Rotational Motion Formula Sheet
All key rotational motion formulas - centre of mass, moment of inertia, torque, angular momentum, rolling. JEE Main & Advanced quick revision.
One-page rapid revision of every formula, standard result, and high-yield trick from the Rotational Motion chapter. Scan before the exam; click through to the full topics when you need the derivation.
Linear vs Rotational Analogues
| Linear | Rotational | Relation |
|---|---|---|
| Displacement $s$ | Angular displacement $\theta$ | $s = r\theta$ |
| Velocity $v$ | Angular velocity $\omega$ | $v = r\omega$ |
| Acceleration $a$ | Angular acceleration $\alpha$ | $a_t = r\alpha$ |
| Mass $m$ | Moment of inertia $I$ | — |
| Force $F$ | Torque $\tau$ | $\vec{\tau} = \vec{r}\times\vec{F}$ |
| Momentum $p = mv$ | Angular momentum $L = I\omega$ | $\vec{L} = \vec{r}\times\vec{p}$ |
| $F = ma$ | $\tau = I\alpha$ | — |
| $KE = \tfrac{1}{2}mv^2$ | $KE = \tfrac{1}{2}I\omega^2$ | — |
Every rotational formula is the linear one with the swap $m \to I$, $v \to \omega$, $a \to \alpha$, $F \to \tau$, $p \to L$. Memorise linear mechanics and you already know rotation.
Rotational Kinematics (constant $\alpha$)
$$\omega = \omega_0 + \alpha t$$$$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$$$$\omega^2 = \omega_0^2 + 2\alpha\theta$$Centre of Mass
$$\boxed{\vec{r}_{cm} = \frac{\sum m_i\vec{r}_i}{\sum m_i} = \frac{1}{M}\int \vec{r}\,dm}$$Component form: $\displaystyle x_{cm} = \frac{\sum m_i x_i}{M},\quad y_{cm} = \frac{\sum m_i y_i}{M},\quad z_{cm} = \frac{\sum m_i z_i}{M}$
| Quantity | Formula | Notes |
|---|---|---|
| Two particles (origin at $m_1$) | $x_{cm} = \dfrac{m_2 d}{m_1 + m_2}$ | COM closer to heavier mass; divides line in ratio $m_2 : m_1$ |
| Velocity of COM | $\vec{v}_{cm} = \dfrac{\sum m_i\vec{v}_i}{M} = \dfrac{\vec{p}_{total}}{M}$ | Total momentum / total mass |
| Acceleration of COM | $\vec{a}_{cm} = \dfrac{\sum m_i\vec{a}_i}{M} = \dfrac{\vec{F}_{ext}}{M}$ | Only external force matters |
| Newton’s 2nd law (system) | $\vec{F}_{ext} = M\vec{a}_{cm}$ | Internal forces cancel |
Standard COM Locations
| Body | COM location |
|---|---|
| Uniform rod (length $L$) | At centre, $L/2$ from either end |
| Semicircular ring (radius $R$) | $\dfrac{2R}{\pi}$ from centre |
| Semicircular disc (radius $R$) | $\dfrac{4R}{3\pi}$ from centre |
| Hollow hemisphere (radius $R$) | $\dfrac{R}{2}$ from base |
| Solid hemisphere (radius $R$) | $\dfrac{3R}{8}$ from base |
| Hollow cone (height $h$) | $\dfrac{h}{3}$ from base |
| Solid cone (height $h$) | $\dfrac{h}{4}$ from base |
If a body is symmetric about an axis, the COM lies on it. Two or more axes of symmetry → COM at their intersection. In an explosion with no external force, the COM keeps following the original parabolic path.
