Rigid Body Dynamics

The moment of inertia $I$ is the rotational analogue of mass: just as mass $m$ measures resistance to linear acceleration ($F = ma$), the moment of inertia $I$ measures resistance to angular acceleration ($\tau = I\alpha$). It depends both on the total mass and on how that mass is distributed relative to the rotation axis.

Formal definition:
$\displaystyle I = \sum_i m_i r_i^2 = \int r^2\,dm$

For $N$ discrete particles, we sum the product of each mass times the square of its distance from the axis. For a continuous body (homogeneous solid), $dm = \rho\,dV$ and $I = \int_V \rho r^2\,dV$ integrated over the entire volume.

Physical intuition for $r^2$: why does the distance appear squared? Three converging reasons:

• Kinematics: at the same angular velocity $\omega$, a mass at distance $r$ has linear speed $v = \omega r$ and kinetic energy $\frac{1}{2}m v^2 = \frac{1}{2}m \omega^2 r^2$. Summing: $K_{rot} = \frac{1}{2}\omega^2 \sum m_i r_i^2 = \frac{1}{2}I\omega^2$, hence $I = \sum m_i r_i^2$.
• Angular momentum: a particle's angular momentum is $\vec{L} = \vec{r} \times \vec{p}$; its magnitude for circular motion is $mvr = m(\omega r)r = m\omega r^2$, giving $L = I\omega$.
• Lever-arm geometry: both the speed ($\propto r$) and the momentum arm ($\propto r$) contribute, making the effect $\propto r^2$.

What it depends on vs. does not:

• $I$ depends on: total mass, shape, mass distribution, choice of rotation axis
• $I$ does NOT depend on: angular velocity, angular acceleration, time, presence of other forces

Physical consequence: at equal mass and radius, a body with mass distributed farther from the axis has larger $I$ and is harder to rotationally accelerate. A ring ($I=mR^2$) is harder to spin than a disk ($I=\frac{1}{2}mR^2$); a hollow sphere ($I=\frac{2}{3}mR^2$) harder than a solid one ($I=\frac{2}{5}mR^2$). Masses far from the axis travel longer arcs for the same rotation angle.

Common moments of inertia (homogeneous bodies):

• Thin rod, ⊥ axis through center: $I = \tfrac{1}{12}mL^2$
• Thin rod, ⊥ axis through end: $I = \tfrac{1}{3}mL^2$
• Solid disk / solid cylinder, central axis: $I = \tfrac{1}{2}mR^2$
• Ring / thin-walled cylinder, central axis: $I = mR^2$
• Solid sphere, diameter axis: $I = \tfrac{2}{5}mR^2$
• Thin hollow sphere, diameter axis: $I = \tfrac{2}{3}mR^2$
• Rectangular plate, ⊥ axis through center: $I = \tfrac{1}{12}m(a^2 + b^2)$

Real-world examples:

• Flywheel: mass concentrated at the periphery to maximize $I$ and store rotational energy, smoothing out combustion engine fluctuations.
• Baseball bat: mass concentrated in the head to increase $I$ and transfer more energy to the ball on impact.
• Tightrope walker: a long, heavy pole increases the system's $I$, slowing angular oscillations and giving more time to correct balance.
• Bicycle: the gyroscopic effect of spinning wheels (non-negligible $I$) contributes to lateral stability.
• Rotating bicycle wheel: once spinning, it resists attempts to change its axis direction (precession).

The parallel axis theorem (Steiner's theorem) allows computing the moment of inertia $I$ about any axis, given $I_{CM}$ about a parallel axis through the center of mass:

$I = I_{CM} + Md^2$

where $d$ is the distance between the two parallel axes and $M$ is the total mass.

Intuitive derivation: shifting the rotation axis by a distance $d$, the mean squared distance of masses from the axis increases by $d^2$ per particle. For $N$ particles:
$I = \sum m_i (\vec{r}_i - \vec{d})\cdot(\vec{r}_i - \vec{d}) = \sum m_i r_i^2 - 2\vec{d}\cdot\sum m_i\vec{r}_i + d^2\sum m_i$
The middle term vanishes because $\sum m_i\vec{r}_i = M\vec{R}_{CM} = 0$ in the CM frame. Thus $I = I_{CM} + Md^2$.

When can it be used? The two axes must be parallel — the theorem does not apply to non-parallel axes. Additionally, $I_{CM}$ must be computed about an axis through the CM.

Typical applications:

• Physical pendulum: a rod swinging about one end has $I = \frac{1}{3}mL^2$ about the pivot, not $\frac{1}{12}mL^2$ about the CM. Steiner gives the correct $I$, hence the period $T = 2\pi\sqrt{I/(mgd_{CM})}$.
• Compound objects: to compute $I$ for a body+axis system, compute $I_{CM}$ of the body and add $Md^2$ if the axis is offset.

