Class 11 Physics Chapter 6: Rotational Motion & Centre of Mass

01 / Concept Mastery

Comprehensive NCERT Guide

Thorough theoretical foundations, pristine stacked KaTeX derivations, clean diagrams, and stepwise worked examples 6.1 to 6.12.

12 SECTIONS

Section Directory

6.1 Rigid Body Motion 6.2 Centre of Mass 6.3 Motion of CM 6.4 Linear Momentum of System 6.5 Vector Cross Product 6.6 Angular Velocity & Acceleration 6.7 Torque & Angular Momentum 6.8 Equilibrium of Rigid Bodies 6.9 Moment of Inertia & Gyration 6.10 Rotational Kinematics 6.11 Dynamics of Rotation 6.12 Fixed Axis Angular Momentum

6.1

NCERT Section

Introduction & Motion of a Rigid Body

Fundamentals

6.1.1 Rigid Body & Types of Motion

An ideal rigid body is one in which the distances between all pairs of constituent particles remain strictly unchanged under external forces. In real extended bodies, deformations are often negligible.

Pure Translation: At any given instant, every particle of the body moves with the exact same velocity ($\mathbf{v}_1 = \mathbf{v}_2 = \dots = \mathbf{v}_n$). The orientation of any line segment inside the body remains constant with time.
Pure Rotation about a Fixed Axis: Every particle of the body moves in a circle whose plane is perpendicular to the fixed axis with its centre lying on the axis. All particles on the axis of rotation remain stationary ($r = 0$). All particles share the same instantaneous angular velocity $\omega$.
Combined Translation and Rotation (Rolling): The motion of the body is a superposition of translation of its centre of mass and rotation about an axis passing through the centre of mass. In pure rolling down an incline, the instantaneous point of contact has zero velocity ($v_\text{contact} = 0$).

Figure 6.1: Comparison of rigid body motions: (a) Pure translation, (b) Rotation about a fixed z-axis, (c) Rolling as a superposition of translation and rotation.

Interactive 3D Pure Rolling on Incline ($v_\text{contact} = 0$)

Dynamics Lab

Rotate incline perspective

Topmost Point Velocity

$v_\text{top} = 2 v_\text{cm}$ (Rose Arrow)

Centre of Mass Velocity

$v_\text{cm} = \omega R$ (Blue Arrow)

Contact Point Velocity

$v_\text{contact} = 0$ (Instantaneous Rest)

6.2

NCERT Section

Centre of Mass (CM)

Mathematical Definition

6.2.1 Two-Particle, N-Particle & Continuous Mass Systems

The Centre of Mass (CM) of a system of particles is the mass-weighted average position of all its constituent particles.

Centre of Mass Master Formulations 1. Two-Particle System on x-axis: $$X_\text{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$ $$\text{If } m_1 = m_2 = m \implies X_\text{cm} = \frac{x_1 + x_2}{2} \quad (\text{Midpoint})$$ 2. General System of $n$ Particles in 3D Space: $$X_\text{cm} = \frac{\sum_{i=1}^n m_i x_i}{M}, \quad Y_\text{cm} = \frac{\sum_{i=1}^n m_i y_i}{M}, \quad Z_\text{cm} = \frac{\sum_{i=1}^n m_i z_i}{M}$$ $$\mathbf{R}_\text{cm} = \frac{\sum_{i=1}^n m_i \mathbf{r}_i}{M} \quad \text{where } M = \sum_{i=1}^n m_i$$ 3. Continuous Mass Distribution: $$\mathbf{R}_\text{cm} = \frac{1}{M} \int \mathbf{r}\, dm \implies X_\text{cm} = \frac{1}{M}\int x\, dm, \quad Y_\text{cm} = \frac{1}{M}\int y\, dm, \quad Z_\text{cm} = \frac{1}{M}\int z\, dm$$

Symmetry Principle: For homogeneous regular bodies (uniform thin rods, rings, discs, spheres, cylinders, rectangular plates), the centre of mass coincides with the geometric centre by reflection symmetry. For an origin chosen at the CM, $\sum m_i \mathbf{r}_i = 0$ or $\int \mathbf{r}\, dm = 0$.

Interactive 3D Centre of Mass of a 3-Particle System

Particle Cluster Lab

Drag to rotate in 3D

Centre of Mass Coordinates $\mathbf{R}_\text{cm} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i}$

(0.00, 0.00) • Total Mass: 3.0 kg

Mass A ($m_1$):

Mass B ($m_2$):

Mass C ($m_3$):

Example 6.1 • CM of Three Particles on an Equilateral Triangle

NCERT Solved

Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are $100\text{ g}$, $150\text{ g}$, and $200\text{ g}$ respectively. Each side of the equilateral triangle is $0.5\text{ m}$ long.

Final Answer: $(X_\text{cm}, Y_\text{cm}) = \left(\frac{5}{18}\text{ m}, \frac{1}{3\sqrt{3}}\text{ m}\right) \approx (0.28\text{ m}, 0.19\text{ m})$

Step 1: Set up coordinates of the three vertices: Let origin $O(0, 0)$ be at the $100\text{ g}$ mass. Point $A(0.5, 0)$ along x-axis holds the $150\text{ g}$ mass. Point $B(0.25, 0.25\sqrt{3})$ holds the $200\text{ g}$ mass. Total mass $M = 100 + 150 + 200 = 450\text{ g}$.

Step 2: Calculate $X_\text{cm}$: $$X_\text{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{100(0) + 150(0.5) + 200(0.25)}{450} = \frac{75 + 50}{450} = \frac{125}{450} = \mathbf{\frac{5}{18}\text{ m}}$$

Step 3: Calculate $Y_\text{cm}$: $$Y_\text{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} = \frac{100(0) + 150(0) + 200(0.25\sqrt{3})}{450} = \frac{50\sqrt{3}}{450} = \frac{\sqrt{3}}{9} = \mathbf{\frac{1}{3\sqrt{3}}\text{ m}}$$

Example 6.2 • Centre of Mass of a Triangular Lamina

NCERT Solved

Find the centre of mass of a uniform triangular lamina.

Conclusion: The centre of mass coincides with the centroid $G$ of the triangle.

Step 1: Slice lamina into parallel strips: Subdivide the triangular lamina $\Delta LMN$ into narrow strips parallel to the base $MN$. By symmetry, the centre of mass of each individual strip lies at its midpoint.

Step 2: Intersection of Medians: The line connecting the midpoints of all such strips forms the median $LP$. Therefore, the CM of the triangle must lie on median $LP$. By identical reasoning with respect to other bases, the CM lies on medians $MQ$ and $NR$.

Step 3: Centroid concurrence: The point of concurrence of the three medians is the centroid $G$, dividing each median in ratio $2:1$.

Example 6.3 • Centre of Mass of an L-Shaped Uniform Lamina

NCERT Solved

Find the centre of mass of a uniform L-shaped lamina of total mass $3\text{ kg}$ formed by 3 identical squares of side $1\text{ m}$.

Final Answer: $(X_\text{cm}, Y_\text{cm}) = \left(\frac{5}{6}\text{ m}, \frac{5}{6}\text{ m}\right)$

Step 1: Divide into 3 unit square elements: Since total mass $= 3\text{ kg}$, each square of area $1\text{ m}^2$ has mass $m_1 = m_2 = m_3 = 1\text{ kg}$. Centres of mass of squares: $C_1(0.5, 0.5)$, $C_2(1.5, 0.5)$, and $C_3(0.5, 1.5)$.

