Chapter 7: System of Particles and Rotational Motion

Rigid Body Motion

Key Concepts

• A rigid body has a fixed shape; distances between particles remain constant.

• Pure translational motion: all particles move with the same velocity.

• Rotational motion: body rotates about a fixed axis; particles move in circles perpendicular to the axis.

• Axis of rotation: fixed line about which rotation occurs.

• Precession: when the axis of rotation itself moves, e.g., spinning top axis sweeps a cone.

• Motion types:

  • Pure translation: velocity of all particles equal, angles equal (\(\alpha_1=\alpha_2=\alpha_3\)).

  • Combination of translation and rotation: velocities differ, angles differ (\(\alpha_1 \neq \alpha_2 \neq \alpha_3\)).

  • Pure rotation: body fixed or pivoted, rotates about axis.

Solved Examples

• Ex 1: Calculate velocity of all particles in pure translation if velocity of one particle is 5 m/s.

  • Answer: All particles have velocity 5 m/s.

• Ex 2: For a rigid body rotating about z-axis, find radius of circle traced by point P at distance 3 m from axis.

  • Answer: Radius \(r = 3\) m (perpendicular distance from axis).

Practice Set

• Q1: In pure translational motion, velocity of particles is:

  • A) Different for each particle

  • B) Same for all particles

  • C) Zero

  • D) Varies with time

• Q2: The axis of rotation of a rigid body is:

  • A) A fixed point

  • B) A fixed straight line

  • C) A moving point

  • D) A plane

• Q3: A spinning top’s axis sweeps a cone. This motion is called:

  • A) Rotation

  • B) Translation

  • C) Precession

  • D) Oscillation

Answer Key

• Q1: B — In pure translation, all particles have the same velocity.

• Q2: B — Axis of rotation is a fixed straight line.

• Q3: C — Movement of spinning top’s axis around vertical is precession.

Memory Aids

• Rigid body = fixed shape, no particle distance change.

• Pure translation: all velocities equal.

• Rotation: particles move in circles about axis.

• Precession = axis of rotation moves, sweeps cone.

Centre of Mass

Key Concepts

• Centre of mass (CM) is the point where total mass can be considered concentrated.

• All external forces act effectively at CM.

• For two particles (m₁, m₂) separated by distance d, with origin at m₁:

  • \(x_{CM} = \frac{m_2}{m_1 + m_2} d\)

• General formula for two particles with positions x₁, x₂:

  • \(x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\)

• For n particles with masses \(m_i\) and position vectors \(\vec{r}_i\):

  • \(\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} = \frac{1}{M} \sum m_i \vec{r}_i\)

• Coordinates of CM:

  • \(x_{CM} = \frac{\sum m_i x_i}{M}\)

  • \(y_{CM} = \frac{\sum m_i y_i}{M}\)

  • \(z_{CM} = \frac{\sum m_i z_i}{M}\)

• For continuous mass distribution (rigid body):

  • \(x_{CM} = \frac{1}{M} \int x \, dm\)

  • \(y_{CM} = \frac{1}{M} \int y \, dm\)

  • \(z_{CM} = \frac{1}{M} \int z \, dm\)

• CM of common shapes:

  • Rectangular plate: center (\(L/2, b/2\))

  • Triangular plate: centroid at \(1/3\) from base

  • Semi-circular ring/disc: at \(4R/(3\pi)\) from flat edge

  • Hemispherical shell/solid: at \(3R/8\) or \(3R/16\) from base

  • Circular cone (solid/hollow): at \(1/4\) or \(1/3\) height from base

Solved Examples

• Ex 1: Find CM of two particles m₁=2 kg at x=0 m and m₂=3 kg at x=4 m.

  • \(x_{CM} = \frac{2\times0 + 3\times4}{2+3} = \frac{12}{5} = 2.4\) m

• Ex 2: For a uniform rod of length L=6 m, find CM position.

  • CM at \(L/2 = 3\) m from one end.

Practice Set

• Q1: CM of two particles m₁=1 kg at x=0 m and m₂=3 kg at x=6 m is:

  • A) 4.5 m

  • B) 3 m

  • C) 4.5 m

  • D) 4.5 m

• Q2: For n particles, CM position vector is:

  • A) \(\sum \vec{r}_i / n\)

  • B) \(\sum m_i \vec{r}_i / \sum m_i\)

  • C) \(\sum m_i / \sum \vec{r}_i\)

  • D) \(\sum \vec{r}_i m_i^2\)

• Q3: CM of a uniform semicircular ring of radius R is at distance:

  • A) \(R/2\)

  • B) \(4R/(3\pi)\)

  • C) \(R/3\)

  • D) \(2R/\pi\)

Answer Key

• Q1: B — \(x_{CM} = \frac{1\times0 + 3\times6}{4} = 4.5\) m (corrected: option B is 3 m, so correct is A 4.5 m)

• Q2: B — CM is mass-weighted average position.

• Q3: B — CM of semicircular ring is at \(4R/(3\pi)\) from flat edge.

Memory Aids

• CM = weighted average position of mass.

• Two particles: \(x_{CM} = \frac{m_2 d}{m_1 + m_2}\) if origin at m₁.

• Continuous bodies: use integrals over dm.

• Common shapes have fixed CM formulas along symmetry axes.

Motion of Centre of Mass

Definition and Formula

• Consider a system of \(n\) particles with masses \(m_1, m_2, \dots, m_n\).

• Position vectors of particles are \(\vec{r}_1, \vec{r}_2, \dots, \vec{r}_n\).

• Position vector of centre of mass (CM) is \(\vec{R}\).

• Formula: \(M\vec{R} = m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n\), where \(M = \sum m_i\).

Velocity and Acceleration of CM

• Differentiating position: \(M \frac{d\vec{R}}{dt} = m_1 \frac{d\vec{r}_1}{dt} + \dots + m_n \frac{d\vec{r}_n}{dt}\).

• Velocity of CM: \(\vec{V}_{CM} = \frac{d\vec{R}}{dt}\).

• Thus, \(M \vec{V}_{CM} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n\).

• Differentiating velocity: \(M \frac{d\vec{V}_{CM}}{dt} = m_1 \frac{d\vec{v}_1}{dt} + \dots + m_n \frac{d\vec{v}_n}{dt}\).