Example — thin rod:

• About CM (⊥ axis): $I_{CM} = \tfrac{1}{12}mL^2$
• About one end (same direction): $I_{end} = I_{CM} + m\cdot(L/2)^2 = \tfrac{1}{12}mL^2 + \tfrac{1}{4}mL^2 = \tfrac{1}{3}mL^2$

Example — disk with offset pivot: a disk of mass $M$ and radius $R$ rotates about an axis parallel to the central axis through a point on its rim. Then $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$. The moment triples compared to the central axis!

Real-world example — hammer: a hammer rotated about its handle has much larger $I$ than when rotated about the CM, because the head is far from the axis. This makes it harder to accelerate the hammer rotationally, but once moving it delivers more powerful strikes.

The rotational analogue of Newton's second law $\vec{F} = m\vec{a}$ is:

$\displaystyle \vec{\tau} = I\vec{\alpha}$

where $\vec{\tau}$ is the torque, $I$ the moment of inertia, and $\vec{\alpha}$ the angular acceleration.

Complete analogies between translation and rotation:

• Force $\vec{F}$ → Torque $\vec{\tau} = \vec{r} \times \vec{F}$
• Mass $m$ → Moment of inertia $I = \int r^2\,dm$
• Acceleration $\vec{a}$ → Angular acceleration $\vec{\alpha}$
• Momentum $\vec{p} = m\vec{v}$ → Angular momentum $\vec{L} = I\vec{\omega}$
• Kinetic energy $K = \frac{1}{2}mv^2$ → $K_{rot} = \frac{1}{2}I\omega^2$
• Work $W = \int \vec{F}\cdot d\vec{s}$ → $W = \int \vec{\tau}\cdot d\vec{\theta}$
• Power $P = \vec{F}\cdot\vec{v}$ → $P = \vec{\tau}\cdot\vec{\omega}$

Torque in detail:
$\tau = rF\sin\theta = r_\perp F$

$r_\perp = r\sin\theta$ is the moment arm: the perpendicular distance from the rotation axis to the line of action of the force. Torque is maximum when the force is perpendicular ($\theta = 90°$), zero when parallel ($\theta = 0°$).

Torque direction (right-hand rule): point your fingers along $\vec{r}$, curl them toward $\vec{F}$, and your thumb points in the direction of $\vec{\tau}$. For planar rotation, torque is out of the page (positive, counterclockwise) or into the page (negative, clockwise).

Everyday examples of torque:

• Door: easier to open pushing far from the hinges (large moment arm) than near them (small arm). The handle is placed as far as possible from the hinges precisely to maximize the lever arm.
• Wrench: a long handle increases the moment arm, producing the same torque with less force. A breaker bar is used for stubborn bolts.
• Bicycle pedals: force is applied perpendicularly to the pedal when horizontal, maximizing $\tau$ and thus the angular acceleration of the chainring.
• Balance scale: equal forces at different distances from the fulcrum produce different torques — the principle of Archimedes' lever.

Work, energy, and power in rotation:
The work done by a constant torque over an angular displacement $\Delta\theta$ is $W = \tau\Delta\theta$, analogous to $W = F\Delta s$. Power is $P = \tau\omega$.

Work-energy theorem for rotation: $W = \int \tau\,d\theta = \int I\alpha\,d\theta = \int I\frac{d\omega}{dt}\,d\theta = \int I\frac{d\theta}{dt}\,d\omega = \int I\omega\,d\omega = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 = \Delta K_{rot}$.

The angular impulse-momentum theorem states that $\int \tau\,dt = \Delta L = I\Delta\omega$, analogous to linear impulse $\int F\,dt = \Delta p$.

A rigid body that rolls without slipping on a surface has a point of contact that is instantaneously at rest relative to the ground. A precise kinematic constraint links translation and rotation:

$v_{CM} = \omega R$,   $a_{CM} = \alpha R$

Where $v_{CM}=\omega R$ comes from (in plain words). Look at the picture. The wheel does two motions at once: it translates (the whole body moves forward at $v_{CM}$) and it rotates about its centre with angular velocity $\omega$ (a rim point moves at $\omega R$ relative to the centre). At the contact point $P$ with the ground the two motions are opposite: translation pushes $P$ forward ($+v_{CM}$), rotation drags it backward ($-\omega R$). "Rolling without slipping" means precisely that $P$ does not slide, i.e. it is at rest: $v_{CM} - \omega R = 0 \Rightarrow v_{CM} = \omega R$. Differentiating in time gives the link between the accelerations, $a_{CM} = \alpha R$.

