System of Particles and Rotational Motion — NEET notes

Centre of Mass

The centre of mass (COM) is the single point where the entire mass of a system can be imagined to be concentrated. Its motion follows Newton's second law for the whole system — even if individual parts move chaotically. NEET tests COM position for discrete particles, continuous bodies, and COM motion under external forces.

R _ cm = m_i r _i m_i | M a _ cm = F _ ext

What is Centre of Mass?

The centre of mass is a unique point where the weighted mean of the position of all particles lies. For a uniform symmetric body, it coincides with the geometric centre.

The centre of mass (COM) of a system is the point where the entire mass of the system can be considered to be concentrated for describing its translational motion. It is the mass-weighted average of the positions of all particles.

• COM is a mathematical point — it may or may not lie inside the body (e.g., COM of a ring is at its centre, where there's no mass)
• For a uniform symmetric body, COM lies at the geometric centre
• COM depends on mass distribution, not on the total mass alone
• The COM of the Earth-Moon system is inside the Earth (about 4600 km from Earth's centre)

COM for Discrete Particles

For a system of point masses, the COM coordinates are calculated as the mass-weighted average of individual coordinates.

X_ cm = m_1 x_1 + m_2 x_2 + + m_n x_n m_1 + m_2 + + m_n = m_i x_i M

Similarly for Y_ cm and Z_ cm . In vector form:

R _ cm = m_i r _i M where M = m_i

Two-particle special case (very common in NEET): For two masses m_1 and m_2 separated by distance d , taking m_1 at origin:

X_ cm = m_2 d m_1 + m_2

The COM divides the line joining the two particles in the inverse ratio of masses: m_1 r_1 = m_2 r_2 , where r_1 and r_2 are distances of the respective masses from the COM.

Quick recall

Two particles of masses 3 kg and 5 kg are at positions x = 2 m and x = 6 m. Where is the COM?

X_ cm = 3 2 + 5 6 3 + 5 = 6 + 30 8 = 36 8 = 4.5 m. The COM is closer to the heavier particle (at 6 m), as expected.

COM for Continuous Bodies

For continuous mass distributions, the summation becomes an integral. NEET requires knowing the COM positions for common uniform shapes (rod, disc, triangle, semicircle, hemisphere).

X_ cm = 1 M x dm ; Y_ cm = 1 M y dm

For uniform bodies, dm = M L dx (rod), M A dA (plate), or M V dV (solid). The integrals simplify because mass is spread evenly.

Motion of the Centre of Mass

The COM moves as if all external forces act on it. Internal forces (explosions, collisions) cannot change COM motion. This is extremely powerful for NEET problem-solving.

Differentiating the COM position equation gives the COM velocity and acceleration:

M V _ cm = m_i v _i = P _ total

M a _ cm = F _ ext (internal forces cancel out!)

• If F _ ext = 0 , then V _ cm = constant (COM moves uniformly or stays at rest)
• An exploding shell in projectile motion: COM continues the SAME parabolic path after explosion
• Two ice skaters pushing each other apart: COM stays at the same place (no external horizontal force)
• M V _ cm = P _ total links COM motion to linear momentum

Quick recall

A man (80 kg) is standing on one end of a 4 m long boat (40 kg) floating on still water. He walks to the other end. How far does the boat move?

No external horizontal force COM stays fixed. Let boat move distance d backward. 80 (4 - d) = (80 + 40) 0 + 40d is wrong approach. Use: m_ man displacement of man = m_ boat displacement of boat (opposite direction). 80(4 - d) = 40d 320 - 80d = 40d d = 320 120 = 8 3 2.67 m.

Torque

Torque ( = r F ) is the rotational analogue of force — it measures how effectively a force can cause rotation. A couple is a special pair of forces that creates pure rotation. NEET tests torque direction (right-hand rule), moment of a couple, and conditions for equilibrium.

