Chapter 7: System of Particles and Rotational Motion

So far we treated bodies as point particles. Real objects rotate, deform and have spatial extent. The trick is to use the centre of mass (CM) to handle translation, and develop a parallel set of laws — torque, angular momentum, moment of inertia — for rotation. The analogy with linear motion is striking and conceptually beautiful.

Concept Map

• 7.1 — Centre of mass: discrete and continuous bodies
• 7.2 — Motion of the CM
• 7.3 — Linear momentum of a system
• 7.4 — Vector (cross) product
• 7.5 — Angular velocity, angular acceleration, $v = \omega r$
• 7.6 — Torque and angular momentum
• 7.7 — Conservation of angular momentum
• 7.8 — Equilibrium of a rigid body; couple, principle of moments
• 7.9 — Centre of gravity vs centre of mass
• 7.10 — Moment of inertia; parallel and perpendicular axis theorems
• 7.11 — MI of standard bodies
• 7.12 — Kinematics and dynamics of rotation
• 7.13 — Rolling motion

7.1 Centre of Mass

Definition (Two-Particle System)

For two particles of masses $m_1, m_2$ at positions $\vec{r}_1, \vec{r}_2$,

$\vec{r}_{\text{cm}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$

The CM lies on the line joining them, closer to the heavier mass.

Generalisation (n Particles)

$\vec{r}_{\text{cm}} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i\vec{r}_i$

Continuous Distribution

$\vec{r}_{\text{cm}} = \frac{1}{M}\int \vec{r}\,dm$

Derivation — CM of a Uniform Rod

A uniform rod of length $L$ and linear density $\lambda$. Take the origin at one end. $dm = \lambda\,dx$.

$x_{\text{cm}} = \frac{1}{\lambda L}\int_0^L x(\lambda\,dx) = \frac{1}{L}\cdot\frac{L^2}{2} = \frac{L}{2}$

CM is at the midpoint, as expected.

Derivation — CM of a Triangular Lamina

A uniform triangular plate has its CM at the centroid — the intersection of the medians, which divides each median in the ratio $2:1$. (Proof: integrate over horizontal strips; the strip width varies linearly with height.)

Derivation — CM of a Semicircular Wire

Wire of radius $R$, linear density $\lambda$, lying in the upper half plane.

Parametrise: $x = R\cos\theta$, $y = R\sin\theta$, $dm = \lambda R\,d\theta$. By symmetry $x_{\text{cm}} = 0$.

$y_{\text{cm}} = \frac{1}{M}\int_0^\pi (R\sin\theta)(\lambda R\,d\theta) = \frac{\lambda R^2}{\lambda\pi R}\int_0^\pi\sin\theta\,d\theta = \frac{R}{\pi}\cdot 2 = \frac{2R}{\pi}$

Derivation — CM of a Semicircular Disc

Surface density $\sigma$, radius $R$, upper half. Use strips $dy$.

$y_{\text{cm}} = \frac{1}{M}\int_0^R y\cdot 2\sigma\sqrt{R^2-y^2}\,dy = \frac{4}{\pi R^2}\cdot\frac{R^3}{3} = \frac{4R}{3\pi}$

Derivation — CM of a Solid Hemisphere

Volume density $\rho$, radius $R$. By symmetry $x_{\text{cm}} = y_{\text{cm}} = 0$. Take strips $dz$ at height $z$; cross-sectional radius $r = \sqrt{R^2-z^2}$.

$z_{\text{cm}} = \frac{1}{M}\int_0^R z\cdot\rho\pi(R^2-z^2)\,dz = \frac{3}{2\pi R^3}\cdot\pi\cdot\frac{R^4}{4} = \frac{3R}{8}$

Special Cases

• Symmetric body (sphere, cube): CM at geometric centre.
• Removed portion: treat as a negative mass and apply the discrete formula.

Worked Example

A uniform disc of radius $R$ has a circular hole of radius $R/2$ cut out, centred at $R/2$ from the disc's centre. Find the CM of the remaining piece.

Treat disc + (negative mass) small disc. Mass of full disc $M$; small disc has mass $M/4$.

$x_{\text{cm}} = \frac{M\cdot 0 - (M/4)(R/2)}{M - M/4} = \frac{-MR/8}{3M/4} = -\frac{R}{6}$

CM is displaced $R/6$ on the side opposite the hole.

