System of Particles and Rotational Motion Notes
Study Notes
Topics
7📖 1. Chapter Overview
Overview
This chapter extends mechanics from a single particle to a system of particles and rigid bodies. In translational motion, every point of a body moves equally, while in rotational motion, different points move in circles about an axis. The centre of mass represents the average position of mass and helps replace a complex system by a single point for translational motion. Rotational motion introduces angular displacement, angular velocity, torque, angular momentum and moment of inertia. Newton’s laws get rotational forms such as τ = Iα. The chapter also covers rolling motion and equilibrium of rigid bodies. For NEET, this is a high-value chapter with frequent questions on centre of mass, torque, moment of inertia, rolling and angular momentum conservation.
- 1Internal forces cannot change motion of centre of mass of a system.
- 2External force changes linear momentum of a system: F_ext = dP/dt.
- 3Torque direction is given by right-hand rule.
- 4Angular momentum is conserved when net external torque is zero.
- 5Moment of inertia is larger when mass is farther from the axis.
- 6For rolling, total kinetic energy is translational plus rotational.
- 7Equilibrium of rigid bodies needs translational and rotational balance simultaneously.
Rotational Dictionary
Replace x by θ, v by ω, a by α, m by I, F by τ and p by L.
COM Shortcut
Centre of mass is the balance point of mass distribution.
Daily Life Example
A door opens easily when pushed near the handle because the lever arm is large, producing greater torque.
NEET Quick Check
If no external torque acts on a rotating system, angular momentum remains conserved even if moment of inertia changes.
Confusing Centre of Mass and Centre of Gravity
They coincide in a uniform gravitational field, but conceptually centre of mass depends on mass distribution while centre of gravity depends on gravitational force distribution.
Using Mass Instead of Moment of Inertia
In rotation, resistance depends not just on mass but also on how far mass is from the axis.
Ignoring Torque Direction
Torque is a vector; clockwise and anticlockwise torques must be given opposite signs in equilibrium.
Position vector of centre of mass of a system of particles.
Variables
R_cm=Position vector of centre of mass
m_i=Mass of ith particle
r_i=Position vector of ith particle
Magnitude of turning effect of a force about a point or axis.
Variables
τ=Torque
r=Distance from axis to point of force application
F=Force
θ=Angle between r and F
⚖️ 2. Centre of Mass
Overview
A system of particles may contain many masses moving differently, but its overall translational motion can be described using a special point called the centre of mass. Centre of mass is the mass-weighted average position of all particles. For two particles, it lies closer to the heavier mass. For multiple particles, coordinates are found using weighted averages. For continuous bodies, summation becomes integration. In symmetric bodies of uniform density, centre of mass lies at the geometric centre. The motion of centre of mass depends only on external forces, not internal forces. This idea simplifies explosions, collisions, recoil and motion of extended bodies in NEET problems.
- 1COM may lie outside the material body, such as in a ring.
- 2Choice of origin changes coordinates of COM but not its physical location.
- 3For equal masses, COM is simple average of positions.
- 4For a uniform rod, COM lies at its midpoint.
- 5For a triangular lamina, COM lies at the centroid.
- 6A body can rotate about its COM while the COM translates like a particle.
COM Location
COM is the mass balance point; heavier side pulls it closer.
Symmetry Trick
If a uniform body has symmetry, COM lies on every symmetry line; their intersection gives COM.
Two Particle Example
Masses 2 kg at x = 0 and 3 kg at x = 10 m have x_cm = (2×0 + 3×10)/5 = 6 m.
Equal Mass Example
Three equal masses at x = 0, 3 and 6 m have x_cm = (0 + 3 + 6)/3 = 3 m.
Assuming COM Always Lies Inside Body
COM can lie outside the material, such as at the centre of a ring or horseshoe-like shape.
Using Simple Average for Unequal Masses
For unequal masses, use weighted average, not ordinary average.
Forgetting External Force Condition
COM moves uniformly only when net external force is zero, not merely when internal forces are present.
Position of COM of two particles on a line.
Variables
x_cm=Centre of mass coordinate
m1, m2=Masses of two particles
x1, x2=Coordinates of particles
Vector formula for COM of a system of particles.