Step 2: Calculate composite coordinates: $$X_\text{cm} = \frac{1(0.5) + 1(1.5) + 1(0.5)}{1 + 1 + 1} = \frac{2.5}{3} = \mathbf{\frac{5}{6}\text{ m}}$$ $$Y_\text{cm} = \frac{1(0.5) + 1(0.5) + 1(1.5)}{3} = \frac{2.5}{3} = \mathbf{\frac{5}{6}\text{ m}}$$ The CM lies symmetrically on the diagonal line $y = x$.

6.3 - 6.4

NCERT Sections

Motion of Centre of Mass & Linear Momentum

Newton's 2nd Law for Systems

6.3.1 Velocity, Acceleration & Momentum Conservation

Differentiating the position vector of CM $\mathbf{R}_\text{cm} = \frac{1}{M}\sum m_i \mathbf{r}_i$ with respect to time yields the velocity and acceleration of the centre of mass:

System Momentum & Dynamics Equations $$\mathbf{V}_\text{cm} = \frac{d\mathbf{R}_\text{cm}}{dt} = \frac{1}{M}\sum_{i=1}^n m_i \mathbf{v}_i \implies \mathbf{P}_\text{total} = M \mathbf{V}_\text{cm}$$ $$\mathbf{A}_\text{cm} = \frac{d\mathbf{V}_\text{cm}}{dt} = \frac{1}{M}\sum_{i=1}^n m_i \mathbf{a}_i \implies M \mathbf{A}_\text{cm} = \sum_{i=1}^n \mathbf{F}_i$$ $$M \mathbf{A}_\text{cm} = \mathbf{F}_\text{ext} \quad \left(\text{since internal forces } \sum \mathbf{F}_\text{int} = 0 \text{ by Newton's 3rd law}\right)$$ $$\frac{d\mathbf{P}_\text{total}}{dt} = \mathbf{F}_\text{ext}$$

Conservation of Total Linear Momentum: When the net external force acting on a system is zero ($\mathbf{F}_\text{ext} = 0$), the total linear momentum is conserved ($\mathbf{P} = \text{constant}$), meaning the velocity of the centre of mass $\mathbf{V}_\text{cm}$ remains strictly constant.

6.5

NCERT Section

Vector Product (Cross Product) of Two Vectors

Mathematical Tool

6.5.1 Definition & Properties of Cross Product

The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is a vector $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ defined by:

Vector Product Formalism $$\mathbf{a} \times \mathbf{b} = (a b \sin\theta)\, \hat{\mathbf{n}}$$ where $\theta$ ($0 \le \theta \le \pi$) is the smaller angle between $\mathbf{a}$ and $\mathbf{b}$, and $\hat{\mathbf{n}}$ is a unit normal vector determined by the Right-Hand Screw Rule. 1. Anti-commutative: $\mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a})$ 2. Self-product: $\mathbf{a} \times \mathbf{a} = 0 \implies \hat{\mathbf{i}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} \times \hat{\mathbf{k}} = 0$ 3. Orthogonal Unit Vectors (Cyclic): $$\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}, \quad \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}, \quad \hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}}$$ $$\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}, \quad \hat{\mathbf{k}} \times \hat{\mathbf{j}} = -\hat{\mathbf{i}}, \quad \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}$$ 4. Determinant Form: $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = (a_y b_z - a_z b_y)\hat{\mathbf{i}} + (a_z b_x - a_x b_z)\hat{\mathbf{j}} + (a_x b_y - a_y b_x)\hat{\mathbf{k}}$$

Example 6.4 • Dot and Cross Product of 3D Vectors

NCERT Solved

Find the scalar and vector products of two vectors $\mathbf{a} = 3\hat{\mathbf{i}} - 4\hat{\mathbf{j}} + 5\hat{\mathbf{k}}$ and $\mathbf{b} = -2\hat{\mathbf{i}} + \hat{\mathbf{j}} - 3\hat{\mathbf{k}}$.

Answers: $\mathbf{a}\cdot\mathbf{b} = -25$ | $\mathbf{a} \times \mathbf{b} = 7\hat{\mathbf{i}} - \hat{\mathbf{j}} - 5\hat{\mathbf{k}}$

Step 1: Compute Scalar Dot Product: $$\mathbf{a}\cdot\mathbf{b} = (3)(-2) + (-4)(1) + (5)(-3) = -6 - 4 - 15 = \mathbf{-25}$$

Step 2: Compute Vector Cross Product via Determinant: $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -4 & 5 \\ -2 & 1 & -3 \end{vmatrix}$$ $$= \hat{\mathbf{i}}[(-4)(-3) - (5)(1)] - \hat{\mathbf{j}}[(3)(-3) - (5)(-2)] + \hat{\mathbf{k}}[(3)(1) - (-4)(-2)]$$ $$= \hat{\mathbf{i}}[12 - 5] - \hat{\mathbf{j}}[-9 + 10] + \hat{\mathbf{k}}[3 - 8] = \mathbf{7\hat{\mathbf{i}} - \hat{\mathbf{j}} - 5\hat{\mathbf{k}}}$$

6.6

NCERT Section

Angular Velocity & Relation with Linear Velocity

Kinematic Connections

6.6.1 Vector Relation $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$

In rotation of a rigid body about a fixed axis, all particles share the same instantaneous angular velocity vector $\boldsymbol{\omega}$, directed along the axis of rotation.

Angular & Linear Velocity Formulas $$\omega = \frac{d\theta}{dt}, \qquad \boldsymbol{\alpha} = \frac{d\boldsymbol{\omega}}{dt}$$ $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$$ $$v = \omega r_\perp \quad (\text{where } r_\perp \text{ is perpendicular distance from the axis of rotation})$$

6.7

NCERT Section

Torque & Angular Momentum

Rotational Dynamics Principles

6.7.1 Torque ($\boldsymbol{\tau}$) & Angular Momentum ($\mathbf{L}$)

Torque & Angular Momentum Formulations 1. Moment of Force (Torque): $$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}, \qquad \tau = r F \sin\theta = r_\perp F = r F_\perp$$ $$\text{SI Unit: } \text{N m}, \quad \text{Dimensions: } [\text{M}\text{L}^2\text{T}^{-2}]$$ 2. Angular Momentum of a Particle: $$\mathbf{l} = \mathbf{r} \times \mathbf{p}, \qquad l = r p \sin\theta = m v r_\perp$$ $$\text{SI Unit: } \text{J s} \text{ or } \text{kg m}^2\text{ s}^{-1}, \quad \text{Dimensions: } [\text{M}\text{L}^2\text{T}^{-1}]$$ 3. Fundamental Dynamic Relation (Newton's 2nd Law for Rotation): $$\frac{d\mathbf{l}}{dt} = \frac{d}{dt}(\mathbf{r} \times \mathbf{p}) = \left(\frac{d\mathbf{r}}{dt} \times \mathbf{p}\right) + \left(\mathbf{r} \times \frac{d\mathbf{p}}{dt}\right) = (\mathbf{v} \times m\mathbf{v}) + (\mathbf{r} \times \mathbf{F}) = 0 + \boldsymbol{\tau} \implies \boldsymbol{\tau} = \frac{d\mathbf{l}}{dt}$$ 4. System of Particles: $$\mathbf{L} = \sum_{i=1}^n \mathbf{l}_i = \sum_{i=1}^n (\mathbf{r}_i \times \mathbf{p}_i) \implies \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_\text{ext}$$

Interactive 3D Vector Cross Product & Torque ($\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$)

Vector Cross Lab

Drag to change 3D angle

Torque Vector Magnitude & Right-Hand Rule Result

|τ| = r F sinθ = (8.0m) × (10N) × sin(90°) = 80.00 N m

Force Angle ($\theta$):

Moment Arm ($r$):

Force ($F$):

Example 6.5 • Torque of a Force About the Origin

NCERT Solved

Find the torque of a force $\mathbf{F} = 7\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ acting on a particle at $\mathbf{r} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$ about the origin.