= r F | = rF

Torque (Moment of Force)

Torque is the cross product of position vector and force. Its magnitude depends on force, distance from axis, and the angle between them. Only the perpendicular component of force causes rotation.

Torque (or moment of force) about a point O is defined as the cross product of the position vector r (from O to the point of application of force) and the force F :

= r F | | = rF

Here is the angle between r and F . The direction of is perpendicular to the plane containing r and F , given by the right-hand rule.

• SI unit: N m (same dimensions as energy, but NEVER called 'Joule' for torque)
• = rF = F r_ where r_ = r is the perpendicular distance (also called moment arm)
• If = 0° or 180° (force along r ) = 0 (no rotation)
• If = 90° (force perpendicular to r ) = rF (maximum torque)
• Torque is an axial vector — its direction is along the axis of rotation

Couple

A couple consists of two equal and opposite forces whose lines of action don't coincide. It produces pure rotation with zero net force. The moment of a couple is the same about every point.

A couple consists of two forces that are: (1) equal in magnitude, (2) opposite in direction, and (3) have different lines of action (separated by a perpendicular distance d ). The net force is zero, but the net torque is NOT zero.

_ couple = F d

where d is the perpendicular distance between the two lines of action (called the arm of the couple).

• Net force from a couple = zero (so COM doesn't accelerate)
• Net torque = F d (produces pure rotation)
• The moment of a couple is independent of the point about which it's calculated — this is unique to couples
• A couple can only be balanced by another couple of equal and opposite moment, NOT by a single force

Quick recall

Two forces of 10 N each act on a body. They are antiparallel and separated by 0.5 m. What is the torque?

= F d = 10 0.5 = 5 N m. The net force is zero (pure rotation), and this torque value is the same regardless of where you calculate it from.

Sign Convention & Direction of Torque

For 2D problems, anticlockwise torques are taken positive and clockwise negative. For 3D, use the right-hand rule. Correct sign convention is critical for equilibrium problems.

In 2D (which covers most NEET problems), the convention is:

• Anticlockwise (ACW) torque arrow positive (out of the page, + k )
• Clockwise (CW) torque arrow negative (into the page, - k )
• For 3D: use the right-hand rule — curl fingers from r to F , thumb gives direction of
• Net torque = sum of all individual torques (with signs)

Equilibrium condition for torques: For a body in rotational equilibrium, the sum of all torques about ANY point must be zero:

= 0 (rotational equilibrium)

Angular Momentum

Angular momentum ( L = r p = I ) is the rotational analogue of linear momentum. When no external torque acts, angular momentum is conserved — explaining why ice skaters spin faster by pulling arms in, why planets speed up near the Sun, and why a spinning top stays upright. Heavy NEET weightage.

L = I | If _ ext = 0, then L = constant

Angular Momentum

Angular momentum of a particle about a point is the cross product of position vector and linear momentum. For a rigid body rotating about a fixed axis, it simplifies to L = I .

The angular momentum of a particle about a point O is:

L = r p = r m v

Magnitude: L = rmv = mv r_ , where r_ is the perpendicular distance from O to the line of motion. Direction: perpendicular to the plane of r and v (right-hand rule).

For a rigid body rotating about a fixed axis with angular velocity :

L = I

• SI unit: kg m ^2 /s (or equivalently, J s)
• Angular momentum is an axial vector — direction along the rotation axis
• For a particle in circular motion: L = mvr (since = 90° , = 1 )
• The relation between torque and angular momentum: = d L dt (rotational analogue of F = d p dt )

Conservation of Angular Momentum

When the net external torque on a system is zero, the total angular momentum is conserved. This leads to spectacular physical phenomena and is one of NEET's most-tested principles.