Common Mistakes

• Forgetting to use signed masses for removed portions.
• Mixing up densities (linear, surface, volume) when integrating.

7.2 Motion of the Centre of Mass

Derivation

Differentiating $M\vec{r}_{\text{cm}} = \sum m_i\vec{r}_i$:

$M\vec{v}_{\text{cm}} = \sum m_i\vec{v}_i = \vec{P}_{\text{total}}$ $M\vec{a}_{\text{cm}} = \sum m_i\vec{a}_i = \sum \vec{F}_i = \vec{F}_{\text{ext}}$

Internal forces cancel pairwise (Newton's third law), so the CM moves as if all mass were concentrated there and only external forces acted on it.

Consequence

If $\vec{F}_{\text{ext}} = 0$, then $\vec{v}_{\text{cm}} =$ constant — the CM coasts.

Worked Example

A grenade flying in a parabolic path explodes mid-flight. Where does the CM go after the explosion?

The CM continues along the original parabolic trajectory, because the explosion produces only internal forces. (Until the first fragment lands and external normal/contact forces start acting.)

Common Mistakes

• Equating CM motion to motion of any particular point.
• Ignoring that internal forces can change individual momenta but never the total.

7.3 Linear Momentum of a System

Definition

$\vec{P} = \sum_i m_i\vec{v}_i = M\vec{v}_{\text{cm}}$

Derivation of the System Equation of Motion

$\frac{d\vec{P}}{dt} = M\vec{a}_{\text{cm}} = \vec{F}_{\text{ext}}$

So Newton's second law applies to the CM with only external forces.

Worked Example

Two skaters of $50$ and $70\,\text{kg}$ stand on smooth ice $4\,\text{m}$ apart. They pull on a rope. Where do they meet?

CM stays put. Let the heavier skater be at origin. CM at $50\cdot 4/(50+70) = 200/120 = 1.67\,\text{m}$ from the heavier skater.

They meet at the CM, $1.67\,\text{m}$ from the $70\,\text{kg}$ skater and $2.33\,\text{m}$ from the $50\,\text{kg}$ one.

Common Mistakes

• Forgetting that pulling forces are internal and CM cannot move.

7.4 Vector (Cross) Product

Definition

For vectors $\vec{A}$ and $\vec{B}$ with angle $\theta$ between them,

$\vec{A}\times\vec{B} = AB\sin\theta\,\hat{n}$

where $\hat{n}$ is perpendicular to both, chosen by the right-hand rule.

Cartesian Form

$\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z\end{vmatrix}$

Key Properties

• Anti-commutative: $\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}$.
• Distributive: $\vec{A}\times(\vec{B}+\vec{C}) = \vec{A}\times\vec{B} + \vec{A}\times\vec{C}$.
• $\vec{A}\times\vec{A} = 0$.
• $\hat{i}\times\hat{j}=\hat{k}$, $\hat{j}\times\hat{k}=\hat{i}$, $\hat{k}\times\hat{i}=\hat{j}$ (cyclic).

Worked Example

$\vec{A} = \hat{i} + 2\hat{j}$, $\vec{B} = 3\hat{i} - \hat{j}$. Find $\vec{A}\times\vec{B}$.

$\vec{A}\times\vec{B} = (0 - 0)\hat{i} - (0 - 0)\hat{j} + (-1 - 6)\hat{k} = -7\hat{k}$

Common Mistakes

• Forgetting the sign or treating cross-product like dot-product.
• Mis-applying the right-hand rule.

7.5 Angular Velocity and Acceleration

Definitions

Angular velocity:

$\vec{\omega} = \frac{d\vec{\theta}}{dt}$

Angular acceleration:

$\vec{\alpha} = \frac{d\vec{\omega}}{dt}$

For a particle in circular motion of radius $r$,

$\vec{v} = \vec{\omega}\times\vec{r}$

with magnitude $v = \omega r$.

Derivation of $v = \omega r$

If $\vec{r}$ has constant magnitude (circular motion),

$\vec{v} = \frac{d\vec{r}}{dt} = \vec{\omega}\times\vec{r}$

In magnitude: $v = \omega r\sin 90^\circ = \omega r$.