Answer: $\boldsymbol{\tau} = 2\hat{\mathbf{i}} + 12\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\text{ N m}$

Step 1: Set up cross product determinant: $$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 7 & 3 & -5 \end{vmatrix}$$

Step 2: Expand along first row: $$\boldsymbol{\tau} = \hat{\mathbf{i}}[(-1)(-5) - (1)(3)] - \hat{\mathbf{j}}[(1)(-5) - (1)(7)] + \hat{\mathbf{k}}[(1)(3) - (-1)(7)]$$ $$= \hat{\mathbf{i}}[5 - 3] - \hat{\mathbf{j}}[-5 - 7] + \hat{\mathbf{k}}[3 + 7] = \mathbf{2\hat{\mathbf{i}} + 12\hat{\mathbf{j}} + 10\hat{\mathbf{k}}\text{ N m}}$$

Example 6.6 • Angular Momentum of Particle Moving with Constant Velocity

NCERT Solved

Show that the angular momentum about any point of a single particle moving with constant velocity remains constant throughout the motion.

Conclusion: Angular momentum $\mathbf{l} = \text{constant}$ in magnitude and direction.

Step 1: Magnitude of angular momentum: Let the particle have mass $m$ and constant velocity $\mathbf{v}$. The magnitude of angular momentum about arbitrary point $O$ is: $$l = r p \sin\theta = m v (r \sin\theta)$$

Step 2: Invariance of perpendicular line of action: $r\sin\theta = d$ is the perpendicular distance from $O$ to the line of motion. Because the particle moves along a straight line with speed $v$, $d$ and $v$ are constant $\implies l = mvd = \text{constant}$.

Step 3: Vector direction: $\mathbf{r} \times \mathbf{v}$ points perpendicular to the plane containing the line of motion and $O$ (fixed direction). Since external force is zero, $\boldsymbol{\tau} = 0 \implies \mathbf{l}$ is strictly conserved.

6.8

NCERT Section

Equilibrium of a Rigid Body

Statics & Moments

6.8.1 Conditions for Mechanical Equilibrium

A rigid body is in complete mechanical equilibrium if and only if both translational and rotational accelerations are zero:

Equilibrium Equations 1. Translational Equilibrium: $\sum \mathbf{F}_i = 0 \iff \sum F_x = 0, \; \sum F_y = 0, \; \sum F_z = 0$ 2. Rotational Equilibrium: $\sum \boldsymbol{\tau}_i = 0 \iff \sum \tau_x = 0, \; \sum \tau_y = 0, \; \sum \tau_z = 0$ 3. Principle of Moments (Lever): $$\text{Load} \times \text{Load Arm} = \text{Effort} \times \text{Effort Arm} \implies F_1 d_1 = F_2 d_2$$ $$\text{Mechanical Advantage (M.A.)} = \frac{F_1}{F_2} = \frac{d_2}{d_1}$$

Example 6.7 • Moment of a Couple is Origin-Independent

NCERT Solved

Show that the moment of a couple (two equal and opposite forces with distinct lines of action) does not depend on the point about which moments are taken.

Conclusion: Moment of couple $\boldsymbol{\tau}_\text{couple} = \mathbf{AB} \times \mathbf{F}$ (independent of origin $O$).

Step 1: Calculate torque about origin $O$: Let force $-\mathbf{F}$ act at point $A(\mathbf{r}_1)$ and force $+\mathbf{F}$ act at point $B(\mathbf{r}_2)$: $$\boldsymbol{\tau} = \mathbf{r}_1 \times (-\mathbf{F}) + \mathbf{r}_2 \times \mathbf{F} = (\mathbf{r}_2 - \mathbf{r}_1) \times \mathbf{F}$$

Step 2: Vector difference: Since $\mathbf{r}_2 - \mathbf{r}_1 = \mathbf{AB}$ (displacement vector from $A$ to $B$), $$\boldsymbol{\tau}_\text{couple} = \mathbf{AB} \times \mathbf{F}$$ This is completely independent of the choice of reference origin $O$.

Example 6.8 • Reactions on a Supported Metal Bar

NCERT Solved

A metal bar $70\text{ cm}$ long and $4.00\text{ kg}$ in mass is supported on two knife-edges placed $10\text{ cm}$ from each end. A $6.00\text{ kg}$ load is suspended at $30\text{ cm}$ from one end. Find the normal reactions $R_1$ and $R_2$ at the knife-edges ($g = 9.8\text{ m s}^{-2}$).

Answers: $R_1 = 54.88\text{ N} \approx 55\text{ N}$ | $R_2 = 43.12\text{ N} \approx 43\text{ N}$

Step 1: Translational Equilibrium: $$R_1 + R_2 - W - W_1 = 0 \implies R_1 + R_2 = (4.00 + 6.00)g = 10g = 98.00\text{ N} \quad \text{--- (1)}$$

Step 2: Rotational Equilibrium about Centre of Gravity $G$ ($35\text{ cm}$ mark): Knife edge $K_1$ is at $25\text{ cm}$ to the left ($K_1 G = 0.25\text{ m}$). Load $P$ ($6\text{ kg}$) is at $30\text{ cm}$ mark ($PG = 0.05\text{ m}$ to left). Knife edge $K_2$ is at $25\text{ cm}$ to the right ($K_2 G = 0.25\text{ m}$). $$-R_1(0.25) + W_1(0.05) + R_2(0.25) = 0 \implies 0.25(R_1 - R_2) = 0.05(6.00g)$$ $$R_1 - R_2 = \frac{0.30g}{0.25} = 1.2g = 11.76\text{ N} \quad \text{--- (2)}$$

Step 3: Solve linear system: Adding (1) and (2): $2R_1 = 109.76 \implies R_1 = \mathbf{54.88\text{ N}}$ Subtracting: $R_2 = 98.00 - 54.88 = \mathbf{43.12\text{ N}}$

Example 6.9 • Ladder Leaning on a Frictionless Wall

NCERT Solved

A $3\text{ m}$ long ladder weighing $20\text{ kg}$ leans on a frictionless wall with its foot $1\text{ m}$ from the wall. Find the reaction forces of the wall and the floor ($g = 9.8\text{ m s}^{-2}$).

Answers: Wall Reaction $F_1 = 34.6\text{ N}$ | Ground Reaction $F_2 = 199.0\text{ N}$ at $\alpha \approx 80^\circ$ to horizontal

Step 1: Geometry & Force Decomposition: Ladder length $L = 3\text{ m}$, foot distance $AC = 1\text{ m}$. Height on wall $BC = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}\text{ m}$. Weight $W = 20 \times 9.8 = 196.0\text{ N}$ acts at midpoint $D$ ($0.5\text{ m}$ from wall horizontally).