From _ ext = d L dt , if _ ext = 0 :

L = I = constant I_1 _1 = I_2 _2

• If I decreases (mass moves closer to axis), increases — and vice versa
• Kepler's second law (equal areas in equal times) is actually conservation of angular momentum for planets
• A diver tucking into a ball spins faster (smaller I ), then opens up to slow down before entering the water
• A spinning neutron star (pulsar) contracts under gravity: I drops massively shoots up to ~700 rev/s

Quick recall

A disc of moment of inertia I is rotating at . A ring of moment of inertia 2I is gently placed on top. What is the new angular velocity?

No external torque L conserved. I = (I + 2I) ' ' = 3 . Note: kinetic energy is NOT conserved here — some is lost to friction between disc and ring.

Torque-Angular Momentum Relation

The rate of change of angular momentum equals the external torque — the rotational version of Newton's second law in its most general form.

_ ext = d L dt

This is the rotational analogue of F = d p dt . Just as force changes linear momentum, torque changes angular momentum. This is actually more fundamental than = I because it works even when I is changing.

Moment of Inertia

Moment of inertia ( I = m_i r_i^2 ) is the rotational analogue of mass — it quantifies how hard it is to change a body's rotational state. Its value depends on mass AND how that mass is distributed relative to the axis. NEET requires memorizing I for standard shapes: ring, disc, rod, sphere, cylinder.

I = m_i r_i^2 = r^2 dm

Moment of Inertia

Moment of inertia measures a body's resistance to angular acceleration. Unlike mass, it depends on the axis of rotation — the same body has different MOI about different axes.

The moment of inertia (MOI) of a system of particles about an axis is:

I = _ i=1 ^ n m_i r_i^2

where r_i is the perpendicular distance of the i -th particle from the axis of rotation. For continuous bodies: I = r^2 dm .

• SI unit: kg m ^2 . Dimensions: [ML^2]
• MOI is always positive and depends on: (1) total mass, (2) distribution of mass relative to axis
• Moving mass farther from the axis INCREASES I (since r^2 grows fast)
• MOI is additive: for a composite body, I_ total = I_1 + I_2 +
• The same body has DIFFERENT MOI about different axes — always specify the axis!

MOI for Standard Shapes

NEET requires instant recall of MOI formulas for ring, disc, solid sphere, hollow sphere, rod, and cylinder about their standard axes.

• Hollow bodies always have LARGER MOI than solid bodies of same mass and size (mass is farther from axis)
• For any body: MOI about centre < MOI about any parallel axis (parallel axis theorem guarantees this)
• Ring and hollow cylinder have the same formula ( MR^2 ) about their own axis
• Disc and solid cylinder have the same formula ( 1 2 MR^2 ) about their own axis

Theorems of MOI

The parallel axis theorem ( I = I_ cm + Md^2 ) and perpendicular axis theorem ( I_z = I_x + I_y ) let you calculate MOI about ANY axis once you know it about one axis. These save you from doing integrals every time. NEET heavily tests both theorems — especially in combination with the standard shapes table.

I = I_ cm + Md^2 | I_z = I_x + I_y (planar bodies only)

Parallel Axis Theorem

The MOI about any axis is equal to MOI about a parallel axis through the COM plus Md^2 , where d is the distance between the two axes. Works for ANY body, ANY shape.

I = I_ cm + Md^2

Here I_ cm is the MOI about an axis through the centre of mass, M is the total mass, and d is the perpendicular distance between the two parallel axes.

• The theorem ONLY works when one axis passes through the COM
• You CANNOT use it between two arbitrary parallel axes directly — go through COM first: I_1 = I_ cm + Md_1^2 and I_2 = I_ cm + Md_2^2
• Since Md^2 0 , the COM axis ALWAYS gives the minimum MOI among all parallel axes
• Works for ANY body (1D, 2D, 3D — rods, plates, spheres, anything)

Quick recall

Find MOI of a uniform disc (mass M , radius R ) about an axis tangent to the rim and perpendicular to the plane.

I_ cm = 1 2 MR^2 (about centre, to plane). Distance from centre to rim = R . So I = 1 2 MR^2 + MR^2 = 3 2 MR^2 .