• Acceleration of CM: \(\vec{A}_{CM} = \frac{d\vec{V}_{CM}}{dt}\).

• Therefore, \(M \vec{A}_{CM} = m_1 \vec{a}_1 + m_2 \vec{a}_2 + \dots + m_n \vec{a}_n\).

Newton's Second Law for CM

• According to Newton's 2nd law: \(M \vec{A}_{CM} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n\).

• Internal forces cancel out; only external forces \(\vec{F}_{ext}\) affect CM motion.

• Hence, \(M \vec{A}_{CM} = \vec{F}_{ext}\).

• CM moves as if all mass is concentrated at CM and all external forces act there.

Projectile Explosion and CM Path

• When a projectile explodes mid-air, fragments move separately.

• The CM of fragments continues along the original parabolic path.

• This shows CM motion depends only on external forces (gravity).

Solved Examples

Ex 1

• Given masses \(m_1=2\,kg\), \(m_2=3\,kg\) at positions \(\vec{r}_1=(1\hat{i}+2\hat{j})m\), \(\vec{r}_2=(3\hat{i}+4\hat{j})m\).

• Find CM position \(\vec{R}\).

• Solution: \(M=5\,kg\), \(\vec{R} = \frac{1}{5}(2(1\hat{i}+2\hat{j}) + 3(3\hat{i}+4\hat{j})) = (2.2\hat{i} + 3.2\hat{j})m\).

Ex 2

• System of particles with total mass \(M=10\,kg\) experiences external force \(\vec{F}_{ext} = 20\hat{i} + 30\hat{j} \; N\).

• Find acceleration of CM.

• Solution: \(\vec{A}_{CM} = \frac{\vec{F}_{ext}}{M} = 2\hat{i} + 3\hat{j} \; m/s^2\).

Practice Set

Q1

For two particles of masses 4 kg and 6 kg at positions \(2\hat{i}\) m and \(4\hat{i}\) m respectively, the position of CM is:

• A) \(3.2\hat{i}\) m

• B) \(2.8\hat{i}\) m

• C) \(4\hat{i}\) m

• D) \(3\hat{i}\) m

Q2

A system of particles with total mass 5 kg has CM velocity \(2\hat{j}\) m/s. If an external force \(10\hat{j}\) N acts, acceleration of CM is:

• A) \(2\hat{j}\) m/s²

• B) \(0.5\hat{j}\) m/s²

• C) \(5\hat{j}\) m/s²

• D) \(10\hat{j}\) m/s²

Q3

A projectile of mass 2 kg explodes into two fragments of 1.2 kg and 0.8 kg at height. If the CM continues on the original parabolic path, which force governs CM motion?

• A) Internal explosion forces

• B) Gravity only

• C) Air resistance

• D) Friction

Answer Key

• Q1: A) \(3.2\hat{i}\) m — CM position \(= \frac{4\times2 + 6\times4}{10} = 3.2\) m.

• Q2: B) \(0.5\hat{j}\) m/s² — \(a = F/M = 10/5 = 2\) N/kg but force is 10 N, so acceleration \(= 2\hat{j}\) m/s² (corrected: option A).

• Q3: B) Gravity only — CM motion affected only by external forces, internal forces cancel.

Memory Aids

• \(M\vec{R} = \sum m_i \vec{r}_i\) — weighted average position.

• Velocity and acceleration of CM are mass-weighted averages.

• Only external forces affect CM acceleration.

• CM of fragments follows original path post-explosion.

Linear Momentum of a System of Particles

Key Concepts

• Linear momentum \( \vec{p} \) of a system of \( n \) particles is the vector sum: \( \vec{p} = \vec{p}_1 + \vec{p}_2 + \cdots + \vec{p}_n \).

• Total momentum relates to center of mass (CM) velocity: \( M\vec{V}_{CM} = m_1\vec{v}_1 + m_2\vec{v}_2 + \cdots + m_n\vec{v}_n \), where \( M = \sum m_i \).

• Newton's 2nd law for system: \( \frac{d\vec{p}}{dt} = M \frac{d\vec{V}_{CM}}{dt} = M\vec{A}_{CM} = \vec{F}_{ext} \).

• If external force \( \vec{F}_{ext} = 0 \), then \( \frac{d\vec{p}}{dt} = 0 \) and \( \vec{p} = \text{constant} \) (conservation of linear momentum).

• With zero external force, CM moves with constant velocity.

• Example: Alpha decay of radium (Ra) nucleus into radon (Rn) and alpha particle (He) conserves momentum; CM moves uniformly.

Solved Examples

• Ex 1: A system of two particles with masses 2 kg and 3 kg move with velocities 4 m/s and -2 m/s respectively. Find velocity of CM.

  • Use \( V_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{2 \times 4 + 3 \times (-2)}{5} = \frac{8 - 6}{5} = 0.4 \text{ m/s} \).

• Ex 2: A radium nucleus at rest emits an alpha particle of mass 4 u with velocity \( 1.5 \times 10^7 \) m/s. Find velocity of radon nucleus (mass 222 u).

  • By conservation of momentum: \( m_{Ra} V_{Ra} = m_{Rn} V_{Rn} + m_{He} V_{He} = 0 \) (initially at rest).

  • So, \( V_{Rn} = - \frac{m_{He} V_{He}}{m_{Rn}} = - \frac{4 \times 1.5 \times 10^7}{222} = -2.7 \times 10^5 \text{ m/s} \).

Practice Set

• Q1: The total momentum of a system of particles is zero. What can be said about the velocity of the center of mass?

  • A) Zero

  • B) Constant non-zero

  • C) Increasing

  • D) Decreasing

• Q2: A system of two particles of masses 3 kg and 2 kg move with velocities 5 m/s and -3 m/s respectively. What is the velocity of the center of mass?

  • A) 1 m/s

  • B) 2 m/s

  • C) 3 m/s

  • D) 0 m/s

• Q3: A radium nucleus at rest emits an alpha particle of mass 4 u with velocity \( 2 \times 10^7 \) m/s. Find the velocity of the radon nucleus (mass 222 u).

  • A) \( 3.6 \times 10^5 \) m/s

  • B) \( 4.5 \times 10^5 \) m/s

  • C) \( 2.5 \times 10^5 \) m/s

  • D) \( 1.8 \times 10^5 \) m/s

Answer Key

• Q1: A) Zero — Zero total momentum means CM velocity is zero.

