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CBSE Class 11 · Physics

System of Particles and Rotational Motion

Official NCERT chapter from Physics Part I–II (book code keph1). ExamMaster notes are original teaching at CBSE Class 11 depth.

This lesson follows the official chapter “System of Particles and Rotational Motion” in Physics Part I–II. The words below are ExamMaster’s teaching, not a paste from the book. Use the NCERT chapter for the classroom sequence; use these notes to hold the idea without copying exercises or figures.

  • CBSE Class 11
  • Medium level
  • 11 concepts

1Centre of mass

The centre of mass is the mass-weighted average position. Two 2 kg and 4 kg on a line at x=0 and x=6 m sit with CM at 4 m. The system can be treated as the total mass at that point for translation. CM is not “the heaviest object”.

Guessing the midpoint 3 m ignores the 4 kg.

Figure. Two masses on one rod sit at 2 kg and 6 kg. The centre of mass splits the rod in the inverse-mass ratio 3:1, so it is three parts from the 2 kg end and one part from the 6 kg end. Boxes name the masses and are not drawn to mass scale; the 3:1 spans along the rod are to scale.

How it works

  1. Write each m and its x (or r)The given.
  2. Form Σ m x / Σ mThe CM.
  3. Read it as the translation-pointThe heading.

2 kg at 0, 4 kg at 6 m

Find the CM on the line.

  • Σ m x0+24=24
  • Σ m6 kg
  • x_cm4 m

Pro tip. Weighted average, not the midpoint.

2Motion of centre of mass

The CM moves as if F_net on the system acted on the total mass there: M a_cm = F_ext. Internal pairs cancel. A mid-air explosion can throw pieces apart while the CM still follows the parabola if gravity is the only external (as taught).

Watching a fragment and calling that the CM is a miss.

Figure. The centre of mass of a system moves as one particle that feels only the external force. The internal pair on m1 and m2 is equal and opposite, so it cannot shift the CM. F_ext and v_cm sit on the CM mark.

How it works

  1. Name F_ext and MThe system.
  2. Write a_cm = F_ext / MThe motion of CM.
  3. Keep internal bangs off this writeThey cancel.

3Linear momentum of a system of particles

Linear momentum of a system is Σ m_i v_i, equal to M v_cm. If F_ext=0, that total p stays. Two skaters pushing off: equal-opposite Δp. System p is not “the fastest piece”.

Adding speeds without masses is a miss.

Figure. The system's linear momentum is the vector sum of the particle momenta, and it equals M times the velocity of the centre of mass. The three arrows are to scale: lengths 2, 4 and 6 in the same units, so P is literally p1 plus p2.

How it works

  1. Write each m v, or M v_cmThe total p.
  2. Ask F_extConserve or not.
  3. Keep the unit kg·m/sThe system.

4Vector product of two vectors

A vector (cross) product a × b is a vector perpendicular to both, size ab sinθ, direction by the taught right-hand rule. Torque and angular momentum use this product. a × a = 0. A raw a b multiply is the wrong product.

Using a · b (a scalar) when a × b was asked is the heading-steal.

Figure. The magnitude of a vector product is the area of the parallelogram the two vectors span: |a × b| = ab sinθ. Here the angle is a true right angle (horizontal a, vertical b), so sinθ = 1 and the area is just ab. The direction is perpendicular to this page and is not drawn.

How it works

  1. Name the two vectors and the angle between themThe given.
  2. Write size ab sinθ and the perp directionThe cross.
  3. Use it next for τ = r × FThe job.

5Angular velocity and its relation with linear velocity

Angular velocity ω is the turn-rate; for a point at distance r on a rigid body, v = ω r (as taught, perpendicular). A 2 rad/s wheel of radius 0.4 m has rim speed 0.8 m/s. ω is not v.

Using v=ω/r is the flip-steal.

Figure. A point at perpendicular distance r from a fixed axis has linear speed v = ω r, and v is perpendicular to the radius. The radius is drawn horizontal and v is drawn vertical, so the right angle is true on the page. No circular rim is drawn.

How it works

  1. Name ω and rThe given.
  2. Write v = ω r for the rim-pointThe link.
  3. Keep ω in rad/s for this writeThe caution.

6Torque and angular momentum

Torque τ = r F sinθ (or r × F) turns a body; angular momentum L = Iω or r × p as taught. A 4 N at 0.5 m perpendicular is 2 N·m. Equilibrium of rotation needs net τ=0 as well as net F=0 for a rigid body.

A force through the axis (r=0 line of action) gives no this torque.

Figure. Torque about O is r times the force perpendicular to the rod: τ = r F. The rod is horizontal and F is vertical, so the lever arm is the rod itself. A dashed line through the axis is a line of action that would give zero torque — force must miss the axis.

How it works

  1. Name r, F, and the angleThe given.
  2. Write τ and, if asked, LThe pair.
  3. See that τ changes L as taught (τ = dL/dt)The link.

