Rotational Motion: Quick Revision

1. Centre of Mass

\mathbf{R} = \dfrac{1}{M}\sum_i m_i\mathbf{r}_i \qquad \mathbf{F}_{ext} = M\mathbf{a}_{cm}

The CM moves exactly as if the whole mass were concentrated there and the whole external force acted there — internal forces always cancel in pairs (Newton’s third law).

2. Torque and Angular Momentum

\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \qquad \mathbf{L} = \mathbf{r}\times\mathbf{p} \qquad \boldsymbol{\tau}_{ext} = \dfrac{d\mathbf{L}}{dt}

Both $\boldsymbol{\tau}$ and $\mathbf{L}$ are defined only relative to a chosen origin. $\tau = rF\sin\theta$ , direction by the right-hand rule.

3. Moment of Inertia — Standard Results

I = \sum_i m_ir_i^2 = MK^2

Ring about central (symmetry) axis: $I = MR^2$
Disc about central axis: $I = \tfrac{1}{2}MR^2$
Solid sphere about a diameter: $I = \tfrac{2}{5}MR^2$
Thin rod about centre, ⊥ to length: $I = \tfrac{1}{12}ML^2$

$r_i$ is the perpendicular distance from the axis; $K$ (radius of gyration) is where all the mass would sit, as a point, to give the same $I$ .

4. Perpendicular and Parallel Axes Theorems

I_z = I_x + I_y \qquad I = I_{cm} + Md^2

Perpendicular axes: planar lamina only; $x,y$ in the plane, $z$ through their intersection, normal to the plane.
Parallel axes: any rigid body; $I_{cm}$ is the minimum over all parallel axes — moving away only adds $Md^2 \ge 0$ .

5. Conservation of Angular Momentum

\mathbf{L} = \text{constant} \quad\text{when}\quad \boldsymbol{\tau}_{ext}=0 \qquad I_1\omega_1 = I_2\omega_2

KE = \tfrac{1}{2}I\omega^2 = \dfrac{L^2}{2I}

Skater pulls arms in: $I$ drops, $\omega$ shoots up to keep $L$ fixed, and $KE$ rises — that extra energy comes from muscular work, not for free.

6. Rolling — Kinetic Energy Split

v_{cm} = \omega R

KE = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}Mv_{cm}^2\left(1+\dfrac{K^2}{R^2}\right)

7. Rolling Down an Incline — Key Result

v_{cm} = \sqrt{\dfrac{2gh}{1+K^2/R^2}}

Smaller $K^2/R^2$ (mass closer to the axis) needs less energy budget to spin up, so more is left for translation — it arrives faster, regardless of mass or radius.
Solid sphere $(K^2/R^2=2/5)$ beats a hollow cylinder $(K^2/R^2=1)$ down the same incline, every time.

Full derivations and worked examples: detailed notes →