System of Particles and Rotational Motion ISC Class 11: Notes

What System of Particles and Rotational Motion is about

Weightage: System of Particles and Rotational Motion belongs to unit 5 of the ISC Class XI Physics syllabus. CISCE weights units 4-6 together (Work, Energy and Power; Motion of System of Particles and Rigid Body; Gravitation) at 17 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Centre of mass of a two-particle system

The centre of mass (CM) is the point that moves as if the whole mass of the system were concentrated there and all external forces acted there. For two particles on a line: $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$; in general ${\vec{r}}_{cm}=\frac{m_1{\vec{r}}_1+m_2{\vec{r}}_2}{m_1+m_2}$. It lies on the line joining them, nearer the heavier particle; distances from the CM are inversely proportional to the masses ($m_1{r}_1=m_2{r}_2$). For many particles, ${\vec{r}}_{cm}=\frac{\sum m_i{\vec{r}}_i}{M}$ (no further detail is needed).

Worked example

Masses of 2 kg and 3 kg are placed at x = 0 and x = 5 m. Find the position of their centre of mass and its distance from each mass.

Try it first, then show the solutionSolution

1. $x_{cm}=\frac{2(0)+3(5)}{5}=3m$
2. Distances: 3 m from the 2 kg mass and 2 m from the 3 kg mass (check: 2 × 3 = 3 × 2).

Answer: x = 3 m; nearer the heavier mass

Motion of the centre of mass and momentum conservation

Differentiating $M{\vec{r}}_{cm}=\sum m_i{\vec{r}}_i$: $M{\vec{v}}_{cm}=\sum m_i{\vec{v}}_i=\vec{P}$, the total momentum. Differentiating again: $M{\vec{a}}_{cm}={\vec{F}}_{ext}$, because internal forces cancel in pairs (third law). So the CM of a system moves like a single particle of mass M under the total external force. If ${\vec{F}}_{ext}=0$, total momentum is conserved and ${\vec{v}}_{cm}$ is constant. Examples: a shell exploding in mid-air: the fragments scatter but their CM continues along the original parabola until a fragment lands; two skaters pushing apart: their CM stays at rest.

Worked example

A 1 kg particle moves at 4 m s⁻¹ along +x and a 3 kg particle at 2 m s⁻¹ along −x. Find the velocity of their centre of mass.

Try it first, then show the solutionSolution

1. $v_{cm}=\frac{1(4)+3(-2)}{1+3}=\frac{-2}{4}$
2. $v_{cm}=-0.5m{s}^{-1}$

Answer: 0.5 m s⁻¹ along −x

Centre of mass of a rigid body; centre of gravity

A rigid body is one in which the distances between particles do not change. For a uniform body with a centre of symmetry, the CM is at that centre: the midpoint of a uniform rod, the centre of a ring, disc or sphere, the intersection of the diagonals of a rectangular plate. The CM need not lie in the material (a ring). The centre of gravity (CG) is the point through which the total weight acts; in a uniform gravitational field it coincides with the CM. A body suspended freely comes to rest with its CG vertically below the point of suspension, which is how the CG of an irregular lamina is found.

Worked example

A uniform rod 1 m long and of mass 1 kg carries a 0.5 kg mass at one end. Find the centre of mass of the system from the other end.

Try it first, then show the solutionSolution

1. Rod's mass acts at its midpoint, 0.5 m from the free end; the 0.5 kg mass is at 1 m.
2. $x_{cm}=\frac{1(0.5)+0.5(1)}{1.5}=0.667m$

Answer: 0.67 m from the free end

Torque, couple and the principle of moments

The turning effect of a force about an axis is its torque (moment of force): $\vec{\tau }=\vec{r}\times \vec{F}$, magnitude $\tau =rF\sin \theta$ = force × perpendicular distance of its line of action from the axis (N m). Torque is greatest when the force is perpendicular to r and zero when it passes through the axis. A couple is a pair of equal, opposite, parallel forces with different lines of action; its torque is F × (perpendicular distance between them), the same about any point, and it causes rotation without translation. Principle of moments: for a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments. Equilibrium of a rigid body needs both $\sum \vec{F}=0$ and $\sum \vec{\tau }=0$, as in the ladder problem.

Worked example

On a see-saw pivoted at its centre, a 30 kg child sits 2 m from the pivot. Where must a 40 kg child sit to balance it?