Perpendicular Axis Theorem

For a planar body lying in the xy -plane, the MOI about the z -axis (perpendicular to the plane) equals the sum of MOIs about two perpendicular axes in the plane. This ONLY works for 2D/flat bodies.

I_z = I_x + I_y

where x , y , z are three mutually perpendicular axes meeting at a point, and the body lies entirely in the xy -plane. The z -axis is perpendicular to the plane of the body.

• Applicable ONLY to planar (2D) bodies: ring, disc, rectangular plate, triangular lamina, etc.
• NOT valid for: solid sphere, solid cylinder (about non-axis directions), solid cone, any 3D body
• For a symmetric planar body (ring, disc): I_x = I_y (by symmetry), so I_z = 2I_x = 2I_y
• This is how we derive MOI of a ring/disc about a diameter: I_z = 2I_ dia I_ dia = I_z/2

Example — Ring about diameter: I_z = MR^2 (perpendicular axis through centre). By symmetry I_x = I_y . So MR^2 = 2I_x I_x = 1 2 MR^2 .

Example — Disc about diameter: I_z = 1 2 MR^2 (perpendicular axis through centre). By symmetry I_x = I_y . So 1 2 MR^2 = 2I_x I_x = 1 4 MR^2 .

Rotational Kinematics

Rotational kinematics describes 'how' a body rotates — angular displacement ( ), angular velocity ( ), and angular acceleration ( ) — without worrying about 'why'. The equations are identical twins of the linear kinematic equations with direct substitutions. NEET loves converting between linear and angular quantities.

= _0 + t | = _0 t + 1 2 t^2 | ^2 = _0^2 + 2

Angular Displacement, Velocity & Acceleration

Every linear kinematic quantity has a rotational counterpart. Angular displacement is measured in radians, angular velocity in rad/s, and angular acceleration in rad/s².

Angular displacement is measured in radians (1 full revolution = 2 rad = 360°). It is a dimensionless ratio: = arc length radius .

Angular velocity = d dt (unit: rad/s). For uniform rotation: = 2 T = 2 f . Angular acceleration = d dt (unit: rad/s ^2 ).

• Angular displacement ( ): unit = radian (dimensionless), 2 rad = 360°
• Angular velocity ( ): unit = rad/s. Related to rpm: = 2 n 60 (where n = rpm)
• Angular acceleration ( ): unit = rad/s ^2
• and are axial vectors — direction along the axis of rotation (right-hand rule)

Rotational Kinematic Equations

For constant angular acceleration, the three kinematic equations have exact rotational counterparts. You already know them from linear motion — just swap symbols.

For constant angular acceleration , the three equations of rotational kinematics are:

= _0 + t

= _0 t + 1 2 t^2

^2 = _0^2 + 2

Quick recall

A wheel starting from rest reaches 120 rad/s in 4 seconds with uniform angular acceleration. Find and total angle turned.

= - _0 t = 120 - 0 4 = 30 rad/s ^2 . = _0 t + 1 2 t^2 = 0 + 1 2 (30)(16) = 240 rad. Number of revolutions = 240 2 38.2 .

Rotational Dynamics

Rotational dynamics answers 'why' a body's rotation changes: = I (Newton's second law for rotation). Rotational KE ( 1 2 I ^2 ) and the work-energy theorem for rotation complete the picture. NEET tests pulley problems with = I , energy methods for rolling, and power in rotation.

= I | KE_ rot = 1 2 I ^2 | W =

= I

The net external torque equals the product of moment of inertia and angular acceleration. This is the fundamental equation governing rotational dynamics, directly analogous to F = ma .

_ net = I

Just as F = ma tells us how force produces linear acceleration, = I tells us how torque produces angular acceleration. This is valid for rotation about a fixed axis when the moment of inertia I is constant.