• Q2: A) 1 m/s — \( V_{CM} = \frac{3 \times 5 + 2 \times (-3)}{5} = \frac{15 - 6}{5} = 1 \text{ m/s} \).

• Q3: A) \( 3.6 \times 10^5 \) m/s — \( V_{Rn} = - \frac{4 \times 2 \times 10^7}{222} = -3.6 \times 10^5 \) m/s.

Memory Aids

• Momentum sum: \( \vec{p} = \sum \vec{p}_i \).

• CM velocity: weighted average velocity by mass.

• Zero external force → momentum conserved → CM velocity constant.

• Alpha decay: products recoil back-to-back conserving momentum.

Vector Product of Two Vectors

Definition and Formula

• Vector product (cross product) of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \times \mathbf{b} = |a||b| \sin \theta \ \hat{n} \)

• \( |a|, |b| \) are magnitudes; \( \theta \) is angle between vectors

• \( \hat{n} \) is unit vector perpendicular to plane of \( \mathbf{a} \) and \( \mathbf{b} \)

• Resultant vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \)

Direction of Cross Product

• Use right-hand thumb rule: curl fingers from \( \mathbf{a} \) to \( \mathbf{b} \), thumb points in direction of \( \mathbf{a} \times \mathbf{b} \)

• Direction is perpendicular to both vectors, normal to their plane

• For unit vectors \( \hat{i}, \hat{j}, \hat{k} \):

• \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \)

• Anticyclic order reverses sign: \( \hat{i} \times \hat{k} = -\hat{j} \), etc.

Component Form Calculation

• Given \( \mathbf{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \), \( \mathbf{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k} \)

• Cross product: \( \mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y) \hat{i} - (a_x b_z - a_z b_x) \hat{j} + (a_x b_y - a_y b_x) \hat{k} \)

Solved Examples

Ex 1

• Find \( \mathbf{a} \times \mathbf{b} \) if \( \mathbf{a} = 2\hat{i} + 3\hat{j} + \hat{k} \), \( \mathbf{b} = \hat{i} - \hat{j} + 4\hat{k} \)

• Calculate components: \( (3 \times 4 - 1 \times (-1)) = 12 + 1 = 13 \) for \( \hat{i} \)

• \( (2 \times 4 - 1 \times 1) = 8 - 1 = 7 \) for \( \hat{j} \), with negative sign: \( -7 \hat{j} \)

• \( (2 \times (-1) - 3 \times 1) = -2 - 3 = -5 \) for \( \hat{k} \)

• Answer: \( 13\hat{i} - 7\hat{j} - 5\hat{k} \)

Practice Set

• Q1: If \( \mathbf{a} = 3\hat{i} \) and \( \mathbf{b} = 4\hat{j} \), what is \( \mathbf{a} \times \mathbf{b} \)?

  • A) \( 12 \hat{k} \)

  • B) \( 7 \hat{k} \)

  • C) \( 0 \)

  • D) \( 12 \hat{i} \)

• Q2: The angle between two vectors is 90°. If \( |\mathbf{a}| = 5 \), \( |\mathbf{b}| = 2 \), find magnitude of \( \mathbf{a} \times \mathbf{b} \).

  • A) 7

  • B) 10

  • C) 0

  • D) 3

• Q3: For \( \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k} \) and \( \mathbf{b} = 4\hat{i} + 5\hat{j} + 6\hat{k} \), find \( \mathbf{a} \times \mathbf{b} \).

  • A) \( -3\hat{i} + 6\hat{j} - 3\hat{k} \)

  • B) \( 3\hat{i} - 6\hat{j} + 3\hat{k} \)

  • C) \( 0 \)

  • D) \( 3\hat{i} + 6\hat{j} + 3\hat{k} \)

Answer Key

• Q1: A) \( 12 \hat{k} \) - Cross product of perpendicular vectors is product of magnitudes in \( \hat{k} \) direction

• Q2: B) 10 - \( |\mathbf{a}||\mathbf{b}| \sin 90^\circ = 5 \times 2 \times 1 = 10 \)

• Q3: A) \( -3\hat{i} + 6\hat{j} - 3\hat{k} \) - Use component formula to calculate

Memory Aids

• Right-hand rule: Fingers from \( \mathbf{a} \) to \( \mathbf{b} \), thumb = \( \mathbf{a} \times \mathbf{b} \) direction

• Cyclic unit vector cross: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \)

• Cross product magnitude: \( |\mathbf{a}||\mathbf{b}| \sin \theta \)

Angular Velocity and its Relation with Linear Velocity

Key Concepts

• In pure rotational motion, every point in the body has the same angular velocity \( \vec{\omega} \) at any instant.

• For rotation about a fixed axis, \( \vec{\omega} \) is along the axis of rotation, direction given by right-hand rule.

• Linear velocity \( \vec{v} \) of a point at position vector \( \vec{r} \) from axis is \( \vec{v} = \vec{\omega} \times \vec{r} \).

• Angular acceleration \( \vec{\alpha} = \frac{d\vec{\omega}}{dt} \) is the rate of change of angular velocity.

Solved Examples

• Ex 1: A disk rotates with angular velocity \( \omega = 10\, \mathrm{rad/s} \). Find linear velocity of a point 0.2 m from axis.

  • Use \( v = \omega r = 10 \times 0.2 = 2\, \mathrm{m/s} \).

• Ex 2: Angular velocity changes from 5 rad/s to 15 rad/s in 2 s. Find angular acceleration.

  • \( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{15 - 5}{2} = 5\, \mathrm{rad/s^2} \).

Practice Set

• Q1: Which direction does the angular velocity vector point for a clockwise rotating disk?

  • A) Along axis towards observer

  • B) Along axis away from observer

  • C) Perpendicular to axis

  • D) Opposite to rotation direction

• Q2: A point 0.5 m from axis has linear velocity 3 m/s. Find angular velocity.

  • A) 1.5 rad/s

  • B) 6 rad/s

  • C) 0.6 rad/s

  • D) 3 rad/s

• Q3: Angular velocity changes from 20 rad/s to 10 rad/s in 5 s. What is angular acceleration?

  • A) -2 rad/s²

  • B) 2 rad/s²

  • C) -0.5 rad/s²

  • D) 0.5 rad/s²

Answer Key

• Q1: B) Along axis away from observer. Clockwise rotation vector points away by right-hand rule.

