Rotational Kinematics & Dynamics

A rigid body is an idealized system of particles in which all mutual pairwise distances $|\mathbf{r}_i - \mathbf{r}_j|$ remain strictly constant in time.

1. The Angular Velocity Vector $\boldsymbol{\omega}$

For a rigid body rotating about an instantaneous axis, its angular velocity vector $\boldsymbol{\omega}$ points along the axis of rotation by the right-hand rule. The linear velocity $\mathbf{v}$ of any point at position $\mathbf{r}$ relative to an origin on the rotation axis is: $$\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$$ Differentiating with respect to time gives total linear acceleration: $$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d\boldsymbol{\omega}}{dt} \times \mathbf{r} + \boldsymbol{\omega} \times \frac{d\mathbf{r}}{dt} = \boldsymbol{\alpha} \times \mathbf{r} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r})$$

• $\mathbf{a}_t = \boldsymbol{\alpha} \times \mathbf{r}$: Tangential acceleration (from angular acceleration $\boldsymbol{\alpha} = \dot{\boldsymbol{\omega}}$).
• $\mathbf{a}_c = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) = -\omega^2 \mathbf{r}_\perp$: Centripetal acceleration pointing perpendicular to the rotation axis.

2. Kinematic Equations for Constant Angular Acceleration $\alpha$

$$\omega(t) = \omega_0 + \alpha t, \quad \theta(t) = \omega_0 t + \frac{1}{2} \alpha t^2, \quad \omega^2 = \omega_0^2 + 2 \alpha \Delta\theta$$

Rotational dynamics represents the rotational counterpart to translational Newtonian mechanics.

1. Torque and Angular Momentum

For a particle acted upon by force $\mathbf{F}$ at position $\mathbf{r}$: $$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \quad (\text{Torque}), \quad \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times (m\mathbf{v}) \quad (\text{Angular Momentum})$$ Differentiating $\mathbf{L}$ with respect to time: $$\frac{d\mathbf{L}}{dt} = \frac{d\mathbf{r}}{dt} \times \mathbf{p} + \mathbf{r} \times \frac{d\mathbf{p}}{dt} = (\mathbf{v} \times m\mathbf{v}) + \mathbf{r} \times \mathbf{F} = \mathbf{0} + \boldsymbol{\tau} = \boldsymbol{\tau}_{\text{net}}$$ Conservation of Angular Momentum: If net external torque vanishes ($\boldsymbol{\tau}_{\text{net}} = \mathbf{0}$): $$\mathbf{L} = \text{constant}$$

2. Central Forces and Kepler's Second Law

A **central force** acts along the line joining the particle to the origin: $\mathbf{F}(\mathbf{r}) = f(r)\hat{\mathbf{r}}$. $$\boldsymbol{\tau} = \mathbf{r} \times (f(r)\hat{\mathbf{r}}) = \mathbf{0} \implies \mathbf{L} = \text{constant}$$ Because $\mathbf{r} \cdot \mathbf{L} = \mathbf{r} \cdot (\mathbf{r} \times \mathbf{p}) = 0$, the particle's trajectory is strictly confined to a fixed plane perpendicular to $\mathbf{L}$. The swept-out area in time $dt$ is $dA = \frac{1}{2}|\mathbf{r} \times d\mathbf{r}| = \frac{1}{2}|\mathbf{r} \times \mathbf{v}| dt$: $$\frac{dA}{dt} = \frac{|\mathbf{L}|}{2m} = \text{constant} \quad (\text{Kepler's Second Law: Equal areas in equal times!})$$

For rotation about a fixed axis (say, the $z$-axis), the kinetic energy is: $$K_{\text{rot}} = \frac{1}{2} \sum_{i=1}^N m_i v_i^2 = \frac{1}{2} \sum_{i=1}^N m_i (r_{\perp, i} \omega)^2 = \frac{1}{2} \left( \sum_{i=1}^N m_i r_{\perp, i}^2 \right) \omega^2 = \frac{1}{2} I \omega^2$$ where the **Moment of Inertia** $I$ is: $$I = \sum_{i=1}^N m_i r_{\perp, i}^2 = \iiint_V r_\perp^2 \rho(\mathbf{r}) dV$$ Fixed-axis equation of motion: $\tau_z = I \alpha_z$.