Try it first, then show the solutionSolution

1. Principle of moments: 30g × 2 = 40g × d
2. d = 60/40 = 1.5 m on the other side

Answer: 1.5 m from the pivot

Rotational kinematics: comparing linear and rotational motion

For rotation about a fixed axis: angular displacement θ (rad), angular velocity $\omega =\frac{d\theta }{dt}$ (rad s⁻¹), angular acceleration $\alpha =\frac{d\omega }{dt}$ (rad s⁻²); every point has the same ω and α, with v = ωr and tangential acceleration $a_t=\alpha r$. For constant α: $\omega ={\omega }_0+\alpha t$; $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$; ${\omega }^2={\omega }_0^2+2\alpha \theta$. Comparison: displacement s ↔ θ; velocity v ↔ ω; acceleration a ↔ α; mass m ↔ moment of inertia I; force F ↔ torque τ; momentum p = mv ↔ L = Iω; F = ma ↔ τ = Iα; KE ½mv² ↔ ½Iω²; work Fs ↔ τθ; power Fv ↔ τω. Convert rpm to rad s⁻¹ by multiplying by 2π/60.

Worked example

A wheel speeds up uniformly from 600 rpm to 1200 rpm in 10 s. Find its angular acceleration and the number of revolutions made in this time.

Try it first, then show the solutionSolution

1. ω₀ = 600 × 2π/60 = 20π rad s⁻¹; ω = 40π rad s⁻¹
2. α = (40π − 20π)/10 = 2π = 6.28 rad s⁻²
3. θ = ½(ω₀ + ω)t = ½(60π)(10) = 300π rad = 150 revolutions

Answer: 6.28 rad s⁻²; 150 revolutions

Moment of inertia and radius of gyration

The moment of inertia of a body about an axis is $I=\sum m_i{r}_i^2$ (kg m²), where $r_i$ is the perpendicular distance of each particle from the axis. It measures rotational inertia and depends on the mass, how the mass is distributed and the position of the axis. The radius of gyration K is the distance from the axis at which the whole mass could be concentrated to give the same I: $I=M{K}^2$, $K=\sqrt{\frac{I}{M}}$. Standard results (formulae only): ring about its axis $M{R}^2$; thin hollow cylinder about its axis $M{R}^2$; disc about its axis $\frac{1}{2}M{R}^2$; solid cylinder about its axis $\frac{1}{2}M{R}^2$; thin rod about a perpendicular axis through its centre $\frac{M{L}^2}{12}$, through one end $\frac{M{L}^2}{3}$; solid sphere about a diameter $\frac{2}{5}M{R}^2$; thin hollow sphere about a diameter $\frac{2}{3}M{R}^2$. Mass far from the axis raises I: a ring has a larger I than a disc of the same mass and radius.

Worked example

Find the moment of inertia and the radius of gyration of a uniform disc of mass 2 kg and radius 0.2 m about its axis.

Try it first, then show the solutionSolution

1. $I=\frac{1}{2}M{R}^2=\frac{1}{2}\times 2\times 0.04=0.04kg{m}^2$
2. $K=\sqrt{\frac{I}{M}}=\sqrt{0.02}=0.141m$ (= $R/\sqrt{2}$)

Answer: 0.04 kg m²; 0.141 m

Theorems of parallel and perpendicular axes (statements)

Parallel axes theorem: the moment of inertia of a body about any axis equals its moment of inertia about a parallel axis through its centre of mass plus Md², where d is the distance between the axes: $I=I_{cm}+M{d}^2$. Perpendicular axes theorem (for a plane lamina only): the moment of inertia about an axis perpendicular to the plane equals the sum of the moments of inertia about two mutually perpendicular axes in the plane that meet where the perpendicular axis crosses it: $I_z=I_x+I_y$. Illustrations: rod about one end: $\frac{M{L}^2}{12}+M{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{3}$; ring about a diameter: by symmetry $I_x=I_y$, so $I_d=\frac{1}{2}M{R}^2$; disc about a diameter: $\frac{1}{4}M{R}^2$. (Proofs are not required.)

Worked example

Find the moment of inertia of a ring of mass M and radius R about a tangent in its own plane.

Try it first, then show the solutionSolution

1. Perpendicular axes: $I_x+I_y=M{R}^2$ and $I_x=I_y$, so about a diameter $I_d=\frac{1}{2}M{R}^2$.
2. The tangent in the plane is parallel to a diameter at distance R: $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Answer: $\frac{3}{2}M{R}^2$

Torque and angular acceleration; rotational kinetic energy

For a rigid body rotating about a fixed axis, $\tau =I\alpha$ (the rotational form of Newton's second law; scalar form only). Rotational kinetic energy: each particle has $\frac{1}{2}{m}_i{v}_i^2=\frac{1}{2}{m}_i{r}_i^2{\omega }^2$, so $K_{rot}=\frac{1}{2}I{\omega }^2$. Work done by a torque turning a body through θ: W = τθ; the work-energy theorem gives $\tau \theta =\frac{1}{2}I{\omega }^2-\frac{1}{2}I{\omega }_0^2$. Power P = τω. A flywheel with a large I stores rotational energy and keeps an engine turning smoothly.