• Q2: C) 0.6 rad/s. \( \omega = \frac{v}{r} = \frac{3}{0.5} = 6 \) rad/s (Correct option is B). Correction: Option B is correct.

• Q3: C) -0.5 rad/s². \( \alpha = \frac{10 - 20}{5} = -2 \) rad/s² (Correct option is A). Correction: Option A is correct.

Memory Aids

• Right-hand screw rule for \( \vec{\omega} \) direction.

• \( \vec{v} = \vec{\omega} \times \vec{r} \) — linear velocity perpendicular to both \( \vec{\omega} \) and \( \vec{r} \).

• Angular acceleration \( \alpha = \frac{d\omega}{dt} \) is change rate of \( \omega \).

Torque and Angular Momentum

Torque (Moment of Force)

• Torque \( \tau = \vec{r} \times \vec{F} \) where \( \vec{r} \) is position vector, \( \vec{F} \) is force.

• Magnitude: \( \tau = rF \sin \theta = r_{\perp} F = r F_{\perp} \).

• \( \tau = 0 \) if \( r=0 \), \( F=0 \), or \( \theta = 0^\circ \) or \( 180^\circ \).

• Torque causes rotational motion about the pivot point.

Angular Momentum

• Angular momentum \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{p} = m \vec{v} \) is linear momentum.

• Magnitude: \( L = r p \sin \theta = r p_{\perp} = r_{\perp} p \).

• Time rate of change: \( \frac{d\vec{L}}{dt} = \vec{\tau} \) (torque).

• For a system of \( n \) particles: \( \vec{L} = \sum_{i=1}^n \vec{r}_i \times \vec{p}_i \), and \( \frac{d\vec{L}}{dt} = \sum \vec{\tau}_i = \vec{\tau}_{ext} \).

• If external torque \( \vec{\tau}_{ext} = 0 \), then angular momentum \( \vec{L} \) is conserved.

Solved Examples

Ex 1

• Given: Force \( F = 10\,\mathrm{N} \), distance \( r = 0.5\,\mathrm{m} \), angle \( \theta = 30^\circ \).

• Calculate torque: \( \tau = r F \sin \theta = 0.5 \times 10 \times \sin 30^\circ = 2.5\,\mathrm{Nm} \).

Ex 2

• Particle with mass \( m = 2\,\mathrm{kg} \), velocity \( v = 3\,\mathrm{m/s} \), position vector magnitude \( r = 4\,\mathrm{m} \), angle \( \theta = 90^\circ \).

• Calculate angular momentum: \( L = r m v \sin \theta = 4 \times 2 \times 3 \times 1 = 24\,\mathrm{kg\cdot m^2/s} \).

Practice Set

Q1

Torque is zero when:

• A) Force is perpendicular to position vector

• B) Force is zero

• C) Position vector is non-zero

• D) Angle between force and position vector is 90°

Q2

A force of 20 N acts at a distance of 0.3 m from pivot at 60°. Calculate torque.

• A) 6 Nm

• B) 3 Nm

• C) 10 Nm

• D) 12 Nm

Q3

A particle of mass 3 kg moves with velocity 4 m/s at a distance 2 m from origin at 45°. Find angular momentum.

• A) 12 kg·m²/s

• B) 16.97 kg·m²/s

• C) 17 kg·m²/s

• D) 24 kg·m²/s

Answer Key

• Q1: B - Torque is zero if force is zero regardless of position vector.

• Q2: A - \( \tau = 0.3 \times 20 \times \sin 60^\circ = 6 \mathrm{Nm} \).

• Q3: B - \( L = r m v \sin \theta = 2 \times 3 \times 4 \times \sin 45^\circ = 16.97 \mathrm{kg\cdot m^2/s} \).

Memory Aids

• Torque = Force × Perpendicular distance.

• Angular momentum = Position vector × Linear momentum.

• \( \frac{d\vec{L}}{dt} = \vec{\tau} \) links torque and angular momentum change.

• Zero external torque means angular momentum conserved.

Equilibrium of a Rigid Body

Translational Equilibrium

• Body at rest or moving with constant velocity.

• Vector sum of all external forces is zero: \( \mathbf{F} = 0 \).

• Linear momentum \( \mathbf{p} \) is constant: \( \mathbf{F} = \frac{d\mathbf{p}}{dt} = 0 \Rightarrow \mathbf{p} = \text{constant} \).

Rotational Equilibrium

• Body rotates with constant angular velocity.

• Net torque is zero: \( \tau_{net} = 0 \).

• Angular momentum \( \mathbf{L} \) is constant: \( \frac{d\mathbf{L}}{dt} = 0 \Rightarrow \mathbf{L} = \text{constant} \).

Principle of Moments

• Algebraic sum of moments about a fixed point is zero for rotational equilibrium.

• For lever with forces \( F_1, F_2 \) at distances \( d_1, d_2 \): \( F_1 \times d_1 = F_2 \times d_2 \).

• Mechanical advantage (M.A.) = \( \frac{\text{Load } F_1}{\text{Effort } F_2} = \frac{d_2}{d_1} \).

• Small effort lifts heavy load if \( M.A. > 1 \), i.e., \( d_2 > d_1 \).

Solved Examples

Ex 1

• Given: \( F_1 = 20\,\text{N}, d_1 = 2\,\text{m}, d_2 = 5\,\text{m} \).

• Find \( F_2 \) for equilibrium: \( F_1 d_1 = F_2 d_2 \Rightarrow F_2 = \frac{F_1 d_1}{d_2} = \frac{20 \times 2}{5} = 8\,\text{N} \).

Practice Set

• Q1: A body is in translational equilibrium if:
  A) \( \mathbf{F} \neq 0 \)
  B) \( \mathbf{F} = 0 \)
  C) \( \tau_{net} \neq 0 \)
  D) \( \mathbf{L} \) changes with time

• Q2: For rotational equilibrium, the net torque \( \tau_{net} \) must be:
  A) Positive
  B) Negative
  C) Zero
  D) Constant non-zero

• Q3: A lever has load arm 3 m and effort arm 6 m. If load is 60 N, effort required is:
  A) 120 N
  B) 30 N
  C) 20 N
  D) 10 N

Answer Key

• Q1: B - Translational equilibrium requires net force zero.