• Use = I for fixed-axis rotation problems (pulleys, hinged rods, flywheels)
• Always check: torques about which axis? MOI about the SAME axis!
• For a pulley with mass: the tension on the two sides is NOT equal (the difference creates = I )
• Sign convention: choose a rotation direction as positive, then torques in that direction are positive

Quick recall

A solid disc (mass 2 kg, radius 0.5 m) has a string wound around it. A 1 kg block hangs from the string. Find of the disc. ( g = 10 m/s ^2 )

For the block: mg - T = ma 10 - T = a . For the disc: TR = 1 2 MR^2 and a = R . So T = 1 2 MR = 1 2 (2)(0.5) = 0.5 . Also a = 0.5 . Substituting: 10 - 0.5 = 0.5 = 10 rad/s ^2 .

Rotational Kinetic Energy

A spinning body has kinetic energy due to its rotation, given by 1 2 I ^2 . For a body that both translates and rotates (like a rolling ball), the total KE is translational + rotational.

KE_ rot = 1 2 I ^2

For a body that both translates (COM moves) and rotates (spins about COM):

KE_ total = 1 2 Mv_ cm ^2 _ translational + 1 2 I_ cm ^2 _ rotational

• Rotational KE is analogous to translational KE ( 1 2 mv^2 ), with m I and v
• For pure rotation (fixed axis): KE = 1 2 I ^2
• For rolling without slipping: use v = R to express total KE in terms of v or alone
• The ratio KE_ rot KE_ total depends only on the shape (through k^2/R^2 )

Work and Power in Rotation

Work done by a torque is , and instantaneous power is . The work-energy theorem also has a rotational version.

W = (constant torque)

P = (instantaneous power)

The work-energy theorem for rotation states:

W_ net = 1 2 I _f^2 - 1 2 I _i^2 = KE_ rot

Rolling Motion

Rolling without slipping combines translation and rotation with the constraint v = R . The total KE of a rolling body is split between translational and rotational parts. On inclines, the body with the smallest k^2/R^2 ratio wins the race. NEET heavily tests rolling on inclines — acceleration, velocity, and time expressions.

v_ cm = R | a = g 1 + k^2/R^2 | v = 2gh 1 + k^2/R^2

Rolling Without Slipping

Pure rolling is when the point of contact has zero velocity relative to the surface. This imposes the constraint v_ cm = R , linking translational and rotational motion.

Rolling without slipping (pure rolling) means the point of contact between the rolling body and the surface is momentarily at rest. This gives the fundamental constraint:

v_ cm = R and a_ cm = R

• Condition for pure rolling: v_ cm = R (no slipping at contact point)
• Velocity at contact point = 0; velocity at top = 2v_ cm
• Static friction provides the torque for rolling (NOT kinetic friction — the contact point doesn't slide)
• Static friction in pure rolling does no work (the point of application has zero velocity)
• If v > R : the body skids forward (like a car on ice). If v < R : the body spins in place (like wheels on mud)

Kinetic Energy of Rolling

A rolling body's total KE has two parts: translational ( 1 2 Mv^2 ) and rotational ( 1 2 I ^2 ). The fraction going to each depends on the body's shape.

KE_ total = 1 2 Mv_ cm ^2 + 1 2 I_ cm ^2 = 1 2 Mv^2 (1 + k^2 R^2 )

Using I = Mk^2 and v = R , we can express the KE split elegantly:

KE_ rot KE_ total = k^2/R^2 1 + k^2/R^2 ; KE_ trans KE_ total = 1 1 + k^2/R^2

Rolling on an Incline

When a body rolls without slipping down an incline, its acceleration is reduced compared to sliding. The acceleration depends on the body's shape through k^2/R^2 . This is one of NEET's most-tested topics in this chapter.