Worked example

A flywheel of moment of inertia 2 kg m² starts from rest under a constant torque of 10 N m. Find its angular velocity and kinetic energy after 4 s, and check by finding the work done by the torque.

Try it first, then show the solutionSolution

1. α = τ/I = 10/2 = 5 rad s⁻²; ω = 5 × 4 = 20 rad s⁻¹
2. $K=\frac{1}{2}\times 2\times 20^2=400J$
3. θ = ½ × 5 × 4² = 40 rad; W = τθ = 10 × 40 = 400 J ✓

Answer: 20 rad s⁻¹; 400 J

Angular momentum and its conservation

The angular momentum of a particle about a point is $\vec{L}=\vec{r}\times \vec{p}$, magnitude $L=rp\sin \theta$ = mvr for circular motion. For a rigid body rotating about a fixed axis, L = Iω (kg m² s⁻¹ or J s). Torque is the rate of change of angular momentum: $\tau =\frac{dL}{dt}$ (scalar form). Law of conservation of angular momentum: if the net external torque on a system is zero, its total angular momentum is constant: $I_1{\omega }_1=I_2{\omega }_2$. Applications: an ice skater or ballet dancer pulls in the arms to spin faster; a diver curls up to turn more somersaults and straightens to slow down before entering the water; a planet moves faster when nearer the Sun (Kepler's second law). When I decreases, ω increases, and the kinetic energy $\frac{L^2}{2I}$ increases: the extra energy comes from the work done by the person's muscles.

Worked example

A skater spinning at 2 rad s⁻¹ with arms outstretched (I = 4 kg m²) pulls her arms in, reducing I to 1.6 kg m². Find her new angular velocity and the change in kinetic energy.

Try it first, then show the solutionSolution

1. $I_1{\omega }_1=I_2{\omega }_2$: $4\times 2=1.6{\omega }_2$, so ${\omega }_2=5rad{s}^{-1}$
2. KE before = ½ × 4 × 4 = 8 J; after = ½ × 1.6 × 25 = 20 J
3. Increase of 12 J, supplied by the work her arms do

Answer: 5 rad s⁻¹; KE rises from 8 J to 20 J

Physics in real life: System of Particles and Rotational Motion

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Centre of Mass and Its Motion
• Moment of Inertia, Torque and Angular Momentum

System of Particles and Rotational Motion formula sheet

Centre of mass

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$M{\vec{v}}_{cm}=\vec{P}$; $M{\vec{a}}_{cm}={\vec{F}}_{ext}$.

Torque

$\vec{\tau }=\vec{r}\times \vec{F},\tau =rF\sin \theta$

N m; zero if F passes through the axis.

Rotational kinematics

$\omega ={\omega }_0+\alpha t;\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2;{\omega }^2={\omega }_0^2+2\alpha \theta$

Constant α; v = ωr.

Moment of inertia and radius of gyration

$I=\sum m_i{r}_i^2=M{K}^2$

Depends on the axis.

Standard moments of inertia

$\text{ring }M{R}^2;\text{disc }\frac{1}{2}M{R}^2;\text{rod (centre) }\frac{M{L}^2}{12};\text{rod (end) }\frac{M{L}^2}{3};\text{solid sphere }\frac{2}{5}M{R}^2;\text{hollow sphere }\frac{2}{3}M{R}^2$

Solid cylinder ½MR², hollow cylinder MR² (about their axes).

Parallel axes theorem

$I=I_{cm}+M{d}^2$

Axes parallel; one through the CM.

Perpendicular axes theorem

$I_z=I_x+I_y$

Plane laminas only.

Rotational dynamics

$\tau =I\alpha ,L=I\omega ,\tau =\frac{dL}{dt}$

Scalar forms.

Rotational energy, work and power

$K=\frac{1}{2}I{\omega }^2=\frac{L^2}{2I},W=\tau \theta ,P=\tau \omega$

Analogues of ½mv², Fs and Fv.

Conservation of angular momentum

$I_1{\omega }_1=I_2{\omega }_2$

When the external torque is zero.

System of Particles and Rotational Motion: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks System of Particles and Rotational Motion

The question patterns to expect, and the method that works for each.

Tricks to remember System of Particles and Rotational Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

• The MoI ladder

  For the same M and R: hollow ring/cylinder (1) > hollow sphere (2/3) > disc/solid cylinder (1/2) > solid sphere (2/5).