Step 2: Equilibrium equations: $$\text{Vertical: } N - W = 0 \implies N = 196.0\text{ N}$$ $$\text{Horizontal: } F_\text{friction} - F_1 = 0 \implies F_\text{friction} = F_1$$ $$\text{Torque about base } A: F_1(2\sqrt{2}) - W(0.5) = 0 \implies F_1 = \frac{0.5 \times 196.0}{2\sqrt{2}} = \frac{98}{2\sqrt{2}} = \mathbf{34.6\text{ N}}$$

Step 3: Resultant ground reaction $F_2$: $$F_2 = \sqrt{F_\text{friction}^2 + N^2} = \sqrt{(34.6)^2 + (196.0)^2} = \sqrt{1197.16 + 38416} = \mathbf{199.0\text{ N}}$$ $$\tan\alpha = \frac{N}{F_\text{friction}} = \frac{196.0}{34.6} = 4\sqrt{2} \approx 5.657 \implies \alpha \approx \mathbf{80^\circ}$$

6.9

NCERT Section

Moment of Inertia & Radius of Gyration

Rotational Inertia

6.9.1 Kinetic Energy of Rotation & Moment of Inertia

The rotational kinetic energy of a rigid body rotating with angular velocity $\omega$ about a fixed axis is:

Rotational Inertia & Energy $$K_\text{rot} = \sum_{i=1}^n \frac{1}{2} m_i v_i^2 = \sum_{i=1}^n \frac{1}{2} m_i (r_i \omega)^2 = \frac{1}{2}\left(\sum_{i=1}^n m_i r_i^2\right)\omega^2 = \frac{1}{2} I \omega^2$$ $$I = \sum_{i=1}^n m_i r_i^2 = \int r^2\, dm \quad (\text{Moment of Inertia, SI: } \text{kg m}^2, \; [\text{M}\text{L}^2])$$ $$I = M k^2 \implies k = \sqrt{\frac{I}{M}} \quad (\text{Radius of Gyration } k)$$

NCERT Table 6.1 • Moments of Inertia of Standard Geometries:

Body Shape	Axis of Rotation	Moment of Inertia ($I$)	Radius of Gyration ($k$)
Thin circular ring (radius $R$)	Perpendicular to plane, at centre	$$M R^2$$	$$R$$
Thin circular ring (radius $R$)	Diameter	$$\frac{1}{2} M R^2$$	$$\frac{R}{\sqrt{2}}$$
Thin uniform rod (length $L$)	Perpendicular to rod, at midpoint	$$\frac{1}{12} M L^2$$	$$\frac{L}{\sqrt{12}}$$
Circular disc (radius $R$)	Perpendicular to disc, at centre	$$\frac{1}{2} M R^2$$	$$\frac{R}{\sqrt{2}}$$
Circular disc (radius $R$)	Diameter	$$\frac{1}{4} M R^2$$	$$\frac{R}{2}$$
Hollow cylinder (radius $R$)	Geometric axis of cylinder	$$M R^2$$	$$R$$
Solid cylinder (radius $R$)	Geometric axis of cylinder	$$\frac{1}{2} M R^2$$	$$\frac{R}{\sqrt{2}}$$
Solid sphere (radius $R$)	Diameter	$$\frac{2}{5} M R^2$$	$$\sqrt{\frac{2}{5}} R$$

Interactive 3D Moment of Inertia & Geometry Inspector

Rotational Inertia Lab

Rotating about fixed gold axis

Moment of Inertia ($I$)

I = M R² = 1.00 M R²

Radius of Gyration ($k = \sqrt{I/M}$)

Radius of Gyration: k = R

Select Rigid Body Geometry:

6.10

NCERT Section

Kinematics of Rotational Motion

Kinematic Analogy

6.10.1 Equations of Uniform Angular Acceleration

Rotational Kinematic Master Equations $$\omega = \omega_0 + \alpha t$$ $$\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$$ $$\omega^2 = \omega_0^2 + 2\alpha (\theta - \theta_0)$$ $$\theta = \left(\frac{\omega_0 + \omega}{2}\right) t$$

Example 6.10 • First Principle Calculus Derivation of $\omega = \omega_0 + \alpha t$

NCERT Solved

Obtain the kinematic relation $\omega = \omega_0 + \alpha t$ from first principles using calculus.

Step 1: Set up differential definition: $$\alpha = \frac{d\omega}{dt} = \text{constant} \implies d\omega = \alpha\, dt$$

Step 2: Definite Integration: $$\int_{\omega_0}^\omega d\omega = \alpha \int_0^t dt \implies [\omega - \omega_0] = \alpha t \implies \mathbf{\omega = \omega_0 + \alpha t}$$

Example 6.11 • Angular Acceleration & Revolutions of Motor Wheel

NCERT Solved

The angular speed of a motor wheel is increased from $1200\text{ rpm}$ to $3120\text{ rpm}$ in $16\text{ seconds}$. (i) What is its angular acceleration? (ii) How many revolutions does the engine make during this time?

Answers: (i) $\alpha = 4\pi\text{ rad s}^{-2} \approx 12.57\text{ rad s}^{-2}$ | (ii) Total Revolutions $= 576\text{ rev}$

Step 1: Convert rpm to $\text{rad s}^{-1}$: $$\omega_0 = \frac{2\pi \times 1200}{60} = 40\pi\text{ rad s}^{-1}, \qquad \omega = \frac{2\pi \times 3120}{60} = 104\pi\text{ rad s}^{-1}$$

Step 2: Calculate angular acceleration: $$\alpha = \frac{\omega - \omega_0}{t} = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} = \mathbf{4\pi\text{ rad s}^{-2}}$$

Step 3: Calculate angular displacement $\theta$ and revolutions: $$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = (40\pi)(16) + \frac{1}{2}(4\pi)(16)^2 = 640\pi + 512\pi = 1152\pi\text{ rad}$$ $$n = \frac{\theta}{2\pi} = \frac{1152\pi}{2\pi} = \mathbf{576\text{ revolutions}}$$

6.11 - 6.12

NCERT Sections

Dynamics & Conservation of Angular Momentum

Dynamical Equivalences

6.11.1 Work, Power, Torque ($\tau = I\alpha$) & Conservation of $L$

Rotational Dynamics Master Equations 1. Work Done by Torque: $dW = \tau\, d\theta \implies W = \int \tau\, d\theta$$ 2. Instantaneous Rotational Power: $P = \frac{dW}{dt} = \tau \frac{d\theta}{dt} = \tau \omega$ 3. Newton's 2nd Law for Rotation: $\tau = I \alpha = I \frac{d\omega}{dt}$ 4. Fixed-Axis Angular Momentum: $L = I \omega \implies \tau_\text{ext} = \frac{dL}{dt}$ 5. Conservation of Angular Momentum: If $\tau_\text{ext} = 0 \implies I_1 \omega_1 = I_2 \omega_2 = \text{constant}$

Example 6.12 • Cord Unwinding from a Flywheel

NCERT Solved

A cord of negligible mass is wound round the rim of a flywheel of mass $20\text{ kg}$ and radius $20\text{ cm}$. A steady pull of $25\text{ N}$ is applied on the cord. (a) Compute the angular acceleration $\alpha$. (b) Find the work done by pull when $2\text{ m}$ is unwound. (c) Find the kinetic energy of the wheel at this point. (d) Compare work and KE.

Answers: (a) $\alpha = 12.5\text{ rad s}^{-2}$ | (b) Work $= 50\text{ J}$ | (c) $K_\text{rot} = 50\text{ J}$ | (d) $W = \Delta K$ (Conserved)

(a) Angular acceleration: Flywheel moment of inertia (solid disc): $$I = \frac{1}{2} M R^2 = \frac{1}{2}(20)(0.20)^2 = 0.4\text{ kg m}^2$$ Applied Torque $\tau = F R = 25 \times 0.20 = 5.0\text{ N m}$ $$\alpha = \frac{\tau}{I} = \frac{5.0}{0.4} = \mathbf{12.5\text{ rad s}^{-2}}$$

(b) Work done by pull: $$W = F \cdot s = 25\text{ N} \times 2.0\text{ m} = \mathbf{50.0\text{ J}}$$

(c) Kinetic energy of the wheel: Angular displacement $\theta = \frac{s}{R} = \frac{2.0}{0.20} = 10\text{ rad}$. $$\omega^2 = \omega_0^2 + 2\alpha\theta = 0 + 2(12.5)(10) = 250\text{ rad}^2\text{ s}^{-2}$$ $$K_\text{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2}(0.4)(250) = \mathbf{50.0\text{ J}}$$

(d) Comparison: $W = K_\text{rot} = 50\text{ J}$. In the absence of friction, all work done by the external cord pull is converted into rotational kinetic energy.