7Equilibrium of a rigid body

Rigid-body equilibrium: F_net=0 and τ_net=0 about a taught axis (any axis, as the lesson said, if both hold). A beam with two supports: the two normals and the weight must both-sum and both-torque to zero. Still-looking is not enough without the two tests.

Balancing F_net only on a rod that can spin about an end is a miss.

Figure. A rigid body is in equilibrium only when both the net force and the net torque vanish. Downward 6 N and 3 N are drawn at true 2:1 length; the support is 9 N up. Torques about O: 6×1 = 3×2. Arrow lengths follow the forces; the 1:2 spans follow the arms.

How it works

  1. Draw every forceThe body.
  2. Set ΣF=0 and Στ=0The two tests.
  3. Pick a convenient axis for τThe method.

8Moment of inertia

Moment of inertia I is the rotational inertia: Σ m r² about the axis (or the taught rod/disc writes). A 3 kg point at 0.4 m has I=0.48 kg·m². I depends on the axis. I is not mass.

Using m r as if it were I is a miss.

Figure. Moment of inertia about an axis is Σ m r². The two boxes are the same mass. The farther mass sits at twice the distance, so its I is four times the nearer one. Distance marks are to scale (r versus 2r). The boxes are the same size because the masses are equal, not because I is.

How it works

  1. Name the axis and each m, rThe given.
  2. Form Σ m r² or the taught shape-writeI.
  3. Keep kg·m²Rotational inertia.

9Kinematics of rotational motion about a fixed axis

Kinematics about a fixed axis mirrors the line: ω = ω0 + αt, θ = ω0 t + ½αt², ω² = ω0² + 2αθ, as taught. 2 rad/s speeding at 0.5 rad/s² for 4 s → ω=4 rad/s. Constant α is the pack’s gate.

Using v=u+at with metres when the unknown was ω is a label-swap.

Figure. Fixed-axis rotation uses the same linear kinematics as a straight-line v–t graph, with ω in place of v and α in place of a. The rising line is ω = 2 + t (α = 1) from ω0 = 2 at t = 0 to ω = 6 at t = 4. The dashed flat line is the α = 0 case that stays at ω0.

How it works

  1. Name ω0, α, t (or θ)The given.
  2. Pick the angular packThe kinematics.
  3. Keep rad, rad/s, rad/s²The labels.

10Dynamics of rotational motion about a fixed axis

Dynamics about a fixed axis: τ_net = I α. A 2 N·m on I=0.5 kg·m² gives α=4 rad/s². This is F=ma’s turn-cousin. A torque that is not the net is a stolen τ.

Using τ=Iω is the v/a mix-up.

Figure. Newton's second law for a fixed axis is τ = I α, the rotation twin of F = m a. Mass resists linear acceleration; moment of inertia resists angular acceleration. The dashed links are the analogy, not extra forces.

How it works

  1. Name τ_net and IThe given.
  2. Write α = τ_net / IThe dynamics.
  3. Then the kinematic pack if a time was askedThe pair.

11Angular momentum in case of rotations about a fixed axis

Angular momentum about a fixed axis is Iω (for the rigid case as taught). If τ_net=0, L stays — a pull-in of arms (I down, ω up) if the lesson used that demo. L-conserve is the gate τ_net=0, not “it is spinning”.

Using p-conserve as if it were L-conserve without the axis-story is a miss.

Figure. About a fixed axis, L = I ω. If the external torque is zero, L is the same number before and after I changes. The two bars are equal (both 12): dropping I from 6 to 2 triples ω from 2 to 6. The bars compare L, not I.

How it works

  1. Name I and ω, or r × pL.
  2. Ask τ_netConserve or not.
  3. If arms pull in, Iω stays so ω risesThe demo.
CM of 2 kg at 0 and 4 kg at 6 m is at
  1. 4 m
  2. 3 m
  3. 6 m

24/6=4, not the midpoint.

Notes

  • Mapped to the official NCERT chapter “System of Particles and Rotational Motion”. Original teaching only — no textbook sentences.
  • Science here is Physics, Chemistry and Biology ideas at this class, never a language or social-science chapter.

Formulas

  • x_cm=Σmx/Σm
  • v=ωr
  • τ=Iα
  • L=Iω (fixed-axis rigid as taught)

Recap

Hold these pegs from the official chapter “System of Particles and Rotational Motion”. The wording is ExamMaster’s teaching, not a textbook recap.

Centre of mass
The centre of mass is the mass-weighted average position.
Motion of centre of mass
The CM moves as if F_net on the system acted on the total mass there: M a_cm = F_ext.
Linear momentum of a system of particles
Linear momentum of a system is Σ m_i v_i, equal to M v_cm.
Vector product of two vectors
A vector (cross) product a × b is a vector perpendicular to both, size ab sinθ, direction by the taught right-hand rule.
Angular velocity and its relation with linear velocity
Angular velocity ω is the turn-rate; for a point at distance r on a rigid body, v = ω r (as taught, perpendicular).
Torque and angular momentum
Torque τ = r F sinθ (or r × F) turns a body; angular momentum L = Iω or r × p as taught.

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