  Use it when: Ranking moments of inertia.

• Rod: 12 at the centre, 3 at the end

  ML²/12 through the middle, ML²/3 through an end (the end is 4 times larger).

  Use it when: Rod problems and parallel-axis checks.

• Linear to rotational dictionary

  s θ, v ω, a α, m I, F τ, p L: swap each letter and the same equations work.

  Use it when: Writing rotational equations quickly.

• Parallel adds, perpendicular sums

  Parallel: add Md². Perpendicular: two in-plane axes add up to the one sticking out.

  Use it when: Choosing the right theorem.

• Pull in, speed up

  Skater rule: I down, ω up, KE up (muscles do work).

  Use it when: Conservation of angular momentum questions.

System of Particles and Rotational Motion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in System of Particles and Rotational Motion

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the wrong moment of inertia formula

  Students mix up ring and disc, or solid and hollow sphere, or use ML²/12 for a rod hinged at its end.

  Wrong working

  Rod hinged at one end: I = ML²/12.

  Right working

  About an end: ML²/3 (parallel axes: ML²/12 + M(L/2)²).

  ✓Do this: Write the body and the axis next to every I you use.

  Usually costs: 1-2 marks.

• ✗Mistake: Applying the perpendicular axes theorem to a solid body

  Students use $I_z=I_x+I_y$ for a sphere or a cylinder.

  Wrong working

  Solid sphere: I about a diameter = ½ × (2/5)MR².

  Right working

  The theorem applies only to flat laminas such as rings and discs; a sphere's I about every diameter is (2/5)MR².

  ✓Do this: Check the body is plane before using the perpendicular axes theorem.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting to convert rpm to rad s⁻¹

  Students put rpm directly into ½Iω² or ω = ω₀ + αt.

  Wrong working

  ω = 3000 rpm, so K = ½ × 0.025 × 3000².

  Right working

  ω = 3000 × 2π/60 = 314 rad s⁻¹, K = 1234 J.

  ✓Do this: Multiply rpm by 2π/60 first.

  Usually costs: 1 mark.

• ✗Mistake: Assuming KE is conserved when angular momentum is

  In skater and turntable problems, students conserve both, which is impossible when I changes.

  Wrong working

  ½I₁ω₁² = ½I₂ω₂², so ω₂ = ω₁√(I₁/I₂).

  Right working

  Only I₁ω₁ = I₂ω₂; KE changes (rises when the skater pulls in, falls when a child jumps on).

  ✓Do this: Conserve L; then compute KE before and after separately.

  Usually costs: 2 marks.

• ✗Mistake: Taking the full distance instead of the perpendicular distance for torque

  When a force acts at an angle, students multiply by r instead of r sin θ.

  Wrong working

  5 N at 30° to a 0.4 m spanner: τ = 0.4 × 5 = 2 N m.

  Right working

  τ = 0.4 × 5 × sin 30° = 1 N m.

  ✓Do this: Draw the line of action and the perpendicular from the axis.

  Usually costs: 1 mark.

Practice questions on System of Particles and Rotational Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of angular momentum is:
   (a) kg m s⁻¹
   (b) kg m² s⁻¹
   (c) kg m² s⁻²
   (d) N m

   Hint

   L = Iω.

   Answer

   (b) (also J s)

2. Q2 · 1 mark · MCQExam level

   For the same mass and radius, which has the largest moment of inertia about its symmetry axis?
   (a) disc
   (b) solid sphere
   (c) ring
   (d) solid cylinder

   Hint

   Mass far from the axis.

   Answer

   (c) ring (MR²)

3. Q3 · 1 mark · Very short answerExam level

   State the principle of moments.

   Hint

   Clockwise and anticlockwise.

   Answer

   For a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point.

4. Q4 · 1 mark · Very short answerExam level

   What are the rotational analogues of mass and of force?

   Hint

   τ = Iα.

   Answer

   Moment of inertia and torque.

5. Q5 · 2 marks · Short answerExam level

   Particles of 2 kg and 1 kg are at (−1, 4) m and (5, 1) m. Find their centre of mass.

   Hint

   Weighted average of each coordinate.

   Answer

   $x_{cm}=\frac{2(-1)+1(5)}{3}=1$ m; $y_{cm}=\frac{2(4)+1(1)}{3}=3$ m: the centre of mass is at (1 m, 3 m).

6. Q6 · 2 marks · Short answerHarder exam level

   A 5 N force is applied at the end of a 0.4 m spanner at 30° to the spanner. Find the torque.