02 / Self-Assessment Lab

Concept Check Questions

Select an option to immediately inspect detailed explanations for every single choice, reinforcing correct principles and pinpointing common traps.

15 MCQS

0 / 15 Answered Score: 0

Q1 Pending

For a rigid body undergoing pure rotational motion about a fixed axis, which physical quantity is identical for all particles of the body at any given instant?

Incorrect. Linear speed $v = \omega r$ depends directly on perpendicular distance $r$ from the axis.

Correct! By definition of a rigid body rotating about a fixed axis, all particles sweep equal angles in equal intervals of time ($d\theta/dt = \omega$).

Incorrect. $a_\text{c} = \omega^2 r$ varies linearly with distance $r$.

Incorrect. Depends on individual particle mass and distance $p = m\omega r$.

Takeaway: In pure rotation about a fixed axis, $\omega$ and $\alpha$ are characteristics of the body as a whole.

Q2 Pending

When a projectile in flight explodes into fragments in mid-air, what is the trajectory of the centre of mass of all fragments (neglecting air resistance)?

Incorrect. Internal explosive forces do not change the total momentum or CM trajectory.

Correct! Explosive forces are strictly internal ($\sum \mathbf{F}_\text{int} = 0$). External gravity $\mathbf{F}_\text{ext} = M\mathbf{g}$ remains unchanged, so $\mathbf{A}_\text{cm} = \mathbf{g}$.

Incorrect. Gravity still acts vertically downward.

Incorrect. Individual pieces scatter, but their mass-weighted average CM follows the original parabola.

Takeaway: Internal forces cancel in pairs and have zero effect on the acceleration of the centre of mass.

Q3 Pending

Which of the following bodies has its centre of mass located entirely outside its material volume?

Incorrect. The CM of a solid sphere is at its solid geometric centre.

Correct! By symmetry, the CM of a ring is at the geometric centre of the hole, where no material exists.

Incorrect. Located at the midpoint of the central axis inside the material.

Incorrect. Lies at the central intersection of spatial diagonals inside the solid.

Takeaway: Centre of mass is a mathematical point and does not require physical matter at its location.

Q4 Pending

If the cross product of two non-zero vectors $\mathbf{A} \times \mathbf{B} = 0$, what can be concluded about the angle $\theta$ between them?

Incorrect. When $\theta = 90^\circ$, $|\mathbf{A} \times \mathbf{B}| = AB$ (maximum).

Correct! $|\mathbf{A} \times \mathbf{B}| = AB\sin\theta = 0 \implies \sin\theta = 0 \implies \theta = 0^\circ \text{ or } 180^\circ$.

Incorrect. $\sin 45^\circ = 1/\sqrt{2} \neq 0$.

Incorrect. $\sin 60^\circ = \sqrt{3}/2 \neq 0$.

Takeaway: The vector cross product vanishes if and only if vectors are parallel or antiparallel.

Q5 Pending

What is the moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ about its diameter?

Incorrect. This is for a solid cylinder or circular disc about its perpendicular central axis.

Incorrect. This is for a thin hollow spherical shell.

Correct! For a uniform solid sphere about any diameter, $I = \frac{2}{5} M R^2$.

Incorrect. This is for a thin circular ring or hollow cylinder.

Takeaway: Memorize Table 6.1: solid sphere $= \frac{2}{5}MR^2$, disc $= \frac{1}{2}MR^2$, ring $= MR^2$.

03 / Textbook Solutions

Complete NCERT Exercises 6.1 – 6.17

Fully solved, stepwise answers with KaTeX math formatting for every textbook exercise question.

17 PROBLEMS

All Questions (6.1-6.17)

Centre of Mass

Vectors & Cross Product

Equilibrium & Levers

Moment of Inertia & Angular Momentum

Ex 6.1 Centre of Mass Location

6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

Prerequisite: ↗ 6.2 Centre of Mass Symmetry

Answer: CM is at geometric centre for all 4 symmetric bodies. No, CM need not lie inside the body (e.g. ring).

Locations: (i) Sphere: At its geometric centre.
(ii) Cylinder: At the midpoint of its central geometric axis of symmetry.
(iii) Ring: At the geometric centre of the ring.
(iv) Cube: At the point of intersection of its spatial body diagonals.

Inside/Outside property: No. In a uniform ring, hollow cylinder, or hollow sphere, the centre of mass lies in empty space where no material exists.

Ex 6.2 CM of HCl Molecule

6.2 In the $\text{HCl}$ molecule, the separation between the nuclei of the two atoms is about $1.27\text{ \AA}$ ($1\text{ \AA} = 10^{-10}\text{ m}$). Find the approximate location of the CM of the molecule, given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom.

Prerequisite: ↗ 6.2 Two-Particle CM Formula

Answer: $X_\text{cm} = 1.235\text{ \AA} = 1.235 \times 10^{-10}\text{ m}$ from the Hydrogen nucleus.

Step 1: Set coordinate system: Place origin at the Hydrogen nucleus ($x_1 = 0$, $m_1 = m_\text{H}$). The Chlorine nucleus is at $x_2 = 1.27\text{ \AA}$ with mass $m_2 = 35.5 m_\text{H}$.

Step 2: Calculate $X_\text{cm}$: $$X_\text{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m_\text{H}(0) + 35.5 m_\text{H}(1.27\text{ \AA})}{m_\text{H} + 35.5 m_\text{H}} = \frac{35.5 \times 1.27}{36.5}\text{ \AA}$$ $$X_\text{cm} = \mathbf{1.235\text{ \AA}} = 1.235 \times 10^{-10}\text{ m} \quad (\text{only } 0.035\text{ \AA} \text{ from the Cl nucleus})$$

Ex 6.3 Child on a Moving Trolley

6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Prerequisite: ↗ 6.3 Internal Forces and CM Velocity

Answer: The speed of the CM remains exactly $V$.

Explanation: The forces exerted between the child and the trolley floor (such as friction and muscular action) are strictly internal forces. Since the horizontal floor is smooth, the net external horizontal force $\mathbf{F}_\text{ext} = 0$. By $\mathbf{F}_\text{ext} = M \mathbf{A}_\text{cm} = 0$, $\mathbf{V}_\text{cm}$ remains unchanged and constant at speed $V$.

Ex 6.4 Triangle Area via Cross Product

6.4 Show that the area of the triangle contained between the vectors $\mathbf{a}$ and $\mathbf{b}$ is one half of the magnitude of $\mathbf{a} \times \mathbf{b}$.

Prerequisite: ↗ 6.5 Magnitude of Cross Product

Proof: $\text{Area}(\Delta) = \frac{1}{2}|\mathbf{a} \times \mathbf{b}| = \frac{1}{2}ab\sin\theta$.