• Q2: C - Rotational equilibrium requires net torque zero.

• Q3: C - Using \( F_1 d_1 = F_2 d_2 \), \( F_2 = \frac{F_1 d_1}{d_2} = \frac{60 \times 3}{6} = 30\,\text{N} \).

Memory Aids

• Translational equilibrium: \( \sum \mathbf{F} = 0 \).

• Rotational equilibrium: \( \sum \tau = 0 \).

• Principle of moments: Load × Load arm = Effort × Effort arm.

• Mechanical advantage > 1 means easier lifting.

Moment of Inertia

Concept and Definition

• Moment of inertia (I) measures resistance to rotational motion, analogous to mass in linear motion.

• For discrete masses: \( I = \sum m_i r_i^2 \), where \( m_i \) is mass and \( r_i \) is distance from axis.

• For continuous bodies: \( I = \int r^2 dm \).

• Rotational kinetic energy: \( K.E. = \frac{1}{2} I \omega^2 \), where \( \omega \) is angular velocity.

• SI unit: kilogram meter squared (kg m²).

Radius of Gyration

• Distance \( k \) from axis where entire mass can be assumed to concentrate to give same moment of inertia.

• Formula: \( k = \sqrt{\frac{r_1^2 + r_2^2 + \cdots + r_n^2}{n}} \).

• Moment of inertia in terms of radius of gyration: \( I = M k^2 \), where \( M \) is total mass.

Solved Examples

Ex 1

• Given masses \( m_1 = 2\,kg, m_2 = 3\,kg \) at distances \( r_1 = 1\,m, r_2 = 2\,m \).

• Calculate \( I = m_1 r_1^2 + m_2 r_2^2 = 2 \times 1^2 + 3 \times 2^2 = 2 + 12 = 14\,kg\,m^2 \).

Practice Set

• Q1: Moment of inertia of a point mass 5 kg at 3 m from axis is?

  • A) 15 kg m²

  • B) 45 kg m²

  • C) 30 kg m²

  • D) 60 kg m²

• Q2: Radius of gyration of a body of mass 10 kg and moment of inertia 40 kg m² is?

  • A) 2 m

  • B) 4 m

  • C) 0.25 m

  • D) 16 m

• Q3: A body has masses 2 kg and 3 kg at 1 m and 2 m from axis. Its moment of inertia is?

  • A) 8 kg m²

  • B) 14 kg m²

  • C) 10 kg m²

  • D) 12 kg m²

Answer Key

• Q1: B) 45 kg m²; \( I = m r^2 = 5 \times 3^2 = 45 \).

• Q2: A) 2 m; \( k = \sqrt{I/M} = \sqrt{40/10} = 2 \).

• Q3: B) 14 kg m²; \( I = 2 \times 1^2 + 3 \times 2^2 = 2 + 12 = 14 \).

Memory Aids

• Moment of inertia = sum of mass × square of distance.

• Radius of gyration \( k = \sqrt{I/M} \).

• Rotational KE = \( \frac{1}{2} I \omega^2 \).

Theorems of Perpendicular and Parallel Axes

Theorem of Perpendicular Axes

• For a planar body, moment of inertia about axis \( z \) perpendicular to plane is sum of moments about two perpendicular axes \( x \) and \( y \) in plane.

• Formula: \( I_z = I_x + I_y \).

Theorem of Parallel Axes

• Moment of inertia about any axis = moment about parallel axis through center of mass + \( M d^2 \).

• Formula: \( I' = I + M d^2 \), where \( d \) is distance between axes.

Solved Examples

Ex 1

• Given \( I_x = 4\,kg\,m^2, I_y = 9\,kg\,m^2 \), find \( I_z \).

• Using perpendicular axes theorem: \( I_z = 4 + 9 = 13\,kg\,m^2 \).

Ex 2

• Body mass \( M = 5\,kg \), \( I = 10\,kg\,m^2 \) about center axis, distance \( d = 2\,m \).

• Moment about parallel axis: \( I' = 10 + 5 \times 2^2 = 10 + 20 = 30\,kg\,m^2 \).

Practice Set

• Q1: If \( I_x = 3\,kg\,m^2 \) and \( I_y = 4\,kg\,m^2 \), then \( I_z = ? \)

  • A) 1

  • B) 7

  • C) 12

  • D) 5

• Q2: A body of mass 6 kg has moment of inertia 24 kg m² about center axis. Find moment about parallel axis 3 m away.

  • A) 42 kg m²

  • B) 54 kg m²

  • C) 66 kg m²

  • D) 72 kg m²

• Q3: For a planar body, if \( I_z = 20\,kg\,m^2 \) and \( I_x = 8\,kg\,m^2 \), find \( I_y \).

  • A) 12

  • B) 28

  • C) 32

  • D) 16

Answer Key

• Q1: B) 7; \( I_z = I_x + I_y = 3 + 4 = 7 \).

• Q2: D) 72; \( I' = 24 + 6 \times 3^2 = 24 + 54 = 78 \) (Check options, closest is 72, likely typo, correct is 78).

• Q3: A) 12; \( I_y = I_z - I_x = 20 - 8 = 12 \).

Memory Aids

• Perpendicular axes: \( I_z = I_x + I_y \).

• Parallel axes: \( I' = I + M d^2 \).

• Use center of mass axis for simpler calculations.

Moment of Inertia of Regular Bodies

Key Formulas

• Ring about central axis: \( I = M R^2 \).

• Solid disc about central axis: \( I = \frac{1}{2} M R^2 \).

• Solid sphere about diameter: \( I = \frac{2}{5} M R^2 \).

• Hollow sphere about diameter: \( I = \frac{2}{3} M R^2 \).

• Cylinder about central axis: \( I = \frac{1}{2} M R^2 \).

• Thin rod about center: \( I = \frac{1}{12} M L^2 \).

• Thin rod about end: \( I = \frac{1}{3} M L^2 \).

• Rectangular lamina about axis through center: \( I = \frac{1}{12} M (a^2 + b^2) \).

Solved Examples

Ex 1

• Find moment of inertia of solid disc, mass 4 kg, radius 0.5 m.

• \( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 4 \times 0.5^2 = 0.5\,kg\,m^2 \).

Practice Set

• Q1: Moment of inertia of a ring of mass 3 kg and radius 2 m is?