   Hint

   rF sin θ.

   Answer

   τ = 0.4 × 5 × sin 30° = 1 N m.

7. Q7 · 2 marks · Short answerHarder exam level

   A wheel starting from rest has a constant angular acceleration of 4 rad s⁻². Find its angular velocity after 5 s and the number of revolutions made.

   Hint

   θ = ½αt².

   Answer

   ω = 20 rad s⁻¹; θ = 50 rad = 7.96 revolutions.

8. Q8 · 2 marks · Short answerHarder exam level

   Find the ratio of the radii of gyration of a thin hollow sphere and a solid sphere of the same mass and radius about their diameters.

   Hint

   K = √(I/M).

   Answer

   $\sqrt{2/3}:\sqrt{2/5}=\sqrt{5}:\sqrt{3}\approx 1.29$.

9. Q9 · 2 marks · Short answerHarder exam level

   A uniform rod of mass 2 kg and length 1 m rotates at 6 rad s⁻¹ about a perpendicular axis through its centre. Find its rotational kinetic energy.

   Hint

   I = ML²/12.

   Answer

   I = 1/6 kg m²; K = ½ × (1/6) × 36 = 3 J.

10. Q10 · 3 marks · Short answerHarder exam level

    A uniform rod of mass 3 kg and length 2 m: find its moment of inertia about a perpendicular axis through its centre, and use the parallel axes theorem to find it about a parallel axis through one end.

    Hint

    d = L/2.

    Answer

    I_cm = 3 × 4/12 = 1 kg m²; I_end = 1 + 3 × 1² = 4 kg m² (= ML²/3 ✓).

11. Q11 · 3 marks · Short answerHarder exam level

    A 2 kg particle moves at 3 m s⁻¹ along +x and a 1 kg particle at 6 m s⁻¹ along +y. Find the velocity of their centre of mass.

    Hint

    Vector weighted average.

    Answer

    ${\vec{v}}_{cm}=\frac{6\hat{i}+6\hat{j}}{3}=2\hat{i}+2\hat{j}$ m s⁻¹; magnitude 2.83 m s⁻¹ at 45° to the x-axis.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) State the relation between torque and angular momentum, and the law of conservation of angular momentum. [2]
    (b) Explain two applications of this law. [2]
    (c) A disc of moment of inertia 0.2 kg m² spins at 10 rad s⁻¹. An identical disc at rest is dropped coaxially onto it and they move together. Find the common angular velocity. [1]

    Hint

    I doubles.

    Answer

    (a) τ = dL/dt; if the external torque is zero, L is constant. (b) A skater pulls in her arms to spin faster (I falls, ω rises); a diver tucks to somersault faster and straightens to slow down. (c) 0.2 × 10 = 0.4 ω, ω = 5 rad s⁻¹ (half the KE is lost).

13. Q13 · 3 marks · Short answerHOTS / competency

    A 4 kg block hangs from a light string wound round a wheel of radius 0.5 m and moment of inertia 2 kg m², mounted on a frictionless horizontal axle. Find the acceleration of the block and the tension in the string. (g = 9.8 m s⁻²)

    Hint

    mg − T = ma and Tr = I(a/r).

    Answer

    $a=\frac{mg}{m+I/r^2}=\frac{39.2}{4+8}=3.27m{s}^{-2}$; T = 4(9.8 − 3.27) = 26.1 N.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Where is the centre of mass of a uniform rod?Show answer

   At its midpoint.

2. 2.What changes the velocity of the centre of mass?Show answer

   Only a net external force.

3. 3.Define torque.Show answer

   The turning effect of a force: τ = r × F, magnitude rF sin θ.

4. 4.MoI of a disc about its axis?Show answer

   ½MR².

5. 5.State the parallel axes theorem.Show answer

   I = I_cm + Md².

6. 6.Rotational KE?Show answer

   ½Iω².

7. 7.Relation between torque and angular momentum?Show answer

   τ = dL/dt.

8. 8.When is angular momentum conserved?Show answer

   When the net external torque is zero.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find the centre of mass of two particles and of a loaded rod?
• ☐Can you explain why an exploding shell's CM keeps its path?
• ☐Can you calculate a torque using the perpendicular distance?
• ☐Can you write the MoI of a ring, disc, rod, solid and hollow sphere and cylinders?
• ☐Can you use the parallel and perpendicular axes theorems?
• ☐Can you solve a flywheel or pulley problem with τ = Iα?
• ☐Can you apply conservation of angular momentum to a skater or turntable?
• ☐Can you list five pairs of linear and rotational analogues?

System of Particles and Rotational Motion: frequently asked questions