Proof: Consider triangle $OAB$ with adjacent sides formed by $\mathbf{a} = \mathbf{OA}$ (base length $a$) and $\mathbf{b} = \mathbf{OB}$ at angle $\theta$. The altitude (height) of the triangle is $h = b\sin\theta$. $$\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} a (b\sin\theta) = \mathbf{\frac{1}{2} |\mathbf{a} \times \mathbf{b}|}$$

Ex 6.5 Scalar Triple Product & Volume

6.5 Show that $\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.

Prerequisite: ↗ 6.5 Cross Product Area Vectors

Proof: $\text{Volume} = |\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})|$.

Proof: The cross product $\mathbf{b} \times \mathbf{c} = \mathbf{A}_\text{base} = (bc\sin\theta)\hat{\mathbf{n}}$ represents the area vector of the base parallelogram perpendicular to the base. The height of the parallelepiped along $\hat{\mathbf{n}}$ is $h = a\cos\phi$, where $\phi$ is the angle between $\mathbf{a}$ and $\hat{\mathbf{n}}$. $$\text{Volume} = \text{Base Area} \times \text{Height} = |\mathbf{b} \times \mathbf{c}| (a\cos\phi) = \mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})$$

Ex 6.6 Components of Angular Momentum

6.6 Find the components along the $x, y, z$ axes of the angular momentum $\mathbf{l}$ of a particle with position $\mathbf{r}(x, y, z)$ and momentum $\mathbf{p}(p_x, p_y, p_z)$. Show that if the particle moves only in the $x-y$ plane, angular momentum has only a $z$-component.

Prerequisite: ↗ 6.7 Definition of Angular Momentum

Components: $l_x = yp_z - zp_y$, $l_y = zp_x - xp_z$, $l_z = xp_y - yp_x$. In $x-y$ plane, $l_x = l_y = 0$.

Step 1: Cross product determinant: $$\mathbf{l} = \mathbf{r} \times \mathbf{p} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} = (y p_z - z p_y)\hat{\mathbf{i}} + (z p_x - x p_z)\hat{\mathbf{j}} + (x p_y - y p_x)\hat{\mathbf{k}}$$

Step 2: Planar restriction: For motion in the $x-y$ plane, $z = 0$ and $p_z = 0$. $$l_x = y(0) - (0)p_y = 0, \quad l_y = (0)p_x - x(0) = 0, \quad \mathbf{l}_z = x p_y - y p_x$$ Thus $\mathbf{l} = (xp_y - yp_x)\hat{\mathbf{k}}$ has strictly only a $z$-component.

Ex 6.7 Angular Momentum of Parallel Opposite Particles

6.7 Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two-particle system is the same whatever be the point about which it is taken.

Prerequisite: ↗ 6.7 Angular Momentum of System

Proof: $\mathbf{L}_\text{total} = mvd\, \hat{\mathbf{n}}$ (constant and independent of reference point).

Proof: Let particle 1 move with momentum $\mathbf{p}_1 = m\mathbf{v}$ along line 1, and particle 2 with $\mathbf{p}_2 = -m\mathbf{v}$ along line 2. About any arbitrary origin $O$: $$\mathbf{L} = \mathbf{r}_1 \times (m\mathbf{v}) + \mathbf{r}_2 \times (-m\mathbf{v}) = (\mathbf{r}_1 - \mathbf{r}_2) \times m\mathbf{v}$$ The vector $\mathbf{r}_1 - \mathbf{r}_2$ is the displacement vector from particle 2 to particle 1. Its perpendicular component across the velocity vector is strictly the fixed separation $d$. $$|\mathbf{L}| = mvd = \text{constant}$$ Direction is fixed by the right-hand rule, invariant under any shift of origin.

Ex 6.8 Non-Uniform Suspended Bar

6.8 A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight. The angles made by strings with the vertical are $36.9^\circ$ and $53.1^\circ$ respectively. The bar is $2\text{ m}$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

Prerequisite: ↗ 6.8 Coplanar Torque Equilibrium

Answer: $d = 0.72\text{ m}$ from the left end.

Step 1: Tension Resolution: Let string 1 make angle $\theta_1 = 36.9^\circ$ ($\sin 36.9^\circ = 0.6, \cos 36.9^\circ = 0.8$) and string 2 make $\theta_2 = 53.1^\circ$ ($\sin 53.1^\circ = 0.8, \cos 53.1^\circ = 0.6$). $$\text{Horizontal Equilibrium: } T_1 \sin 36.9^\circ = T_2 \sin 53.1^\circ \implies 0.6 T_1 = 0.8 T_2 \implies T_1 = \frac{4}{3} T_2$$ $$\text{Vertical Equilibrium: } T_1 \cos 36.9^\circ + T_2 \cos 53.1^\circ = W \implies 0.8\left(\frac{4}{3}T_2\right) + 0.6 T_2 = W \implies \frac{5}{3} T_2 = W \implies T_2 = \frac{3}{5}W, \; T_1 = \frac{4}{5}W$$

Step 2: Rotational Equilibrium about Centre of Gravity: $$(T_1 \cos 36.9^\circ) d = (T_2 \cos 53.1^\circ)(2 - d) \implies \left(\frac{4}{5}W \times 0.8\right) d = \left(\frac{3}{5}W \times 0.6\right)(2 - d)$$ $$0.64 d = 0.36(2 - d) = 0.72 - 0.36 d \implies 1.00 d = 0.72 \implies \mathbf{d = 0.72\text{ m}}$$

Ex 6.9 Car Wheel Ground Reaction Forces

6.9 A car weighs $1800\text{ kg}$. The distance between front and back axles is $1.8\text{ m}$. Its centre of gravity is $1.05\text{ m}$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel ($g = 9.8\text{ m s}^{-2}$).

Prerequisite: ↗ 6.8 Equilibrium of Parallel Forces

Answers: Force on each front wheel $= 3675\text{ N}$ | Force on each back wheel $= 5145\text{ N}$

Step 1: Total normal reactions: Let total reaction on two front wheels be $R_\text{f}$ and on two back wheels be $R_\text{b}$. Total weight $W = 1800 \times 9.8 = 17640\text{ N}$. $$R_\text{f} + R_\text{b} = 17640\text{ N} \quad \text{--- (1)}$$

Step 2: Rotational Equilibrium about Back Axle: $$R_\text{f} \times 1.8 = W \times (1.8 - 1.05) = 17640 \times 0.75 = 13230 \implies R_\text{f} = \frac{13230}{1.8} = 7350\text{ N}$$ $$R_\text{b} = 17640 - 7350 = 10290\text{ N}$$

Step 3: Force per wheel (2 wheels per axle): $$\text{Each Front Wheel} = \frac{R_\text{f}}{2} = \frac{7350}{2} = \mathbf{3675\text{ N}}$$ $$\text{Each Back Wheel} = \frac{R_\text{b}}{2} = \frac{10290}{2} = \mathbf{5145\text{ N}}$$

Ex 6.10 Hollow Cylinder vs Solid Sphere

6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Prerequisite: ↗ 6.9 & 6.11 Torque $\tau = I\alpha$

Answer: The solid sphere will acquire a greater angular speed.

Step 1: Compare Moments of Inertia: $$I_\text{hollow cylinder} = M R^2, \qquad I_\text{solid sphere} = \frac{2}{5} M R^2$$ Clearly, $I_\text{solid sphere} < I_\text{hollow cylinder}$.

Step 2: Compare Angular Accelerations: $$\alpha = \frac{\tau}{I} \implies \alpha_\text{sphere} = \frac{\tau}{\frac{2}{5}MR^2} = 2.5 \left(\frac{\tau}{MR^2}\right) > \alpha_\text{cylinder} = \frac{\tau}{MR^2}$$ Since $\omega = \alpha t$, the solid sphere acquires a higher angular speed in time $t$.