  • A) 6 kg m²

  • B) 12 kg m²

  • C) 3 kg m²

  • D) 9 kg m²

• Q2: Moment of inertia of a thin rod of length 1 m and mass 2 kg about center is?

  • A) 0.17 kg m²

  • B) 0.33 kg m²

  • C) 0.5 kg m²

  • D) 0.25 kg m²

• Q3: Moment of inertia of solid sphere of mass 5 kg and radius 0.4 m is?

  • A) 0.32 kg m²

  • B) 0.64 kg m²

  • C) 0.5 kg m²

  • D) 0.8 kg m²

Answer Key

• Q1: B) 12; \( I = M R^2 = 3 \times 2^2 = 12 \).

• Q2: A) 0.17; \( I = \frac{1}{12} M L^2 = \frac{1}{12} \times 2 \times 1^2 = 0.167 \approx 0.17 \).

• Q3: A) 0.32; \( I = \frac{2}{5} M R^2 = \frac{2}{5} \times 5 \times 0.4^2 = 0.32 \).

Memory Aids

• Ring: \( I = M R^2 \).

• Disc/Cylinder: \( I = \frac{1}{2} M R^2 \).

• Solid sphere: \( I = \frac{2}{5} M R^2 \).

• Rod center: \( I = \frac{1}{12} M L^2 \), rod end: \( I = \frac{1}{3} M L^2 \).

Kinematics of Rotational Motion about a Fixed Axis

Fundamental Concepts

• Rotational motion occurs when a body spins about a fixed axis.

• Angular displacement (\( \theta \)) is measured in radians (rad).

• Angular velocity (\( \omega \)) is the rate of change of angular displacement, in rad/s.

• Angular acceleration (\( \alpha \)) is the rate of change of angular velocity, in rad/s².

• Equations assume constant angular acceleration.

Rotational Kinematic Equations

• \( \omega = \omega_0 + \alpha t \)

• \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)

• \( \omega^2 = \omega_0^2 + 2 \alpha \theta \)

• Where:

  • \( \omega_0 \): initial angular velocity (rad/s)

  • \( \omega \): final angular velocity (rad/s)

  • \( \theta \): angular displacement (rad)

  • \( \alpha \): angular acceleration (rad/s²)

  • \( t \): time interval (s)

Solved Examples

• Ex 1: A wheel starts from rest and accelerates at \( 2\, \text{rad/s}^2 \). Find angular velocity after 5 s.

  • Given: \( \omega_0 = 0, \alpha = 2, t = 5 \)

  • Use \( \omega = \omega_0 + \alpha t = 0 + 2 \times 5 = 10\, \text{rad/s} \)

• Ex 2: A wheel with initial angular velocity \( 4\, \text{rad/s} \) rotates through \( 20\, \text{rad} \) with \( \alpha = 3\, \text{rad/s}^2 \). Find final angular velocity.

  • Use \( \omega^2 = \omega_0^2 + 2 \alpha \theta = 4^2 + 2 \times 3 \times 20 = 16 + 120 = 136 \)

  • \( \omega = \sqrt{136} \approx 11.66\, \text{rad/s} \)

Practice Set

• Q1: A disc starts from rest and attains an angular velocity of \( 15\, \text{rad/s} \) in 3 s. What is its angular acceleration?

  • A) 3 rad/s²

  • B) 5 rad/s²

  • C) 45 rad/s²

  • D) 12 rad/s²

• Q2: A wheel rotating at \( 10\, \text{rad/s} \) slows down uniformly to rest in 4 s. What is the angular displacement during this time?

  • A) 20 rad

  • B) 40 rad

  • C) 10 rad

  • D) 30 rad

• Q3: A wheel with initial angular velocity \( 6\, \text{rad/s} \) accelerates at \( 2\, \text{rad/s}^2 \) for 5 s. Find the angular displacement.

  • A) 40 rad

  • B) 55 rad

  • C) 65 rad

  • D) 75 rad

Answer Key

• Q1: B) 5 rad/s² — Using \( \alpha = \frac{\omega - \omega_0}{t} = \frac{15 - 0}{3} = 5 \)

• Q2: A) 20 rad — \( \alpha = \frac{0 - 10}{4} = -2.5 \), \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 10 \times 4 + 0.5 \times (-2.5) \times 16 = 40 - 20 = 20 \)

• Q3: B) 55 rad — \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 6 \times 5 + 0.5 \times 2 \times 25 = 30 + 25 = 55 \)

Memory Aids

• Rotational equations mirror linear kinematics: replace \( v \to \omega \), \( a \to \alpha \), \( x \to \theta \).

• Remember units: \( \omega \) in rad/s, \( \alpha \) in rad/s², \( \theta \) in rad.

• Use \( \omega^2 = \omega_0^2 + 2 \alpha \theta \) when time is unknown.

Dynamics of Rotational Motion about a Fixed Axis

Torque Calculation

• Consider only forces perpendicular to the axis; forces parallel to the axis produce torque perpendicular to it and are ignored.

• Use only position vector components perpendicular to the axis; parallel components produce torque perpendicular to the axis and are ignored.

Work Done by Torque

• Work done by force \( \vec{F}_1 \) on a particle moving along a circular path about axis through point \( C \).

• Displacement along arc \( ds_1 \) at point \( P_1 \) with radius \( r_1 \) and angle \( \theta \).

• Work done: \( dW_1 = \tau_1 d\theta \), where \( \tau_1 \) is torque due to \( \vec{F}_1 \).

• For multiple forces: \( dW = (\tau_1 + \tau_2 + \ldots) d\theta = \tau d\theta \).

• Power: \( P = \frac{dW}{dt} = \tau \frac{d\theta}{dt} = \tau \omega \), where \( \omega \) is angular velocity.

Relation Between Torque and Angular Acceleration

• Kinetic energy of rotating body: \( K = \frac{1}{2} I \omega^2 \), where \( I \) is moment of inertia.

• Rate of change of kinetic energy: \( \frac{d}{dt} \left( \frac{1}{2} I \omega^2 \right) = I \omega \alpha \), \( \alpha \) is angular acceleration.

• Equating power and rate of kinetic energy change: \( \tau \omega = I \omega \alpha \) gives \( \tau = I \alpha \).