For a body rolling without slipping down an incline of angle :

a = g 1 + k^2 R^2

v = 2gh 1 + k^2 R^2

t = 1 2h g (1 + k^2 R^2 )

where h is the vertical height of the incline. The minimum friction required for pure rolling on an incline:

_ min = 1 + R^2/k^2

• Rolling acceleration is ALWAYS less than g (the sliding case) — rotation 'uses up' some of the PE
• The race result depends ONLY on k^2/R^2 (the shape), NOT on mass or radius
• Solid sphere always wins, ring always loses — regardless of their sizes
• A frictionless sliding object ( k^2/R^2 = 0 ) would beat ALL rolling objects
• The condition _s _ min must be satisfied for pure rolling; otherwise the body slides

Quick recall

A solid cylinder and a hollow cylinder of the same mass and radius are released simultaneously from the top of an incline. Which reaches the bottom first?

Solid cylinder ( k^2/R^2 = 1/2 ) has greater acceleration than hollow cylinder ( k^2/R^2 = 1 ). So the solid cylinder reaches the bottom first. The result is independent of mass and radius.

Equilibrium of Rigid Bodies

A rigid body is in mechanical equilibrium when both the net force AND net torque are zero — translational + rotational equilibrium simultaneously. Classic NEET problems involve ladders, beams, seesaws, and hanging signs. Choosing a smart pivot point simplifies calculations dramatically.

F = 0 AND = 0 (about ANY point)

Conditions for Equilibrium

Translational equilibrium requires zero net force (COM doesn't accelerate). Rotational equilibrium requires zero net torque (no angular acceleration). A rigid body needs BOTH for complete mechanical equilibrium.

For a rigid body to be in complete mechanical equilibrium, two conditions must be satisfied simultaneously:

F = 0 (translational equilibrium — COM doesn't accelerate)

= 0 (rotational equilibrium — no angular acceleration)

In component form (2D problems): F_x = 0 , F_y = 0 , and = 0 . These three equations allow you to solve for up to three unknowns.

• Translational equilibrium alone does NOT guarantee rotational equilibrium (and vice versa)
• A couple creates zero net force but non-zero torque — the body rotates but doesn't translate
• The torque condition = 0 must hold about every point if the body is in equilibrium — but you only need to check ONE (choose wisely!)
• For equilibrium on an incline: resolve forces parallel and perpendicular to the surface

Stable, Unstable & Neutral Equilibrium

Equilibrium is classified by what happens when the body is slightly displaced. Stable = returns, unstable = moves away, neutral = stays wherever displaced. The key factor is the position of the centre of gravity.

When a body in equilibrium is slightly displaced:

• Stable equilibrium: Body returns to original position. COM is at the lowest point. Potential energy is minimum. Example: ball at bottom of a bowl
• Unstable equilibrium: Body moves further away. COM is at the highest point. Potential energy is maximum. Example: ball balanced on top of a hill
• Neutral equilibrium: Body remains in new position. COM height doesn't change. Potential energy is constant. Example: ball on a flat surface

Ladder & Beam Problems

A ladder leaning against a smooth wall, a uniform beam with loads, or a hanging sign — these are the standard equilibrium problems. The method is always: identify forces, pick a smart pivot, and use = 0 .

The standard ladder problem: A uniform ladder of mass M and length L leans against a smooth (frictionless) wall at angle with the floor. The floor has friction. Find the forces and the minimum to prevent slipping.

Forces on the ladder: (1) Weight Mg downward at COM ( L/2 from base), (2) Normal from wall N_w horizontal at top, (3) Normal from floor N_f upward at base, (4) Friction f horizontal at base.

Applying equilibrium:

F_x = 0: f = N_w

F_y = 0: N_f = Mg

Taking torques about the base (to eliminate f and N_f ):

_ base = 0: N_w L = Mg L 2

N_w = Mg 2 _ min = f N_f = N_w Mg = 1 2

• Smooth wall = no friction at wall (only normal force N_w perpendicular to wall)
• Rough floor = both normal ( N_f ) and friction ( f ) act at the base
• The ladder's weight acts at its COM (midpoint for a uniform ladder)
• If a person stands on the ladder, add their weight at their position — then re-do the torque equation