Ex 6.11 Rotating Solid Cylinder KE & L

6.11 A solid cylinder of mass $20\text{ kg}$ rotates about its axis with angular speed $100\text{ rad s}^{-1}$. The radius of the cylinder is $0.25\text{ m}$. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

Prerequisite: ↗ 6.9 & 6.12 $K_\text{rot} = \frac{1}{2}I\omega^2, L = I\omega$

Answers: $K_\text{rot} = 3125\text{ J}$ | $L = 62.5\text{ J s}$ (or $\text{kg m}^2\text{ s}^{-1}$)

Step 1: Moment of Inertia of solid cylinder: $$I = \frac{1}{2} M R^2 = \frac{1}{2}(20\text{ kg})(0.25\text{ m})^2 = 10 \times 0.0625 = \mathbf{0.625\text{ kg m}^2}$$

Step 2: Rotational Kinetic Energy: $$K_\text{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2}(0.625)(100)^2 = \frac{1}{2}(0.625)(10000) = \mathbf{3125\text{ J}}$$

Step 3: Angular Momentum: $$L = I \omega = (0.625)(100) = \mathbf{62.5\text{ J s}}$$

Ex 6.12 Turntable & Conservation of $L$

6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40\text{ rev/min}$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? (Assume no friction).
(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy. How do you account for this increase?

Prerequisite: ↗ 6.12 Conservation of Angular Momentum

Answers: (a) $\omega_2 = 100\text{ rev/min}$ | (b) $K_2 = 2.5 K_1$ (Increase sourced from child's internal muscular work)

(a) Conservation of Angular Momentum: Since $\tau_\text{ext} = 0$, $I_1 \omega_1 = I_2 \omega_2$. Given $I_2 = \frac{2}{5} I_1$: $$\omega_2 = \left(\frac{I_1}{I_2}\right) \omega_1 = \left(\frac{5}{2}\right)(40\text{ rpm}) = \mathbf{100\text{ rev/min}}$$

(b) Ratio of Rotational Kinetic Energies: $$\frac{K_2}{K_1} = \frac{\frac{1}{2}I_2 \omega_2^2}{\frac{1}{2}I_1 \omega_1^2} = \frac{I_2}{I_1} \left(\frac{\omega_2}{\omega_1}\right)^2 = \left(\frac{2}{5}\right)\left(\frac{5}{2}\right)^2 = \frac{5}{2} = \mathbf{2.5}$$ The kinetic energy increases by a factor of $2.5$. The source of this additional kinetic energy is the internal muscular work done by the child in pulling his arms inward against centrifugal reaction.

Ex 6.13 Hollow Cylinder Pulled by Rope

6.13 A rope of negligible mass is wound round a hollow cylinder of mass $3\text{ kg}$ and radius $40\text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30\text{ N}$? What is the linear acceleration of the rope? (No slipping).

Prerequisite: ↗ 6.9 & 6.11 $\tau = I\alpha, a = R\alpha$

Answers: $\alpha = 25\text{ rad s}^{-2}$ | Linear acceleration $a = 10\text{ m s}^{-2}$

Step 1: Moment of Inertia of Hollow Cylinder: $$I = M R^2 = (3\text{ kg})(0.40\text{ m})^2 = 3 \times 0.16 = \mathbf{0.48\text{ kg m}^2}$$

Step 2: Applied Torque and Angular Acceleration: $$\tau = F R = 30\text{ N} \times 0.40\text{ m} = 12.0\text{ N m}$$ $$\alpha = \frac{\tau}{I} = \frac{12.0}{0.48} = \mathbf{25\text{ rad s}^{-2}}$$

Step 3: Linear acceleration of rope: $$a = R \alpha = 0.40\text{ m} \times 25\text{ rad s}^{-2} = \mathbf{10.0\text{ m s}^{-2}}$$

Ex 6.14 Engine Rotational Power

6.14 To maintain a rotor at a uniform angular speed of $200\text{ rad s}^{-1}$, an engine needs to transmit a torque of $180\text{ N m}$. What is the power required by the engine?

Prerequisite: ↗ 6.11 Rotational Power $P = \tau\omega$

Answer: Power $P = 36\text{ kW} = 3.6 \times 10^4\text{ W}$

Calculation: $$P = \tau \omega = 180\text{ N m} \times 200\text{ rad s}^{-1} = \mathbf{36,000\text{ W}} = \mathbf{36\text{ kW}}$$

Ex 6.15 Disc with Circular Cutout

6.15 From a uniform disk of radius $R$, a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Prerequisite: ↗ 6.2 Negative Mass Method for CM

Answer: $X_\text{cm} = -\frac{R}{6}$ (located at distance $R/6$ opposite to the centre of the hole).

Step 1: Negative Mass Principle: Let original uniform complete disc have mass $M$ and area $\pi R^2$. Surface mass density $\sigma = \frac{M}{\pi R^2}$. The removed circular hole has radius $r = R/2 \implies \text{Area } A_1 = \pi(R/2)^2 = \frac{\pi R^2}{4}$. Mass of removed part $m_1 = \frac{M}{4}$, with centre at $x_1 = +R/2$.

Step 2: Center of mass calculation: Treat the cutout as a negative mass $-m_1 = -M/4$: $$X_\text{cm} = \frac{M(0) - m_1 x_1}{M - m_1} = \frac{-\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)}{M - \frac{M}{4}} = \frac{-\frac{MR}{8}}{\frac{3M}{4}} = \mathbf{-\frac{R}{6}}$$ The centre of gravity shifts by distance $\mathbf{R/6}$ along the symmetry axis opposite to the cutout.

Ex 6.16 Metre Stick & Coins Balance

6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5\text{ g}$, are put one on top of the other at the $12.0\text{ cm}$ mark, the stick is found to be balanced at $45.0\text{ cm}$. What is the mass of the metre stick?

Prerequisite: ↗ 6.8 Principle of Moments for Levers

Answer: Mass of metre stick $M = 66.0\text{ g} = 0.066\text{ kg}$

Step 1: Identify positions and lever arms relative to fulcrum ($45.0\text{ cm}$): The centre of gravity of the metre stick is at $50.0\text{ cm}$. Lever arm for stick's weight $M$: $d_\text{stick} = 50.0 - 45.0 = 5.0\text{ cm}$. Combined mass of two coins: $m = 2 \times 5\text{ g} = 10\text{ g}$. Lever arm for coins: $d_\text{coins} = 45.0 - 12.0 = 33.0\text{ cm}$.

Step 2: Rotational Equilibrium about fulcrum: $$m \cdot d_\text{coins} = M \cdot d_\text{stick} \implies (10\text{ g})(33.0\text{ cm}) = M (5.0\text{ cm})$$ $$M = \frac{330}{5.0} = \mathbf{66.0\text{ g}}$$

Ex 6.17 Oxygen Molecule Angular Velocity

6.17 The oxygen molecule has a mass of $5.30 \times 10^{-26}\text{ kg}$ and a moment of inertia of $1.94 \times 10^{-46}\text{ kg m}^2$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500\text{ m s}^{-1}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Prerequisite: ↗ 6.9 Rotational vs Translational Kinetic Energy

Answer: $\omega \approx 6.75 \times 10^{12}\text{ rad s}^{-1}$

Step 1: Calculate Translational Kinetic Energy: $$K_\text{trans} = \frac{1}{2} M v^2 = \frac{1}{2}(5.30 \times 10^{-26}\text{ kg})(500\text{ m s}^{-1})^2 = \frac{1}{2}(5.30 \times 10^{-26})(2.5 \times 10^5) = 6.625 \times 10^{-21}\text{ J}$$

Step 2: Relate to Rotational Kinetic Energy: $$K_\text{rot} = \frac{2}{3} K_\text{trans} = \frac{2}{3}(6.625 \times 10^{-21}) = 4.417 \times 10^{-21}\text{ J}$$

Step 3: Solve for $\omega$: $$K_\text{rot} = \frac{1}{2} I \omega^2 \implies \omega = \sqrt{\frac{2 K_\text{rot}}{I}} = \sqrt{\frac{2 \times 4.417 \times 10^{-21}}{1.94 \times 10^{-46}}} = \sqrt{\frac{8.834 \times 10^{-21}}{1.94 \times 10^{-46}}}$$ $$\omega = \sqrt{4.5536 \times 10^{25}} = \sqrt{45.536 \times 10^{24}} \approx \mathbf{6.75 \times 10^{12}\text{ rad s}^{-1}}$$

04 / Rapid Reference

Chapter Summary & Formulas

Centre of mass coordinates, rotational kinematics, angular torque/momentum, parallel/perpendicular axes theorems, moment of inertia tables, and rolling kinematics.