You can see it in the picture: the centre moves at $v_{CM}$, the top point at twice that ($2v_{CM}$: translation and rotation in the same direction), the bottom point is at rest. The speed grows linearly from bottom (0) to top ($2v_{CM}$).
Why is this constraint important? When a body rolls without slipping, the contact point has zero velocity relative to the ground: the CM's translational velocity is exactly canceled by the rotational velocity of the point at the bottom. This ensures that static friction (non-dissipative) maintains the constraint without sliding.

Important — static friction is essential: without static friction, an object would slide without rotating (like a book on ice). In pure rolling, static friction does no work (the point of application is instantaneously at rest), but it is indispensable for transferring angular momentum and maintaining $\omega = v_{CM}/R$.

Total kinetic energy: the body possesses both translational (CM motion) and rotational (about CM) kinetic energy:
$K_{tot} = K_{tr} + K_{rot} = \tfrac{1}{2}mv_{CM}^2 + \tfrac{1}{2}I_{CM}\omega^2$

Substituting $\omega = v_{CM}/R$:
$K_{tot} = \tfrac{1}{2}mv_{CM}^2\left(1 + \dfrac{I_{CM}}{mR^2}\right)$

Shape comparison — rolling down an incline:

• Solid sphere ($I = \frac{2}{5}mR^2$): $K_{tot} = \frac{7}{10}mv^2$. 71.4% translational, 28.6% rotational. Accelerates fastest.
• Solid cylinder ($I = \frac{1}{2}mR^2$): $K_{tot} = \frac{3}{4}mv^2$. 66.7% translational, 33.3% rotational.
• Ring ($I = mR^2$): $K_{tot} = mv^2$. Only 50% translational — the slowest shape.

Why does the sphere always win? For the same lost potential energy ($mgh$), the energy splits between translation and rotation. The smaller $I/(mR^2)$, the smaller the fraction "stolen" by rotation, the higher the linear speed at the bottom. The sphere has the smallest $I/(mR^2) = 2/5$ among common shapes.

Force-based solution (alternative method). On a body rolling down an incline of angle $\theta$ act three forces (see figure): the weight $mg$ (vertical), the normal reaction $N$ (perpendicular to the plane) and static friction $F_{fr}$ (along the plane, pointing up: it is the force that prevents sliding and spins the body). Two equations are written — one for translation along the plane, one for rotation about the centre:

• translation: $\sum F_{\parallel} = mg\sin\theta - F_{fr} = ma_{CM}$ (the weight component along the plane pulls down, friction holds back);
• rotation: $\sum \tau_{CM} = F_{fr}R = I_{CM}\alpha = I_{CM}a_{CM}/R$ (only friction gives torque about the centre $O$: $N$ and $mg$ pass through $O$, so zero lever arm).

Combining the two and using the constraint $a_{CM}=\alpha R$ gives: $a_{CM} = \dfrac{g\sin\theta}{1 + I_{CM}/(mR^2)}$, identical to the energy result.

When slipping occurs: if static friction is insufficient (too smooth a surface), the body partially slides, friction becomes kinetic and dissipates energy. The motion is no longer pure rolling, and the final speed is lower.

Real-world examples:

• Vehicle wheels: static friction between tire and asphalt is essential for pure rolling. Hard braking that locks the wheels causes sliding and loss of control.
• Billiards: a ball struck with a "cue below center" may initially slide and then transition to pure rolling once static friction has sufficiently altered the rotation.
• Trains: steel wheels on steel rails have a low coefficient of static friction, but it is sufficient for pure rolling under normal conditions.

The angular momentum (moment of momentum) is the rotational analogue of linear momentum. For a rigid body rotating about a fixed axis:

$\vec{L} = I\vec{\omega}$

More generally, for a single particle: $\vec{L} = \vec{r} \times \vec{p} = m\vec{r} \times \vec{v}$, with magnitude $L = mvr\sin\phi$, where $\phi$ is the angle between $\vec{r}$ and $\vec{v}$.

Newton's Second Law for rotation (general form):
$\displaystyle \vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$

The net external torque on a system equals the time derivative of its angular momentum — analogous to $\vec{F}_{ext} = d\vec{p}/dt$.

