2.1 Centre of Mass

The centre of mass (CM) is the mass-weighted geometric mean position of a system. For a uniform, symmetric body, the CM coincides with the geometric centre. Non-symmetric results to memorise:

• Semicircular ring: $y_{cm} = 2R/\pi \approx 0.637R$ from diameter
• Semicircular disc: $y_{cm} = 4R/3\pi \approx 0.424R$ from diameter
• Uniform hemisphere: $y_{cm} = 3R/8 = 0.375R$ from flat base
• Triangular lamina: $y_{cm} = h/3$ from base

The CM of a composite body is found by treating each sub-body as a point mass at its own CM.

2.2 Moment of Inertia and Theorems

$I = \sum m_i r_i^2$ depends on the choice of rotation axis, not just the shape. Standard bodies and axes must be memorised (see Table A in the session). The parallel axis theorem $I = I_{cm} + Md^2$ is valid for all bodies and is the most-used theorem. The perpendicular axis theorem $I_z = I_x + I_y$ is restricted to flat bodies. Typical NEET application: find I of a disc about a tangent in its plane → use perpendicular axis theorem to get $I_{diameter} = MR^2/4$, then parallel axis theorem to add $MR^2$ → result $5MR^2/4$.

2.3 Torque and Rotational Newton's Law

$\tau = r \times F$, magnitude $rF\sin\theta$. Unit N m = kg $m^{2}$ $s^{-2}$. The net torque on a body equals the rate of change of its angular momentum ($\tau = dL/dt$) or equivalently $\tau = I\alpha$ for a rigid body with fixed axis. Power delivered by torque: $P = \tau\omega$.

2.4 Angular Momentum and Conservation

$L = I\omega$ for a rigid body; $L = mvr\sin\theta$ for a point particle. When $\tau_{net} = 0$, $\Delta L = 0$, so $I\omega = \text{constant}$. Applications: figure skaters, divers tucking, rotating platforms. Note that kinetic energy changes when I changes, even with L conserved: $KE = L^2/(2I)$, so KE increases when I decreases.

2.5 Rolling Motion

Rolling without slipping: $v_{cm} = \omega R$; acceleration $a_{cm} = \alpha R$. Kinetic energy $= \frac{1}{2}mv^2(1 + K^2/R^2)$. For pure rolling on a horizontal surface with no external force, no friction is needed. On an incline, static friction provides the torque needed for angular acceleration. The incline acceleration formula $a = g\sin\theta/(1 + K^2/R^2)$ immediately gives the winner of any rolling race: minimise $K^2/R^2$.

2.6 Linear–Rotational Analogues

Every linear formula has a rotational twin (see Table B in the session). This symmetry is very reliable for dimensional analysis and MCQ elimination.