Tangential and Radial Acceleration

For variable $\omega$,

$\vec{a} = \frac{d\vec{v}}{dt} = \vec{\alpha}\times\vec{r} + \vec{\omega}\times\vec{v}$

$\vec{a}_t = \vec{\alpha}\times\vec{r}, \quad |\vec{a}_t| = \alpha r$ $\vec{a}_c = \vec{\omega}\times(\vec{\omega}\times\vec{r}), \quad |\vec{a}_c| = \omega^2 r$

Worked Example

A wheel spins up from rest with $\alpha = 2\,\text{rad/s}^2$. After $5\,\text{s}$, the rim ($r = 0.5\,\text{m}$) has tangential acceleration?

$\omega = 10\,\text{rad/s}$, $a_t = \alpha r = 1\,\text{m/s}^2$. Centripetal: $a_c = \omega^2 r = 50\,\text{m/s}^2$.

Common Mistakes

• Treating $\omega$ as a scalar in 3D problems.
• Confusing $a_t$ with $a_c$.

7.6 Torque and Angular Momentum

Definition (Torque)

$\vec{\tau} = \vec{r}\times\vec{F}$

with magnitude $\tau = rF\sin\theta = F\cdot r_\perp$, where $r_\perp$ is the moment arm.

Definition (Angular Momentum)

For a single particle,

$\vec{L} = \vec{r}\times\vec{p} = m\vec{r}\times\vec{v}$

Derivation (Rotational Newton's Law)

Differentiate $\vec{L} = \vec{r}\times\vec{p}$:

$\frac{d\vec{L}}{dt} = \vec{v}\times\vec{p} + \vec{r}\times\frac{d\vec{p}}{dt}$

The first term vanishes ($\vec{v}\parallel\vec{p}$). So

$\boxed{\frac{d\vec{L}}{dt} = \vec{\tau}}$

This is the angular analog of $d\vec{p}/dt = \vec{F}$.

For a System

Total angular momentum $\vec{L} = \sum_i \vec{r}_i\times\vec{p}_i$. Internal torques cancel pairwise, so

$\frac{d\vec{L}}{dt} = \vec{\tau}_{\text{ext}}$

Worked Example

A particle of $2\,\text{kg}$ moves with $\vec{v} = 3\hat{i}\,\text{m/s}$ at position $\vec{r} = 2\hat{j}\,\text{m}$. Find $\vec{L}$ about the origin.

$\vec{L} = m\vec{r}\times\vec{v} = 2(2\hat{j}\times 3\hat{i}) = 12(\hat{j}\times\hat{i}) = -12\hat{k}\;\text{kg m}^2/\text{s}$

Common Mistakes

• Forgetting that $\vec{L}$ depends on the choice of reference point.
• Adding torques from internal forces (they cancel).

7.7 Conservation of Angular Momentum

Statement

If $\vec{\tau}_{\text{ext}} = 0$, then $\vec{L} =$ constant.

Examples

• Spinning ice skater pulling in arms: $I$ decreases, $\omega$ increases (since $L = I\omega$ is fixed).
• Planetary orbits (Kepler's second law): gravity is central, so $\tau = 0$ about the Sun, hence $L$ is conserved, and the radius vector sweeps equal areas in equal times.
• Diver tucking to spin faster.

Worked Example

A skater with arms outstretched ($I_1 = 10\,\text{kg m}^2$) spins at $2\,\text{rad/s}$. She pulls her arms in, $I_2 = 4\,\text{kg m}^2$. Final $\omega$?

$I_1\omega_1 = I_2\omega_2 \quad \Rightarrow \quad \omega_2 = \frac{10\cdot 2}{4} = 5\,\text{rad/s}$

Common Mistakes

• Confusing conservation of $L$ with conservation of KE (which is not conserved here — the skater does internal work).

7.8 Equilibrium of a Rigid Body

Conditions

1. Translational equilibrium: $\sum \vec{F} = 0$.
2. Rotational equilibrium: $\sum \vec{\tau} = 0$ about any point.

If both hold, the body is in mechanical equilibrium.

Couple

Two equal and opposite forces with different lines of action form a couple. Net force $= 0$, net torque $\neq 0$. Torque of couple = $F\cdot d$, where $d$ is the perpendicular distance between the lines.

Principle of Moments

For a balanced lever:

$F_1 d_1 = F_2 d_2$

where $d_1, d_2$ are perpendicular distances from the pivot to the lines of action.

Mechanical Advantage

$\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{d_{\text{effort}}}{d_{\text{load}}}$

Worked Example

A uniform rod of $4\,\text{kg}$ and length $2\,\text{m}$ rests horizontally on two supports at its ends. Find the reaction at each support.

