancel out pairwise, leaving only net external forces to determine the acceleration of the centre of mass ($F_{ext} = M A_{cm}$).
5. Where does the centre of mass of a uniform triangular lamina lie?
Explanation: For a uniform triangular lamina, the centre of mass is located at its geometric centroid, which is the point of intersection of its medians.
6. A shell explodes in mid-air into fragments. Ignoring air resistance, what happens to the path of its centre of mass?
Explanation: Internal forces during an explosion do not affect the centre of mass motion. Since gravity (external force) remains unchanged, the centre of mass continues its uninterrupted parabolic projectile path.
7. What is the unit of the position vector $R_{cm}$ in the centre of mass formula?
Explanation: The position vector represents a spatial length, so its SI unit is metres (m).
8. If mass is distributed uniformly along a semicircle of radius $R$, where is its centre of mass located?
Explanation: For a uniform semicircular arc (wire/ring), $Y_{cm} = \frac{2R}{\pi}$. (Note: for a semicircular disc it is $\frac{4R}{3\pi}$).
9. Under what condition is the velocity of the centre of mass constant?
Explanation: If $F_{ext} = 0$, then $A_{cm} = 0$, which implies that the velocity of the centre of mass ($V_{cm}$) remains constant.
10. Which of the following objects has its centre of mass located *outside* its physical body?
Explanation: A uniform circular ring is hollow at its geometric center, meaning its centre of mass lies at the empty center outside the material of the ring.