Conservation of angular momentum:
If the net external torque is zero ($\sum\vec{\tau}_{ext} = 0$), the total angular momentum is conserved:

$\vec{L} = \text{constant}$   or   $I_1\omega_1 = I_2\omega_2$

Key examples:

• Figure skater: starts spinning with arms extended (large $I$, small $\omega$). Pulling arms in reduces $I$ → $\omega$ increases ($L$ constant). Re-extending arms slows the spin. This is the most dramatic visualization of $L$ conservation.
• Diver: in the air (negligible external torque), $L$ is constant. Tucking (reducing $I$) increases $\omega$, allowing more somersaults. Opening before impact increases $I$ and slows rotation.
• Gyroscope and precession: a spinning gyroscope with angular momentum $\vec{L}$ subject to torque $\vec{\tau}$ perpendicular to $\vec{L}$ does not fall, but precesses with angular velocity $\bm{\Omega} = \tau / L$. The direction of $\bm{\Omega}$ follows $\vec{\tau} = \bm{\Omega} \times \vec{L}$.
• Collapsing star: when a massive star collapses into a neutron star, the radius shrinks dramatically → $I$ decreases → $\omega$ increases enormously. Neutron stars can rotate at hundreds of Hz (pulsars).
• Cooling fan: if the motor fails and blades break off, angular momentum is conserved and the rotor speeds up to compensate for the lost mass at the ends.

Gyroscope precession — why it doesn't fall.
We'd expect an inclined gyroscope (or a spinning top) to fall under its weight. Instead its axis slowly rotates about the vertical: this is precession. The reason is entirely in $\vec{\tau} = d\vec{L}/dt$.
The weight $Mg$, applied at distance $d$ from the pivot, produces a torque $\tau = Mgd\sin\theta$ perpendicular to $\vec{L}$ (which points along the spin axis). A torque perpendicular to a vector changes only its direction, not its magnitude: so $\vec{L}$ neither grows nor shrinks, it rotates, and its tip traces a circle — precession. At each instant $d\vec{L} = \vec{\tau}\,dt$ pushes $\vec{L}$ sideways (never downward), so the axis turns instead of falling. The precession rate is $\Omega = \tau/L = Mgd/(I\omega)$. Importantly, $\Omega$ is inversely proportional to $\omega$ — the faster the gyroscope spins, the slower (and more stable) the precession. That is why a slow top wobbles visibly while a fast one stays almost upright.

Conservation in three dimensions: $\vec{L}$ is a vector; its conservation implies both magnitude and direction remain constant. This explains why a spinning top stays upright: its angular momentum aligns with the spin axis, and conservation of $\vec{L}$ maintains the axis orientation.

A body is in static equilibrium if it is at rest and remains so: no linear or angular acceleration. The necessary and sufficient conditions are:

1. $\displaystyle \sum\vec{F} = 0$   (translational equilibrium: net force zero)
2. $\displaystyle \sum\vec{\tau} = 0$   (rotational equilibrium: net torque zero)

Step-by-step problem-solving strategy:

1. Free-body diagram (FBD): identify all forces acting on the body (weight, support reactions, tensions, friction).
2. Choose a coordinate system: define $x$ and $y$ axes for force components.
3. Choose a pivot for torques: the ideal pivot is a point where many unknown forces act — their torques will be zero and they drop out of $\sum\tau = 0$. Typically a hinge, pin, or contact point.
4. Write the equations: $\sum F_x = 0$, $\sum F_y = 0$, $\sum \tau_O = 0$ (with consistent sign convention).
5. Solve the system: if unknowns do not exceed equations, the problem is isostatic.

Important: the condition $\sum\tau = 0$ holds about any pivot. Choosing the pivot strategically simplifies calculations, but any choice leads to the same result. You can verify the solution by recomputing torques about a different point.

Center of Mass and Stability:

• Stable equilibrium: CM below the support point. A small perturbation raises the CM → potential energy increases → the system returns. Example: pendulum at the bottom, a chair on four legs.
• Unstable equilibrium: CM above the support point. A small perturbation lowers the CM → potential energy decreases → the system moves away. Example: pencil balanced on its tip, inverted pendulum.
• Neutral equilibrium: CM remains at the same height for any displacement. Example: sphere on a horizontal plane, a cylinder on an inclined plane at the critical angle.

Energy analysis of stability: equilibrium is stable if $U$ (potential energy) has a local minimum, unstable if $U$ has a local maximum, neutral if $U$ is constant. For small angular displacements $\theta$, $U \approx \frac{1}{2}k\theta^2$ produces simple harmonic motion (oscillations) about the stable position.

Hyperstatic (indeterminate) problems: if unknowns exceed the 3 available equations (in 2D: $\sum F_x = 0$, $\sum F_y = 0$, $\sum\tau = 0$), the problem is statically indeterminate. Additional equations from material deformation are needed (beyond this chapter's scope). Example: a beam on three supports.

Real-world examples:

• Leaning ladder: the classic equilibrium problem with ground friction and a smooth wall. If friction is insufficient, the ladder slips.
• Crane: a counterweight is positioned so that the load's torque is balanced, maintaining $\sum\tau = 0$ about the wheels.
• Wheelbarrow: the wheel acts as a fulcrum; upward wheel force and downward load force balance with the operator's applied force.
• Buildings and bridges: support reactions must balance weights and external loads to ensure static equilibrium of the entire structure.