Variables
R_cm=Position vector of centre of mass
m_i=Mass of ith particle
r_i=Position vector of ith particle
🚀 3. Linear Momentum of Systems
Overview
The linear momentum of a system is the vector sum of momenta of all particles. It is also equal to total mass multiplied by velocity of the centre of mass. Newton’s second law for a system states that the rate of change of total momentum equals the net external force. Internal forces occur in equal and opposite pairs, so they cancel in the total momentum equation and cannot change the total momentum of an isolated system. Therefore, when net external force is zero, linear momentum is conserved. This explains recoil of a gun, explosion of bodies and basic rocket propulsion. In rockets, gases expelled backward give the rocket forward momentum.
- 1Momentum conservation is vector conservation; use signs carefully.
- 2Internal forces may change individual particle momenta but not total system momentum.
- 3Recoil velocity is opposite to bullet velocity.
- 4In an explosion initially at rest, total final momentum remains zero.
- 5Rocket propulsion does not require air; it works by momentum conservation.
- 6Centre of mass velocity remains constant for an isolated system.
System Momentum
Whole system momentum changes only when outside pushes it.
Recoil Rule
Something goes forward, something must go backward if total momentum was zero.
Recoil Example
A 5 kg gun fires a 0.05 kg bullet at 200 m/s. Recoil velocity V = -(0.05×200)/5 = -2 m/s.
Explosion Example
A body at rest explodes into two equal fragments. If one moves at +10 m/s, the other moves at -10 m/s.
Forgetting Vector Signs
Momentum conservation must include direction; opposite velocities have opposite signs.
Counting Internal Forces as External
For the selected system, forces between its parts are internal and cancel in total momentum calculation.
Thinking Rocket Needs Air
Rocket propulsion works in space because it is based on momentum conservation, not pushing against air.
Total momentum is vector sum of momenta of all particles.
Variables
P=Total linear momentum of system
m_i=Mass of ith particle
v_i=Velocity of ith particle
Total momentum equals total mass times velocity of centre of mass.
Variables
M=Total mass of system
V_cm=Velocity of centre of mass
Net external force determines rate of change of system momentum and acceleration of COM.
Variables
F_ext=Net external force
dP/dt=Rate of change of total momentum
a_cm=Acceleration of centre of mass
🔄 4. Torque & Angular Momentum
Overview
Torque, also called moment of force, measures the ability of a force to produce rotation about an axis. Its magnitude is τ = rF sinθ, where r is the position vector from axis to point of application and θ is the angle between r and F. Torque direction is given by the right-hand rule. A couple consists of two equal and opposite parallel forces with different lines of action and produces pure rotation. Angular momentum is rotational momentum, defined as L = r × p for a particle and L = Iω for a rigid body rotating about a fixed axis. Net external torque equals rate of change of angular momentum. If external torque is zero, angular momentum is conserved.
- 1Torque is maximum when force is perpendicular to position vector.
- 2Torque is zero if force passes through the axis of rotation.
- 3Right-hand rule gives direction of torque and angular momentum.
- 4A couple has zero resultant force but non-zero torque.
- 5A spinning skater rotates faster by bringing arms inward because moment of inertia decreases.
- 6Planetary motion conceptually follows angular momentum conservation under central gravitational force.
Torque
Torque needs force and arm: bigger handle, easier rotation.
Angular Momentum
No external twist means spin quantity stays fixed.
Torque Example
A 20 N force acts perpendicular at 0.5 m from a hinge. Torque = 0.5×20 = 10 N m.
Skater Example
If a skater reduces moment of inertia from 4 to 2 kg m², angular speed doubles to conserve Iω.
Using Distance Instead of Perpendicular Distance
Torque uses perpendicular distance from axis to line of action, not always the direct distance.
Forgetting Direction
Torque and angular momentum are vectors; use clockwise and anticlockwise signs or right-hand rule.
Applying Angular Momentum Conservation with External Torque
Angular momentum is conserved only when net external torque is zero.
Torque is the cross product of position vector and force.
Variables
τ=Torque vector
r=Position vector from axis to point of force application
F=Force vector
Magnitude of torque depends on force and perpendicular distance from axis.
Variables
θ=Angle between r and F
lever arm=Perpendicular distance from axis to line of action of force
Torque produced by a couple equals one force multiplied by perpendicular separation.
Variables
F=Magnitude of either force
d=Perpendicular distance between lines of action
🛞 5. Moment of Inertia
Overview
Moment of inertia is the rotational analogue of mass. It measures how difficult it is to change the rotational motion of a body about a given axis. Unlike mass, moment of inertia depends not only on the amount of matter but also on how far the matter is distributed from the axis. Mathematically, I = Σmr². Radius of gyration is the distance at which the whole mass could be imagined to be concentrated to give the same moment of inertia. The parallel axis theorem shifts moment of inertia from a centre of mass axis to a parallel axis. The perpendicular axis theorem applies to plane laminae. NEET frequently asks standard moments of inertia of ring, disc, rod, cylinder and sphere.