By symmetry, each support carries half the weight: $R_1 = R_2 = 20\,\text{N}$.

Worked Example

A $40\,\text{kg}$ child sits at the $1\,\text{m}$ mark of a seesaw. Where must a $60\,\text{kg}$ adult sit (other side) to balance the seesaw whose pivot is at the centre?

$40\cdot 1 = 60\cdot d \Rightarrow d = 2/3\,\text{m}$

Common Mistakes

• Choosing a poor reference point for torque (always choose to eliminate an unknown).
• Forgetting that for equilibrium, the choice of axis is arbitrary — pick what is convenient.

7.9 Centre of Gravity vs Centre of Mass

Definitions

• Centre of mass: depends only on the mass distribution.
• Centre of gravity (CG): point where the resultant gravitational force acts.

When They Coincide

If the gravitational field is uniform (constant $g$), CG = CM. For most laboratory-scale objects this holds.

When They Differ

For very large bodies (e.g., a mountain), $g$ varies appreciably across the body, so CG is shifted toward the region of stronger gravity.

Common Mistakes

• Assuming CG and CM are always identical — they only coincide in uniform $g$.

7.10 Moment of Inertia

Definition

For a discrete system about an axis,

$I = \sum_i m_i r_i^2$

For a continuous body,

$I = \int r^2\,dm$

where $r$ is the perpendicular distance from the axis.

Radius of Gyration

$k = \sqrt{\frac{I}{M}} \quad \Rightarrow \quad I = Mk^2$

Perpendicular Axis Theorem (planar lamina)

For a flat body in the $xy$-plane,

$I_z = I_x + I_y$

Proof: $I_z = \int(x^2+y^2)\,dm = \int x^2\,dm + \int y^2\,dm = I_y + I_x$

(Here $I_x$ is about the $x$-axis, which uses the $y$-coordinate, etc.)

Parallel Axis Theorem

If $I_{\text{cm}}$ is the MI about an axis through the CM and $I$ is the MI about a parallel axis at distance $d$,

$I = I_{\text{cm}} + Md^2$

Proof: Let the CM be at the origin. Then

$I = \int|\vec{r}-\vec{d}|^2\,dm = \int(r^2 - 2\vec{r}\cdot\vec{d} + d^2)\,dm$

$= I_{\text{cm}} - 2\vec{d}\cdot\int\vec{r}\,dm + Md^2 = I_{\text{cm}} + Md^2$

since $\int\vec{r}\,dm = 0$ by definition of CM.

Worked Example

A rod of length $L$, mass $M$. $I_{\text{cm}}$ (through centre, perpendicular to rod) $= ML^2/12$. MI about one end?

$I = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{3}$

Common Mistakes

• Applying perpendicular-axis theorem to 3D bodies (it works only for laminae).
• Using parallel-axis theorem with a non-CM axis as the reference.

7.11 Moment of Inertia of Standard Bodies

Uniform Rod about Centre, Perpendicular to Length

$I = \int_{-L/2}^{L/2} x^2\lambda\,dx = \frac{\lambda L^3}{12} = \frac{ML^2}{12}$

Uniform Ring about Central Axis (Perpendicular to Plane)

All mass at distance $R$:

$I = MR^2$

By perpendicular-axis theorem, MI about a diameter: $I_{\text{diam}} = MR^2/2$.

Uniform Disc about Central Axis

Divide into concentric rings of radius $r$, thickness $dr$, mass $dm = \sigma\cdot 2\pi r\,dr$.

$I = \int_0^R r^2 (2\pi\sigma r\,dr) = 2\pi\sigma\cdot\frac{R^4}{4} = \frac{MR^2}{2}$

since $M = \sigma\pi R^2$.

By perpendicular-axis theorem: about a diameter $I_{\text{diam}} = MR^2/4$.

Solid Sphere about Diameter

(Standard result, derivation involves spherical-shell integration)

$I = \frac{2}{5}MR^2$

Hollow Spherical Shell about Diameter

$I = \frac{2}{3}MR^2$

Solid Cylinder about its Axis

$I = \frac{1}{2}MR^2$ (same as disc — the axial dimension doesn't matter).