Systematic Derivations for Standard Geometries

• Thin Uniform Rod of Mass $M$, Length $L$ (About Center): $$I_{\text{cm}} = \int_{-L/2}^{L/2} x^2 \left(\frac{M}{L}\right) dx = \frac{M}{L} \left[ \frac{x^3}{3} \right]_{-L/2}^{L/2} = \frac{1}{12} M L^2$$
• Thin Uniform Rod (About One End): $$I_{\text{end}} = \int_0^L x^2 \left(\frac{M}{L}\right) dx = \frac{1}{3} M L^2$$
• Uniform Solid Cylinder / Disc of Mass $M$, Radius $R$ (About Cylindrical Axis): $$I = \int_0^R r^2 \left(\frac{2M}{R^2} r dr\right) = \frac{2M}{R^2} \left[ \frac{r^4}{4} \right]_0^R = \frac{1}{2} M R^2$$
• Uniform Solid Sphere of Mass $M$, Radius $R$ (About Any Diameter): $$I = \frac{2}{5} M R^2$$
• Thin Spherical Shell of Mass $M$, Radius $R$: $$I = \frac{2}{3} M R^2$$

Two fundamental theorems permit calculating moments of inertia about arbitrary axes without re-evaluating triple volume integrals.

1. Parallel Axis Theorem (Steiner’s Theorem)

The moment of inertia $I$ about any axis parallel to an axis passing through the center of mass at perpendicular distance $d$ is: $$I = I_{\text{cm}} + M d^2$$ Proof: Let the CM be the origin $\mathbf{R}_{\text{cm}} = \mathbf{0}$. The distance of mass element $dm$ to the parallel axis is $\mathbf{r}' = \mathbf{r} - \mathbf{d}$: $$I = \int (r')^2 dm = \int (\mathbf{r} - \mathbf{d}) \cdot (\mathbf{r} - \mathbf{d}) dm = \int r^2 dm - 2\mathbf{d} \cdot \int \mathbf{r} dm + d^2 \int dm$$ Since $\int \mathbf{r} dm = M \mathbf{R}_{\text{cm}} = \mathbf{0}$: $$I = I_{\text{cm}} + M d^2 \quad \blacksquare$$

2. Perpendicular Axis Theorem (Planar Laminae)

For a thin flat planar sheet lying entirely in the $xy$-plane ($z = 0$): $$I_z = I_x + I_y$$ Proof: $$I_x = \int y^2 dm, \quad I_y = \int x^2 dm$$ $$I_z = \int (x^2 + y^2) dm = \int x^2 dm + \int y^2 dm = I_x + I_y \quad \blacksquare$$

Planar rigid body motion combines translational motion of the center of mass with rotation about the center of mass.

1. Decomposition of Kinetic Energy (Chasles’ Theorem)

The total kinetic energy of a rolling body is: $$K_{\text{total}} = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{2} I_{\text{cm}} \omega^2$$

2. Pure Rolling Without Slipping

For a circular body of radius $R$ rolling without slipping along a surface: $$v_{\text{cm}} = R \omega, \quad a_{\text{cm}} = R \alpha$$ The contact point is instantaneously at rest ($v_{\text{contact}} = 0$). Static friction does zero mechanical work!

3. The Great Incline Race

For a body with $I_{\text{cm}} = c M R^2$ rolling down an incline of angle $\theta$: Applying energy conservation: $$Mgh = \frac{1}{2} M v_{\text{cm}}^2 + \frac{1}{2} (cMR^2) \left(\frac{v_{\text{cm}}}{R}\right)^2 = \frac{1}{2} M(1 + c) v_{\text{cm}}^2$$ Differentiating with respect to distance down slope gives linear acceleration: $$a_{\text{cm}} = \frac{g \sin \theta}{1 + c} = \frac{g \sin \theta}{1 + \frac{I_{\text{cm}}}{MR^2}}$$ Ranking:

1. Solid Sphere ($c = 2/5 = 0.40$): $a = 0.714 g \sin\theta$ (Fastest!)
2. Solid Cylinder / Disc ($c = 1/2 = 0.50$): $a = 0.667 g \sin\theta$
3. Spherical Shell ($c = 2/3 = 0.67$): $a = 0.600 g \sin\theta$
4. Hollow Hoop ($c = 1.00$): $a = 0.500 g \sin\theta$ (Slowest!)

4. Condition for Rolling Without Slipping

Static friction force required is: $$f_s = \frac{Mg \sin \theta}{1 + \frac{MR^2}{I_{\text{cm}}}} \le \mu_s Mg \cos \theta \implies \mu_s \ge \frac{\tan \theta}{1 + \frac{MR^2}{I_{\text{cm}}}}$$