- 1Moment of inertia is not unique for a body; it changes with axis.
- 2Parallel axis theorem uses distance between parallel axes.
- 3Perpendicular axis theorem is valid only for plane laminae.
- 4For same mass and radius, ring has greater moment of inertia than disc.
- 5Hollow bodies generally have greater moment of inertia than solid bodies of same mass and radius.
- 6Moment of inertia has SI unit kg m².
Ring vs Disc
Ring keeps all mass at rim, so it has bigger I than disc.
Axis Theorems
Parallel adds Md²; perpendicular adds two in-plane moments.
Parallel Axis Example
A rod has I_cm = 1/12 ML². About one end, I = 1/12 ML² + M(L/2)² = 1/3 ML².
Radius of Gyration Example
For a ring, I = MR² = MK², so radius of gyration K = R.
Using Wrong Axis Formula
Moment of inertia formula depends on the axis; same body has different I for different axes.
Applying Perpendicular Axis Theorem to 3D Bodies
Perpendicular axis theorem is only for plane laminae.
Forgetting Square Dependence
Moment of inertia depends on r², so distant mass contributes strongly.
Sum of each mass element times square of perpendicular distance from rotation axis.
Variables
I=Moment of inertia
m_i=Mass element
r_i=Perpendicular distance from axis
K is the distance at which total mass may be concentrated to give the same I.
Variables
M=Total mass
K=Radius of gyration
🔁 6. Rotational Kinematics
Overview
Rotational kinematics describes rotation without discussing its cause. Angular displacement measures the angle swept by a radius vector, angular velocity is its rate of change and angular acceleration is the rate of change of angular velocity. For constant angular acceleration, equations similar to linear motion apply: ω = ω0 + αt, θ = ω0t + 1/2αt² and ω² = ω0² + 2αθ. Linear and angular quantities are connected by s = rθ, v = rω and a_t = rα. Period is time for one revolution and frequency is revolutions per second. Rolling motion begins when translation and rotation combine, and pure rolling has v_cm = Rω.
- 1All points of a rigid body have same angular velocity about a fixed axis.
- 2Linear speed is larger for points farther from the axis.
- 3Angular quantities may be positive or negative depending on rotation direction.
- 4Uniform circular motion has constant ω but non-zero centripetal acceleration.
- 5Tangential acceleration changes speed; centripetal acceleration changes direction.
- 6In rolling, the point of contact is instantaneously at rest for pure rolling.
Equation Conversion
Convert linear to rotational by replacing s→θ, v→ω and a→α.
Rolling Condition
No slipping means arc unrolled equals distance moved: s = Rθ.
Angular Speed Example
A wheel makes 2 revolutions per second. ω = 2πf = 4π rad/s.
Rolling Example
A wheel of radius 0.5 m rolls without slipping at angular speed 10 rad/s. v_cm = Rω = 5 m/s.
Using Degrees in Equations
Rotational equations require angular displacement in radians.
Assuming Same Linear Speed for All Points
In a rotating rigid body, angular speed is same but linear speed depends on radius.
Confusing Tangential and Centripetal Acceleration
Tangential acceleration changes speed; centripetal acceleration changes direction.
Rate of change of angular displacement.
Variables
ω=Angular velocity
θ=Angular displacement
t=Time
Rate of change of angular velocity.
Variables
α=Angular acceleration
ω=Angular velocity
Valid for constant angular acceleration.
Variables
ω0=Initial angular velocity
ω=Final angular velocity
α=Constant angular acceleration
θ=Angular displacement
⚙️ 7. Rotational Dynamics
Overview
Rotational dynamics studies the causes of rotational motion. Just as net force produces linear acceleration, net torque produces angular acceleration according to τ = Iα for fixed-axis rotation. Work done in rotating a body is related to torque and angular displacement, and power in rotational motion is P = τω. A rotating rigid body has rotational kinetic energy 1/2 Iω². Rolling without slipping combines translational kinetic energy of the centre of mass and rotational kinetic energy about the centre of mass. This explains wheels, pulleys and rolling bodies on inclines. In pulley systems, tension may produce torque if the pulley has moment of inertia, changing the acceleration compared with a massless pulley.
- 1Net torque, not individual torque, decides angular acceleration.