Cylinder about a Diameter through CM

$I = \frac{1}{4}MR^2 + \frac{1}{12}ML^2$

Quick Reference Table

Common Mistakes

• Memorising without understanding — the radius of gyration tells you the characteristic distance of mass from axis.
• Using disc formula for a cylinder about a diameter through CM (wrong — has an additional $ML^2/12$ term).

7.12 Kinematics and Dynamics of Rotation

Rotational Kinematics (constant $\alpha$)

By direct analogy with linear motion:

$\omega = \omega_0 + \alpha t$ $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$ $\omega^2 = \omega_0^2 + 2\alpha\theta$

Rotational Dynamics

$\tau = I\alpha$

This is derived from $\vec{\tau} = d\vec{L}/dt$ with $L = I\omega$ (constant $I$).

Work and KE in Rotation

Work done by torque:

$W = \int \tau\,d\theta$

Rotational kinetic energy:

$K_{\text{rot}} = \tfrac{1}{2}I\omega^2$

Power in Rotation

$P = \tau\omega$

Analogy Table

Worked Example

A flywheel of $I = 0.5\,\text{kg m}^2$ is acted on by a constant torque of $2\,\text{N m}$ for $4\,\text{s}$ from rest. Final $\omega$ and KE?

$\alpha = \tau/I = 4\,\text{rad/s}^2, \quad \omega = 16\,\text{rad/s}$

$K = \tfrac{1}{2}I\omega^2 = 0.5\cdot 0.5\cdot 256 = 64\,\text{J}$

Common Mistakes

• Forgetting that $I$ is about a specific axis.
• Equating angular and linear quantities directly (e.g., $\omega$ in degrees vs radians).

7.13 Rolling Motion

Pure Rolling Condition

A body rolls without slipping if the contact point is instantaneously at rest:

$v_{\text{cm}} = R\omega$

Differentiating:

$a_{\text{cm}} = R\alpha$

KE of a Rolling Body

$K = \tfrac{1}{2}M v_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2$

Using $\omega = v_{\text{cm}}/R$ and $I = Mk^2$,

$K = \tfrac{1}{2}Mv^2\left(1 + \frac{k^2}{R^2}\right)$

Acceleration on an Inclined Plane (Derivation)

A body of mass $M$, MI $I = Mk^2$ rolls down an incline of angle $\theta$ without slipping.

Forces along incline: $Mg\sin\theta$ down, friction $f$ up.

Translational: $Mg\sin\theta - f = Ma$.

Rotational (about CM): $fR = I\alpha = Mk^2\alpha$.

Constraint: $a = R\alpha$.

From the rotational equation: $f = Mk^2 a/R^2$.

Substitute:

$Mg\sin\theta - \frac{Mk^2 a}{R^2} = Ma$

$a = \frac{g\sin\theta}{1 + k^2/R^2}$

Special Cases

• Solid sphere ($k^2/R^2 = 2/5$): $a = (5/7)g\sin\theta$.
• Solid cylinder ($k^2/R^2 = 1/2$): $a = (2/3)g\sin\theta$.
• Hollow sphere ($k^2/R^2 = 2/3$): $a = (3/5)g\sin\theta$.
• Ring/hollow cylinder ($k^2/R^2 = 1$): $a = (1/2)g\sin\theta$.

Order of arrival (fastest first): solid sphere > solid cylinder > hollow sphere > ring. The body with smaller $k^2/R^2$ wins, because less KE goes into rotation.

Minimum Friction for Pure Rolling

From $f = Mk^2 a/R^2$ and $a = g\sin\theta/(1+k^2/R^2)$,

$f = \frac{Mg\sin\theta\cdot k^2/R^2}{1 + k^2/R^2}$

For pure rolling we need $f \le \mu Mg\cos\theta$:

$\mu_{\min} = \frac{(k^2/R^2)\tan\theta}{1 + k^2/R^2}$

Worked Example

A solid sphere rolls down a $30^\circ$ incline. Find its acceleration ($g = 10$).

$a = \frac{5\cdot 10\cdot 0.5}{7} = \frac{25}{7} \approx 3.57\,\text{m/s}^2$

Common Mistakes

• Assuming friction is kinetic in pure rolling — it is static and can be less than $\mu_s N$.
• Forgetting that friction does no work in pure rolling.

Solved Problems

Problem 1

Three particles of $1, 2, 3\,\text{kg}$ are at $(0,0), (4,0), (0,3)$. Find the CM.