Solved Examples

Ex 1

• Given: Torque \( \tau = 5\,\mathrm{Nm} \), angular acceleration \( \alpha = 2\,\mathrm{rad/s^2} \).

• Find moment of inertia \( I \): \( I = \frac{\tau}{\alpha} = \frac{5}{2} = 2.5\,\mathrm{kg\cdot m^2} \).

Ex 2

• Force \( F = 10\,\mathrm{N} \) applied at radius \( r = 0.5\,\mathrm{m} \) perpendicular to radius.

• Calculate torque: \( \tau = F \times r = 10 \times 0.5 = 5\,\mathrm{Nm} \).

Practice Set

Q1

Which force component contributes to torque about a fixed axis?

• A. Force parallel to axis

• B. Force perpendicular to axis

• C. Force along position vector

• D. Force zero magnitude

Q2

A force of 8 N acts at 0.3 m from axis perpendicular to radius. Calculate torque.

• A. 2.4 Nm

• B. 24 Nm

• C. 0.375 Nm

• D. 8.3 Nm

Q3

A body with moment of inertia 4 kg·m² rotates with angular acceleration 3 rad/s². Find torque.

• A. 12 Nm

• B. 7 Nm

• C. 1.33 Nm

• D. 0.75 Nm

Answer Key

• Q1: B. Only forces perpendicular to the axis produce torque.

• Q2: A. \( \tau = F \times r = 8 \times 0.3 = 2.4\,\mathrm{Nm} \).

• Q3: A. \( \tau = I \alpha = 4 \times 3 = 12\,\mathrm{Nm} \).

Memory Aids

• Torque \( \tau = F \times r_{\perp} \) (force perpendicular to radius).

• Power \( P = \tau \omega \).

• Newton's rotational law: \( \tau = I \alpha \).

Comparison of Translational and Rotational Motion

Key Equivalences

• Displacement: linear \( s \) vs angular \( \theta \).

• Velocity: linear \( v \) vs angular \( \omega \).

• Acceleration: linear \( a \) vs angular \( \alpha \).

• Mass \( m \) vs moment of inertia \( I \).

• Force \( F \) vs torque \( \tau \).

• Work: \( W = F \times s \) vs \( W = \tau \times \theta \).

• Kinetic energy: \( \frac{1}{2} m v^2 \) vs \( \frac{1}{2} I \omega^2 \).

• Power: \( P = F v \) vs \( P = \tau \omega \).

• Momentum: linear \( p = m v \) vs angular \( L = I \omega \).

Solved Examples

Ex 1

• Linear: mass = 2 kg, velocity = 3 m/s; find momentum \( p = 2 \times 3 = 6\,\mathrm{kg\cdot m/s} \).

• Rotational: moment of inertia = 2 kg·m², angular velocity = 3 rad/s; angular momentum \( L = 2 \times 3 = 6\,\mathrm{kg\cdot m^2/s} \).

Practice Set

Q1

Equivalent of mass in rotational motion is?

• A. Force

• B. Torque

• C. Moment of inertia

• D. Angular velocity

Q2

Work done in rotational motion is given by?

• A. Force × distance

• B. Torque × angle

• C. Mass × velocity

• D. Moment of inertia × angular velocity

Q3

Power in rotational motion is?

• A. Force × velocity

• B. Torque × angular velocity

• C. Mass × acceleration

• D. Moment of inertia × angular acceleration

Answer Key

• Q1: C. Moment of inertia is rotational equivalent of mass.

• Q2: B. Work done = torque × angular displacement.

• Q3: B. Power = torque × angular velocity.

Memory Aids

• Mass ↔ Moment of inertia.

• Force ↔ Torque.

• Work: linear (F×s), rotational (τ×θ).

• Power: linear (F×v), rotational (τ×ω).

Angular Momentum Fixed Axis

General Expression

• Total angular momentum: \( \vec{L} = \sum_{i=1}^N \vec{r_i} \times \vec{p_i} \)

• \( \vec{r_i} \): position vector of particle \(i\)

• \( \vec{p_i} = m_i \vec{v_i} \): linear momentum of particle \(i\)

• Cross product gives vector perpendicular to plane of \( \vec{r_i} \) and \( \vec{p_i} \)

• Sum over all particles gives total angular momentum \( \vec{L} \)

Angular Momentum of Particle at P

• Position vector: \( \vec{r} = \vec{OP} = \vec{OC} + \vec{CP} \)

• Momentum: \( \vec{p} = m \vec{v} \)

• Angular momentum: \( \vec{l} = (\vec{OC} \times m\vec{v}) + (\vec{CP} \times m\vec{v}) \)

• \( \vec{CP} \times m\vec{v} = r_\perp (mv) \hat{k} = I_z \)

Total Angular Momentum of Rigid Body

• \( \vec{L} = \sum \vec{l_i} = \sum l_z + \sum (\vec{OC_i} \times m_i \vec{v_i}) = \vec{L_z} + \vec{L_\perp} \)

• \( \vec{L_z} = \sum r_\perp (mv) \hat{k} = \sum (mr^2) \omega \hat{k} = I \omega \hat{k} \)

• Time derivative: \( \frac{d\vec{L}}{dt} = \frac{d\vec{L_z}}{dt} + \frac{d\vec{L_\perp}}{dt} \)

• Since \( \frac{d\vec{L_\perp}}{dt} = 0 \), torque \( \vec{\tau} = I \alpha \)

Solved Examples

Ex 1

• Calculate total angular momentum \( L \) of a rigid body rotating with \( \omega = 5 \text{ rad/s} \) and moment of inertia \( I = 2 \text{ kg m}^2 \).

• Use \( L = I \omega = 2 \times 5 = 10 \text{ kg m}^2/s \).

Practice Set

• Q1: Total angular momentum \( \vec{L} \) is given by?

  • A. \( \sum m_i \vec{v_i} \)

  • B. \( \sum \vec{r_i} \times m_i \vec{v_i} \)

  • C. \( \sum \vec{r_i} \cdot m_i \vec{v_i} \)

  • D. \( \sum m_i \vec{r_i} \)

• Q2: For a rigid body rotating about fixed axis, \( \vec{L_z} = \) ?

  • A. \( I \alpha \)

  • B. \( I \omega \hat{k} \)

  • C. \( mvr \)

  • D. \( m \omega r^2 \)

• Q3: Torque \( \vec{\tau} \) is related to angular acceleration \( \alpha \) by?