100% SYLLABUS

6.1 & 6.2 • Centre of Mass

Centre of Mass Coordinates & Motion Dynamics

The centre of mass is a point where the entire mass of a system can be assumed to be concentrated for describing its translational motion.

Two-Particle System: $X_{\text{com}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
N-Particle System Vector: $\vec{R}_{\text{com}} = \frac{\sum m_i \vec{r}_i}{M}$ (where $M = \sum m_i$)
Continuous Mass Distribution: $X_{\text{com}} = \frac{1}{M}\int x dm$

Linear Momentum of System: $\vec{P}_{\text{total}} = M \vec{V}_{\text{com}}$.
Newton's Second Law for System: $\vec{F}_{\text{ext}} = M \vec{A}_{\text{com}}$.
Condition: If the net external force is zero ($\vec{F}_{\text{ext}} = 0$), the centre of mass moves with a constant velocity ($\vec{V}_{\text{com}} = \text{constant}$). Internal forces cannot change the motion of the COM.

6.3 & 6.7 • Kinematic Analogy

Linear vs. Angular Analogy Cheat Sheet

Linear Motion (1D)	Rotational Motion (Fixed Axis)	Analogue Equation / Link
Displacement $x$	Angle $\theta$	$s = \theta R$
Velocity $v = \frac{dx}{dt}$	Angular Velocity $\omega = \frac{d\theta}{dt}$	$v = \omega R$
Acceleration $a = \frac{dv}{dt}$	Angular Acceleration $\alpha = \frac{d\omega}{dt}$	$a_t = \alpha R$
Mass $M$	Moment of Inertia $I = \sum m_i r_i^2$	$I = M k^2$ ($k$: radius of gyration)
Force $F = Ma$	Torque $\tau = I\alpha$	$\vec{\tau} = \vec{r} \times \vec{F}$ (or $\tau = rF\sin\theta$)
Momentum $p = Mv$	Angular Momentum $L = I\omega$	$\vec{L} = \vec{r} \times \vec{p}$ (or $L = rp\sin\theta$)
Kinetic Energy $K = \frac{1}{2}Mv^2$	Rotational KE $K_{\text{rot}} = \frac{1}{2}I\omega^2$	$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}}$

6.4 - 6.6 • Rotational Dynamics

Torque, Angular Momentum & Conservation Laws

Torque ($\vec{\tau}$): Turning effect of force. $\vec{\tau} = \vec{r} \times \vec{F}$.
$\tau = r F \sin\theta = F \cdot r_{\perp}$ (where $r_{\perp}$ is the perpendicular distance from axis to force line).
Angular Momentum ($\vec{L}$): Moment of linear momentum. $\vec{L} = \vec{r} \times \vec{p}$.
$L = r p \sin\theta = I\omega$.
Newton's Second Law in Rotation: $$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$$

Conservation of Angular Momentum: If net external torque is zero ($\vec{\tau}_{\text{ext}} = 0$), then $\vec{L} = \text{constant}$. Thus, $$I_1 \omega_1 = I_2 \omega_2$$
Exam Mnemonic: A spinning dancer pulling arms inward decreases moment of inertia $I$, causing angular velocity $\omega$ to increase instantly.

6.8 • Inertia Constants

Moments of Inertia ($I$) & Theorems

Two Fundamental Theorems

Theorem of Parallel Axes: $I_z = I_{\text{com}} + M d^2$ (where $d$ is distance between parallel axes). Valid for any body shape.
Theorem of Perpendicular Axes: $I_z = I_x + I_y$. Strict Limit: Valid only for planar bodies (laminas) in the $xy$-plane.

Moment of Inertia Formulas (Mass $M$, Radius $R$)

Body Shape	Moment of Inertia ($I$)	Radius of Gyration ($k^2/R^2$)
Thin Ring / Hollow Cylinder (central axis)	$I = MR^2$	$1.0$
Thin Disc / Solid Cylinder (central axis)	$I = \frac{1}{2}MR^2$	$0.5$
Solid Sphere (axis through center)	$I = \frac{2}{5}MR^2$	$0.4$
Hollow Spherical Shell (axis through center)	$I = \frac{2}{3}MR^2$	$0.67$
Thin Rod of length $L$ (axis perpendicular through center)	$I = \frac{1}{12}ML^2$	$L^2/(12R^2)$

6.9 • Rolling Kinematics

Pure Rolling on Flat and Inclined Surfaces

Pure rolling is a combination of translation of COM and rotation about COM. The contact point is at instantaneous rest ($v = 0$).

Pure Rolling Condition: $v_{\text{com}} = \omega R$
Total Kinetic Energy of Rolling: $$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 \left(1 + \frac{k^2}{R^2}\right)$$
Acceleration Down an Inclined Plane ($\theta$): $$a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}}$$

Race Down Inclined Plane: The body with the smallest $k^2/R^2$ ratio has the greatest acceleration. Thus, a solid sphere ($0.4$) reaches the bottom first, followed by solid cylinder ($0.5$), hollow sphere ($0.67$), and ring ($1.0$) last.

NCERT Official

Points to Ponder & Exam Tips

The centre of mass of a body does not need to lie inside the body (e.g. ring, hollow sphere).
To find torque or angular momentum, the position vector $\vec{r}$ must be measured from a chosen origin. Changing the origin changes $\vec{r}$ and thus $\vec{\tau}$ and $\vec{L}$.
Moment of inertia is not a constant for a body; it depends entirely on the location and orientation of the axis of rotation.
For a body in equilibrium, both net force $\sum \vec{F} = 0$ (translational equilibrium) and net torque $\sum \vec{\tau} = 0$ (rotational equilibrium) must be zero.

05 / Timed Exam Prep

Chapter Mastery Tests

Simulated multi-tier exams covering conceptual reasoning, numerical applications, and challenging competitive problems.

3 LEVELS

1. The centre of mass of two particles of masses $2\text{ kg}$ and $3\text{ kg}$ separated by $1\text{ m}$ is at what distance from the $2\text{ kg}$ mass?

2. What are the dimensional units of torque ($\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$)?

3. If a rigid body is in rotational equilibrium, the net torque acting on it must be:

1. A wheel accelerates uniformly from rest to $100\text{ rad s}^{-1}$ in $10\text{ s}$. The total angular displacement in this interval is:

2. A uniform disc has radius of gyration $k$ about its diameter equal to:

1. A solid cylinder of mass $M$ and radius $R$ is rotated by pulling a string wound around it with force $F$. The angular acceleration is:

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