Rotational Motion

A rigid body is an idealized system in which the distances between all constituent particles remain fixed. Its motion can always be decomposed into a translation of the centre of mass (CM) plus a rotation about the CM.

Centre of Mass of a system of particles: $$\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$$

For a continuous body: $$\vec{r}_{cm} = \frac{1}{M}\int \vec{r}\,dm$$

Properties of CM:

• The CM moves as if all the mass were concentrated at that point and all external forces acted on it: $M\vec{a}_{cm} = \vec{F}_{\text{ext}}$.
• Internal forces never affect motion of the CM.
• For symmetric homogeneous bodies, CM lies at the geometric centre.

Locations of CM for Common Bodies:

Torque (Moment of Force): The rotational analogue of force: $$\vec{\tau} = \vec{r} \times \vec{F}$$ where $\vec{r}$ is the position vector of the point of application relative to the reference axis. Magnitude: $\tau = rF\sin\theta = F \cdot d$, where $d = r\sin\theta$ is the moment arm (perpendicular distance from axis to line of action of force). SI unit: N·m. Direction: by right-hand rule.

Angular Momentum: The rotational analogue of linear momentum: $$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$$ For a rigid body rotating with angular velocity $\vec{\omega}$ about a fixed axis: $$L = I\omega$$ where $I$ is the moment of inertia. SI unit of $L$: kg·m²/s.

Newton's Second Law for Rotation: $$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$$

For a body rotating about a fixed axis with constant moment of inertia: $\tau = I\alpha$, where $\alpha = d\omega/dt$ is angular acceleration.

Conservation of Angular Momentum: If $\vec{\tau}_{\text{ext}} = 0$, then $\vec{L}$ is conserved. A classic example: a spinning skater pulls in her arms, decreasing $I$, so $\omega$ increases to keep $L = I\omega$ constant.

Couple: A pair of equal and opposite forces with non-collinear lines of action. Net force is zero; net torque is non-zero, equal to (force) × (perpendicular distance between lines).

Equilibrium of a Rigid Body requires: $$\sum \vec{F}_{\text{ext}} = 0 \quad \text{(translational equilibrium)}$$ $$\sum \vec{\tau}_{\text{ext}} = 0 \quad \text{(rotational equilibrium)}$$ Both must hold; either alone is insufficient.

Moment of Inertia (MI) is the rotational analogue of mass. For a system of particles about a given axis: $$I = \sum m_i r_i^2$$ For a continuous rigid body: $$I = \int r^2 \,dm$$ where $r$ is the perpendicular distance from the axis. SI unit: kg·m². Dimensions: $[ML^2]$.

Radius of Gyration ($k$): $$I = M k^2 \implies k = \sqrt{I/M}$$ The radius at which the entire mass, if concentrated, would have the same MI.

Moments of Inertia of Common Bodies:

Parallel Axis Theorem: If $I_{cm}$ is the MI about an axis through the CM and $I$ about a parallel axis at distance $d$ away: $$I = I_{cm} + M d^2$$

This applies to any axis parallel to one through the CM. Useful for finding $I$ about non-central axes.

Perpendicular Axis Theorem (for plane laminar bodies only): If $I_x$ and $I_y$ are MIs about two perpendicular axes lying in the plane of the lamina, and $I_z$ is the MI about an axis perpendicular to the lamina through the same point: $$I_z = I_x + I_y$$

Example application: For a disc, $I_z = MR^2/2$ (perpendicular to plane). By symmetry $I_x = I_y$, and so $I_x = I_y = MR^2/4$ (diameter).

Kinetic Energy of Rotation: $$K_{\text{rot}} = \frac{1}{2}I\omega^2$$ Analogous to $\frac{1}{2}mv^2$. Total KE of a rigid body in general motion: $$K_{\text{total}} = \frac{1}{2}M v_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$

This is the König theorem for rigid bodies.

Rotational Kinematics: For a rigid body rotating about a fixed axis with constant angular acceleration $\alpha$, the equations parallel linear motion:

Relations between linear and angular quantities for a point at distance $r$ from axis: $v = \omega r$, $a_t = \alpha r$ (tangential), $a_c = \omega^2 r$ (centripetal).

Dynamics: $$\tau = I\alpha, \quad K_{\text{rot}} = \frac{1}{2}I\omega^2, \quad L = I\omega$$ Power delivered by torque: $P = \tau\omega$. Work by torque: $W = \int \tau\,d\theta$.

Work-Energy Theorem for Rotation: $$W_{\text{net}} = \Delta K_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_0^2$$

Conservation of Angular Momentum: Holds when net external torque is zero. Important applications:

• Ice skater pulling in arms: $I_1 \omega_1 = I_2 \omega_2$; smaller $I$ means larger $\omega$.
• Diving and gymnastics: athletes tuck to spin faster.
• Neutron stars: collapse from large stars conserves $L$, leading to extremely high $\omega$ (millisecond pulsars).
• Spinning top precession: gravitational torque on a tilted top causes the spin axis to trace a cone (gyroscopic precession).

Comparison of Translation vs Rotation:

Pure Rolling (rolling without slipping): The contact point of the rolling body is instantaneously at rest relative to the surface. The velocity of CM and angular velocity are related by: $$v_{cm} = \omega R$$ Differentiating: $a_{cm} = \alpha R$ (assuming $R$ constant).

Kinetic Energy of a Rolling Body: $$K = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}M v_{cm}^2 + \frac{1}{2}I_{cm}\omega^2 = \frac{1}{2}M v^2\left(1 + \frac{I_{cm}}{MR^2}\right)$$

Define the MI parameter $\beta = I_{cm}/(MR^2)$, which depends on the body's shape:

• Solid sphere: $\beta = 2/5$
• Hollow sphere: $\beta = 2/3$
• Solid cylinder/disc: $\beta = 1/2$
• Hollow cylinder/ring: $\beta = 1$

Then $K = \frac{1}{2}Mv^2(1 + \beta)$.

Velocity Distribution in a Rolling Body: Combining translation $v_{cm}\hat{i}$ and rotation $\omega = v_{cm}/R$:

• Top point: $v_{\text{top}} = v_{cm} + \omega R = 2v_{cm}$
• Bottom (contact): $v_{\text{bottom}} = v_{cm} - \omega R = 0$
• Centre: $v_{cm}$

Rolling on an Inclined Plane (smooth rolling, no slipping):

For a body of radius $R$, mass $M$, MI parameter $\beta$, on incline angle $\theta$: $$a = \frac{g\sin\theta}{1 + \beta}$$

By energy conservation (descending height $h$): $$\frac{1}{2}Mv^2(1+\beta) = Mgh \implies v = \sqrt{\frac{2gh}{1+\beta}}$$

Hierarchy of Rolling Speeds on Incline: Same height, smaller $\beta$ means greater final velocity:

• Solid sphere fastest (β = 2/5)
• Solid disc
• Hollow sphere
• Ring slowest (β = 1)

Friction in Pure Rolling: For rolling without slipping on an incline, static friction acts at the contact point (no kinetic friction since contact point is at rest). The minimum coefficient of friction needed for pure rolling down an incline: $$\mu_{\min} = \frac{\beta \tan\theta}{1 + \beta}$$

If $\mu < \mu_{\min}$, the body slips while rolling (kinetic friction acts) and energy is dissipated.

Slipping vs Rolling:

• Slipping: $v_{cm} > \omega R$ or $v_{cm} < \omega R$. Slip velocity at contact point is non-zero. Kinetic friction acts opposing slip.
• Pure rolling: $v_{cm} = \omega R$. Static friction (may be zero or non-zero).