$x_{\text{cm}} = \frac{1\cdot 0 + 2\cdot 4 + 3\cdot 0}{6} = \frac{8}{6} = \frac{4}{3}$ $y_{\text{cm}} = \frac{1\cdot 0 + 2\cdot 0 + 3\cdot 3}{6} = \frac{9}{6} = \frac{3}{2}$

CM at $(4/3, 3/2)$.

Problem 2

A wheel of $I = 2\,\text{kg m}^2$ rotates at $10\,\text{rad/s}$. A constant retarding torque of $1\,\text{N m}$ brings it to rest. Time?

$\alpha = -0.5\,\text{rad/s}^2$. $t = \omega/\vert \alpha\vert = 20\,\text{s}$.

Problem 3

Two children of $20$ and $30\,\text{kg}$ sit at the ends of a $4\,\text{m}$ uniform plank of $5\,\text{kg}$ supported at the centre. Find the imbalance torque.

CM of plank at centre — contributes no torque.

Torque from $20\,\text{kg}$ at $2\,\text{m}$ left: $20\cdot 10\cdot 2 = 400\,\text{N m}$ ccw. Torque from $30\,\text{kg}$ at $2\,\text{m}$ right: $30\cdot 10\cdot 2 = 600\,\text{N m}$ cw.

Net: $200\,\text{N m}$ clockwise.

Problem 4

A solid disc of $0.5\,\text{kg}$, $R = 0.1\,\text{m}$ rotates at $20\,\text{rad/s}$. Find KE.

$I = MR^2/2 = 0.0025\,\text{kg m}^2$. $K = 0.5\cdot 0.0025\cdot 400 = 0.5\,\text{J}$.

Problem 5

A solid sphere and a ring both of radius $R$ and mass $M$ are released from rest at the top of an incline of length $L$ and angle $\theta$. Find the ratio of their speeds at the bottom.

$v^2 = 2aL$, $a = g\sin\theta/(1 + k^2/R^2)$.

Sphere: $a_s = (5/7)g\sin\theta \Rightarrow v_s^2 = (10/7)gL\sin\theta$.

Ring: $a_r = (1/2)g\sin\theta \Rightarrow v_r^2 = gL\sin\theta$.

$v_s/v_r = \sqrt{10/7}$.

Problem 6

A platform of $I = 5\,\text{kg m}^2$ rotates at $\omega_0 = 4\,\text{rad/s}$. A child of $20\,\text{kg}$ jumps on at radius $1\,\text{m}$. Final $\omega$?

By conservation of $L$:

$5\cdot 4 = (5 + 20\cdot 1^2)\omega \Rightarrow \omega = 20/25 = 0.8\,\text{rad/s}$

Problem 7

Find the MI of a uniform thin disc of $M, R$ about a tangent in its plane.

Through CM about diameter: $I_d = MR^2/4$. By parallel axis: $I = MR^2/4 + MR^2 = 5MR^2/4$.

JEE/NEET Edge Cases

1. Slipping while rolling: if friction is insufficient, $v_{\text{cm}} \ne R\omega$ and you must handle translation and rotation independently.
2. Spool problem with thread pulled at various angles can roll either way depending on the lever arm relative to the contact point — count torques about the contact point.
3. Angular momentum about a moving point has a subtle correction term; usually safer to use a fixed point or the CM.
4. A body acted on by impulsive force off-centre — gives both linear and angular impulse; subsequent motion involves both translation and rotation.
5. Compound pendulum: $T = 2\pi\sqrt{I/(Mgd)}$ where $d$ is distance from pivot to CM.

Quick Recap

• CM: $\vec{r}_{\text{cm}} = \sum m_i\vec{r}_i/M$; CM moves under external forces only.
• $\vec{\tau} = \vec{r}\times\vec{F}$, $\vec{L} = \vec{r}\times\vec{p}$, $d\vec{L}/dt = \vec{\tau}$.
• $L$ is conserved if external torque is zero.
• Equilibrium: $\sum F = 0$ AND $\sum \tau = 0$.
• $I = \sum m_i r_i^2$; perpendicular- and parallel-axis theorems.
• Rotational dynamics mirrors translational ($\tau \leftrightarrow F$, $I \leftrightarrow m$, $\omega \leftrightarrow v$).
• Rolling: $v = R\omega$, $K = \tfrac{1}{2}Mv^2(1 + k^2/R^2)$, $a = g\sin\theta/(1+k^2/R^2)$.

Formula Sheet