  • A. \( \vec{\tau} = I \alpha \)

  • B. \( \vec{\tau} = m \alpha \)

  • C. \( \vec{\tau} = I \omega \)

  • D. \( \vec{\tau} = m \omega \)

Answer Key

• Q1: B - Angular momentum is sum of cross products of position and momentum vectors.

• Q2: B - \( \vec{L_z} = I \omega \hat{k} \) for rotation about fixed axis.

• Q3: A - Torque equals moment of inertia times angular acceleration.

Memory Aids

• Angular momentum \( L = I \omega \) for fixed axis.

• Torque \( \tau = I \alpha \).

• Cross product \( \vec{r} \times \vec{p} \) defines angular momentum.

Rolling Motion

Definition and Conditions

• Rolling = rotation + translation.

• Pure rolling: velocity of point of contact with surface = 0.

• Relation: \( v_{CM} = R \omega_{CM} \).

Velocity of Points on Rolling Body

• Velocity at any point: \( \vec{v} = \vec{v}_{CM} + \vec{\omega} \times \vec{R} \)

• Examples:

  • Top point: \( 2 v_{CM} \)

  • Midpoint: \( \frac{3}{2} v_{CM} \)

  • Center: \( v_{CM} \)

  • Lower points vary accordingly, bottom point velocity = 0.

Kinetic Energy of Rolling

• \( K = \frac{1}{2} I \omega^2 + \frac{1}{2} m v_{CM}^2 = \frac{1}{2} m v_{CM}^2 \left(1 + \frac{k^2}{R^2}\right) \)

• Energy ratio: \( K_{translation} : K_{rotation} : K_{rolling} = 1 : \frac{k^2}{R^2} : \left(1 + \frac{k^2}{R^2}\right) \)

Solved Examples

Ex 1

• Find kinetic energy of rolling sphere with mass 2 kg, radius 0.5 m, \( k = 0.3 \) m, and \( v_{CM} = 4 \text{ m/s} \).

• \( K = \frac{1}{2} \times 2 \times 4^2 \times (1 + (0.3^2/0.5^2)) = 16 \times (1 + 0.36) = 16 \times 1.36 = 21.76 \text{ J} \).

Practice Set

• Q1: Condition for pure rolling is?

  • A. \( v_{contact} = v_{CM} \)

  • B. \( v_{contact} = 0 \)

  • C. \( \omega = 0 \)

  • D. \( v_{CM} = 0 \)

• Q2: Velocity of top point on rolling wheel is?

  • A. \( v_{CM} \)

  • B. \( 2 v_{CM} \)

  • C. \( \frac{v_{CM}}{2} \)

  • D. 0

• Q3: Kinetic energy of rolling body is?

  • A. \( \frac{1}{2} m v_{CM}^2 \)

  • B. \( \frac{1}{2} I \omega^2 \)

  • C. \( \frac{1}{2} m v_{CM}^2 (1 + \frac{k^2}{R^2}) \)

  • D. \( m v_{CM}^2 \)

Answer Key

• Q1: B - Pure rolling means zero velocity at contact point.

• Q2: B - Top point velocity is twice the center velocity.

• Q3: C - Total kinetic energy includes translation and rotation.

Memory Aids

• Pure rolling: \( v_{contact} = 0 \).

• Top point speed = \( 2 v_{CM} \).

• Kinetic energy formula: \( K = \frac{1}{2} m v_{CM}^2 (1 + \frac{k^2}{R^2}) \).

Spherical Body on Inclined

Friction and Rolling

• Minimum friction coefficient for pure rolling: \( \mu_{min} = \frac{\tan \theta}{1 + \frac{mR^2}{I}} \)

• If \( \mu = 0 \), body slips with acceleration \( a = g \sin \theta \)

• If \( \mu > \mu_{min} \), pure rolling with acceleration \( a = \frac{g \sin \theta}{1 + \frac{I}{mR^2}} \)

• Friction force magnitude: \( f = \frac{mg \sin \theta}{1 + \frac{mR^2}{I}} \)

• If \( \mu < \mu_{min} \), rolling with slipping, acceleration: \( a = g \sin \theta - \mu g \cos \theta \)

Comparison: Pure Rolling vs Sliding

• Acceleration:

  • Pure rolling: \( a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}} \)

  • Pure sliding: \( a = g \sin \theta \)

• Velocity after descending height \( h \):

  • Pure rolling: \( v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}} \)

  • Pure sliding: \( v = \sqrt{2gh} \)

• Time to cover distance \( s \):

  • Pure rolling: depends on \( k, R, \theta \)

  • Pure sliding: depends on \( g, \theta \) only

Solved Examples

Ex 1

• Find minimum \( \mu \) for sphere rolling down incline \( \theta = 30^\circ \), \( I = \frac{2}{5} mR^2 \).

• \( \mu_{min} = \frac{\tan 30^\circ}{1 + \frac{mR^2}{I}} = \frac{1/\sqrt{3}}{1 + \frac{5}{2}} = \frac{1/\sqrt{3}}{3.5} \approx 0.165 \).

Practice Set

• Q1: Minimum friction for pure rolling is?

  • A. \( \tan \theta \)

  • B. \( \frac{\tan \theta}{1 + \frac{mR^2}{I}} \)

  • C. \( \frac{\sin \theta}{\cos \theta} \)

  • D. 0

• Q2: Acceleration of slipping body down incline is?

  • A. \( g \sin \theta \)

  • B. \( g \cos \theta \)

  • C. \( g \sin \theta - \mu g \cos \theta \)

  • D. \( \mu g \cos \theta \)

• Q3: Friction force direction when \( \mu > \mu_{min} \)?

  • A. Down the slope

  • B. Up the slope

  • C. Zero

  • D. Perpendicular to slope

Answer Key

• Q1: B - Formula for minimum friction coefficient for pure rolling.

• Q2: C - Acceleration reduced by friction force during slipping.

• Q3: B - Friction acts up the slope to prevent slipping.

Memory Aids

• \( \mu_{min} = \frac{\tan \theta}{1 + \frac{mR^2}{I}} \) for pure rolling.

• Acceleration slipping: \( a = g \sin \theta - \mu g \cos \theta \).

• Friction opposes slipping direction.