Rotational Motion

Rotational motion extends the laws of mechanics to rigid bodies spinning about an axis.
The centre of mass (CM) of a system of particles is defined as $x_{cm}$ = (sum of $m_{i} \; x_{i}$) / (sum of $m_{i}$) [$M^{0} \; L^{1} \; T^{0}$] (m).
For a continuous body, $x_{cm}$ = integral(x dm) / integral(dm). Standard CM positions include: uniform rod (L/2 from either end), uniform disc (geometric centre), triangular lamina (h/3 from base), semicircular ring (2R/$\pi$ from centre), and semicircular disc (4R/$3\pi$ from centre).

Torque is the rotational analogue of force: $\tau$ = r x F, with magnitude $\tau$ = rF sin($\theta$) [$M^{1} \; L^{2} \; T^{-2}$] (N m), where r is the position vector (m), F is the force (N), and $\theta$ is the angle between r and F. Direction is determined by the right-hand rule.
Angular momentum L = I $\omega$ = r x p [$M^{1} \; L^{2} \; T^{-1}$] (kg $m^{2}$/s), where I is the moment of inertia (kg $m^{2}$) and $\omega$ is the angular velocity (rad/s).
Newton's second law for rotation: $\tau$ = dL/dt.
When the net external torque is zero, angular momentum is conserved: I $\omega$ = constant. This is why an ice skater spinning with arms extended (larger I) rotates slowly, and pulling arms in (smaller I) increases $\omega$.

The moment of inertia I = sum($m_{i} \; r_{i}^{2}$) [$M^{1} \; L^{2} \; T^{0}$] (kg $m^{2}$) depends on mass distribution relative to the rotation axis.
Key values: ring about its axis I = $MR^{2}$, disc about its axis I = $\frac{1}{2}$ $MR^{2}$, solid sphere about diameter I = $\frac{2}{5}$ $MR^{2}$, hollow sphere about diameter I = $\frac{2}{3}$ $MR^{2}$, rod about centre I = $\frac{ML^{2}}{12}$, rod about one end I = $\frac{ML^{2}}{3}$.
The radius of gyration K = $\sqrt{I/M}$ [$M^{0} \; L^{1} \; T^{0}$] (m), so I = $MK^{2}$.

The Parallel Axis Theorem states I = $I_{cm}$ + $Md^{2}$, where d is the distance between the parallel axes. This works for ANY body.
The Perpendicular Axis Theorem states $I_{z}$ = $I_{x}$ + $I_{y}$, but applies ONLY to planar (2D/flat) bodies. It cannot be used for spheres, cylinders, or any 3D body.

Rolling without slipping combines translational and rotational motion under the constraint $v_{cm}$ = $\omega$ R (velocity of the contact point is zero).
Total kinetic energy = $\frac{1}{2}$ $mv^{2}$ + $\frac{1}{2}$ I $\omega$² = $\frac{1}{2}$ $mv^{2}$ (1 + $\frac{K^{2}}{R^{2}}$).
For different bodies rolling on an incline, acceleration a = g sin($\theta$) / (1 + $\frac{K^{2}}{R^{2}}$).
The body with the smallest $\frac{K^{2}}{R^{2}}$ has the greatest acceleration and reaches the bottom first: solid sphere ($\frac{K^{2}}{R^{2}}$ = $\frac{2}{5}$) beats disc ($\frac{1}{2}$) beats hollow sphere ($\frac{2}{3}$) beats ring (1), regardless of mass or radius.

Solved Numerical 1: A uniform disc of mass M = 2 kg and radius R = 0.5 m. Find I about an axis tangent to the disc and in its plane. Step 1: I about a diameter (in-plane axis through centre) using perpendicular axis theorem.
$I_{z}$ = $I_{x}$ + $I_{y}$, and by symmetry $I_{x}$ = $I_{y}$ = $I_{d}$ (diameter).
$I_{z}$ = $\frac{MR^{2}}{2}$ (about axis perpendicular to disc through centre).
So $2I_{d}$ = $\frac{MR^{2}}{2}$, giving $I_{d}$ = $\frac{MR^{2}}{4}$. Step 2: Apply parallel axis theorem.
$I_{tangent}$ = $I_{d}$ + $MR^{2}$ = $\frac{MR^{2}}{4}$ + $MR^{2}$ = $\frac{5MR^{2}}{4}$.
$I_{tangent}$ = 5(2)(0.5)^$\frac{2}{4}$ = 5(2)(0.25)/4 = 5 x 0.$\frac{5}{4}$ = 0.625 kg $m^{2}$.
Dimensional check: [kg][$m^{2}$] = [$M^{1} \; L^{2} \; T^{0}$].

Solved Numerical 2: A disc ($\frac{K^{2}}{R^{2}}$ = $\frac{1}{2}$) and a ring ($\frac{K^{2}}{R^{2}}$ = 1), both of mass 1 kg and radius 0.2 m, roll down from height h = 2 m (g = 10 m/$s^{2}$).
Using energy conservation: mgh = $\frac{1}{2}$ $mv^{2}$(1 + $\frac{K^{2}}{R^{2}}$).
For disc: v = $\sqrt{2gh / (1 + 1/2}$) = $\sqrt{2 x 10 x 2 / 1.5}$ = $\sqrt{26.67}$ = 5.16 m/s.
For ring: v = $\sqrt{2gh / (1 + 1}$) = $\sqrt{2 x 10 x 2 / 2}$ = $\sqrt{20}$ = 4.47 m/s. The disc reaches faster with higher velocity.

Solved Numerical 3: Ice skater: $I_{initial}$ = 6 kg $m^{2}$, $\omega_{initial}$ = 2 rev/s = $4\pi$ rad/s.
Arms pulled in: $I_{final}$ = 2 kg $m^{2}$.
Conservation: $I_{i} \; \omega_{i}$ = $I_{f} \; \omega_{f}$.
$\omega_{f}$ = (6 x $4\pi$)/2 = $12\pi$ rad/s = 6 rev/s.
$\text{KE}_{initial}$ = $\frac{1}{2}$ $I_{i} \; \omega_{i}^{2}$ = $\frac{1}{2}$ (6)($4\pi$)² = 3 x $16\pi^{2}$ = $48\pi^{2}$ J.
$\text{KE}_{final}$ = $\frac{1}{2}$ (2)($12\pi$)² = $144\pi^{2}$ J.
Ratio $\frac{\text{KE}_{f}}{\text{KE}_{i}}$ = $144\pi^{2}$ / $48\pi^{2}$ = 3. Kinetic energy triples (energy comes from internal muscular work).

The key testable concept is moment of inertia calculations using the parallel and perpendicular axis theorems and the rolling race on an incline (solid sphere reaches bottom first because it has the smallest $\frac{K^{2}}{R^{2}}$ ratio).