- 2Moment of inertia plays the role of mass in rotation.
- 3If moment of inertia is larger, angular acceleration is smaller for the same torque.
- 4Rolling friction may be absent in ideal pure rolling, but static friction can provide torque.
- 5For rolling down an incline, acceleration depends on I/MR².
- 6A massive pulley has rotational kinetic energy and cannot be treated like a massless pulley.
Rotation Newton Law
Force makes mass accelerate; torque makes moment of inertia angularly accelerate.
Rolling Energy
Rolling body pays energy in two accounts: translation plus rotation.
Torque-Acceleration Example
A wheel with I = 2 kg m² experiences torque 10 N m. Angular acceleration α = τ/I = 5 rad/s².
Rolling Energy Example
A solid cylinder has I = 1/2 MR². Its rolling kinetic energy is 1/2 Mv² + 1/4 Mv² = 3/4 Mv².
Using τ = Iα with Wrong Axis
Moment of inertia and torque must be calculated about the same axis.
Treating Massive Pulley as Massless
If pulley has moment of inertia, tensions on two sides are generally unequal.
Forgetting Rotational KE in Rolling
A rolling body has both translational and rotational kinetic energy.
Net torque about a fixed axis equals moment of inertia times angular acceleration.
Variables
τ_net=Net torque
I=Moment of inertia
α=Angular acceleration
Kinetic energy due to rotation about an axis.
Variables
K_rot=Rotational kinetic energy
ω=Angular speed
Work done by torque during angular displacement; for constant torque W = τθ.
Variables
W=Work done
τ=Torque
θ=Angular displacement
🏛️ 8. Equilibrium of Rigid Bodies
Overview
A rigid body is in equilibrium when it has neither translational acceleration nor angular acceleration. Therefore, two conditions must be satisfied: net external force must be zero and net external torque about any point must be zero. Translational equilibrium prevents linear motion, while rotational equilibrium prevents angular motion. Static equilibrium is the special case where the body remains at rest. Torque analysis is essential for ladders, beams, seesaws, doors and balances. Centre of gravity is the point where the weight of a body may be considered to act. Equilibrium can be stable, unstable or neutral depending on how the body responds to small displacement from its position.
- 1Torque can be calculated about any point in equilibrium problems.
- 2Choosing the pivot at an unknown force can eliminate that force from torque equation.
- 3For a rigid body, two equal and opposite non-collinear forces can produce rotation.
- 4Centre of gravity and centre of mass coincide in uniform gravitational field.
- 5A wide base and low centre of gravity increase stability.
- 6For static equilibrium, both linear and angular velocities are zero.
Rigid Body Equilibrium
No slide and no spin: ΣF = 0 and Στ = 0.
Torque Balance
Choose pivot where unknown forces act to remove them from torque equation.
Stability
Low centre of gravity and wide base make a body harder to topple.
Seesaw Example
A 300 N child sits 2 m from pivot. A 600 N child balances by sitting 1 m from pivot because 300×2 = 600×1.
Stability Example
A bus is more likely to topple when its centre of gravity is high and base width is small.
Checking Only Force Balance
A rigid body may have zero net force but still rotate if net torque is non-zero.
Using Wrong Lever Arm
Lever arm is perpendicular distance from pivot to line of action of force.
Taking Torque About Different Points Incorrectly
In equilibrium, torque balance can be written about any one point, but all torques in that equation must be about the same point.
Vector sum of all external forces must be zero.
Variables
ΣF=Net external force
Algebraic sum of torques about any point must be zero.
Variables
Στ=Net external torque
Formula Sheet
10Position vector of centre of mass of a system of particles.
Variables
R_cm=Position vector of centre of mass
m_i=Mass of ith particle
r_i=Position vector of ith particle
Magnitude of turning effect of a force about a point or axis.
Variables
τ=Torque
r=Distance from axis to point of force application
F=Force
θ=Angle between r and F
Rotational inertia of a system about a given axis.
Variables
I=Moment of inertia
m_i=Mass element
r_i=Perpendicular distance from axis
Rotational form of Newton’s second law for fixed-axis rotation.
Variables
τ=Net torque
I=Moment of inertia
α=Angular acceleration
Position of COM of two particles on a line.
Variables
x_cm=Centre of mass coordinate
m1, m2=Masses of two particles
x1, x2=Coordinates of particles
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NEET PYQs — System of Particles and Rotational Motion
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A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre C as shown in figure. The moment of inertia of the ring about an axis yy′ will be:
The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is:
A sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is:
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