Moment of Inertia
$$\boxed{I = \sum m_i r_i^2 = \int r^2\,dm}$$Radius of gyration: $\displaystyle I = Mk^2 \;\Rightarrow\; \boxed{k = \sqrt{\tfrac{I}{M}}}$
Standard Moments of Inertia
| Body | Axis | $I$ | $k$ |
|---|---|---|---|
| Point mass | Distance $r$ | $mr^2$ | $r$ |
| Rod (length $L$) | Through centre, ⊥ | $\dfrac{ML^2}{12}$ | $\dfrac{L}{2\sqrt{3}}$ |
| Rod (length $L$) | Through end, ⊥ | $\dfrac{ML^2}{3}$ | $\dfrac{L}{\sqrt{3}}$ |
| Ring (radius $R$) | Through centre, ⊥ plane | $MR^2$ | $R$ |
| Ring (radius $R$) | About diameter | $\dfrac{MR^2}{2}$ | — |
| Disc (radius $R$) | Through centre, ⊥ plane | $\dfrac{MR^2}{2}$ | $\dfrac{R}{\sqrt{2}}$ |
| Disc (radius $R$) | About diameter | $\dfrac{MR^2}{4}$ | — |
| Solid cylinder (radius $R$) | Along axis | $\dfrac{MR^2}{2}$ | $\dfrac{R}{\sqrt{2}}$ |
| Solid cylinder | Through centre, ⊥ length | $M\left(\dfrac{R^2}{4} + \dfrac{L^2}{12}\right)$ | — |
| Hollow cylinder ($R_1, R_2$) | Along axis | $\dfrac{M(R_1^2 + R_2^2)}{2}$ | — |
| Solid sphere (radius $R$) | Any diameter | $\dfrac{2MR^2}{5}$ | $\sqrt{\tfrac{2}{5}}R$ |
| Hollow sphere (radius $R$) | Any diameter | $\dfrac{2MR^2}{3}$ | $\sqrt{\tfrac{2}{3}}R$ |
| Rectangular plate ($a\times b$) | Along side $a$, through centre | $\dfrac{Mb^2}{12}$ | — |
| Rectangular plate ($a\times b$) | ⊥ to plate, through centre | $\dfrac{M(a^2 + b^2)}{12}$ | — |
Disc $=\tfrac{1}{2}$, Rod-centre $=\tfrac{1}{12}$, I (rod-end) $=\tfrac{1}{3}$, Solid sphere $=\tfrac{2}{5}$, Hollow sphere $=\tfrac{2}{3}$. For equal mass and radius, hollow always has larger $I$ than solid. Rod about end $= 4\times$ rod about centre.
Axis Theorems
$$\boxed{I = I_{cm} + Md^2} \qquad \text{(Parallel axis — any rigid body)}$$$$\boxed{I_z = I_x + I_y} \qquad \text{(Perpendicular axis — planar bodies only)}$$| Theorem | Condition | Key point |
|---|---|---|
| Parallel axis | Axes must be parallel; $d$ = ⊥ distance from COM axis | $I \ge I_{cm}$ always; minimum at COM |
| Perpendicular axis | Planar lamina only (disc, ring, plate); 3 axes meet at a point | Symmetric lamina: $I_x = I_y \Rightarrow I_z = 2I_x$ |
Useful derived results (from the theorems):
| Body | Axis | $I$ |
|---|---|---|
| Disc | Tangent ⊥ to plane | $\dfrac{3MR^2}{2}$ |
| Disc | Tangent in plane (edge) | $\dfrac{5MR^2}{4}$ |
| Ring | Tangent in plane | $\dfrac{3MR^2}{2}$ |
| Square plate (side $a$) | ⊥ through centre | $\dfrac{Ma^2}{6}$ |
| Square plate (side $a$) | ⊥ through one corner | $\dfrac{2Ma^2}{3}$ |
Torque
$$\boxed{\vec{\tau} = \vec{r}\times\vec{F}} \qquad \tau = rF\sin\theta = r_\perp F = rF_\perp$$| Quantity | Formula | Notes |
|---|---|---|
| Newton’s 2nd law (rotation) | $\boxed{\tau_{net} = I\alpha}$ | Rotational analogue of $F = ma$ |
| Rotational equilibrium | $\sum\vec{\tau} = 0$ | Holds about any axis; pair with $\sum\vec{F} = 0$ |
| Torque due to gravity | $\tau_g = r_{cm}\,Mg\sin\theta$ | Weight acts at COM |
| Maximum torque | $\tau_{max} = rF$ | When $\theta = 90^\circ$ |
Torque is zero when $F = 0$, or $r = 0$ (force through axis), or $\vec{F}\parallel\vec{r}$ ($\sin\theta = 0$). Sign convention: CCW positive, CW negative.
Take torques about the point where the unknown forces act — their torque vanishes and the equation simplifies instantly.
Angular Momentum
$$\boxed{\vec{L} = \vec{r}\times\vec{p} = \vec{r}\times m\vec{v}} \qquad L = mvr\sin\theta$$$$\boxed{L = I\omega} \qquad \boxed{\vec{\tau} = \frac{d\vec{L}}{dt}}$$| Concept | Formula | Notes |
|---|---|---|
| Particle (straight line / circular) | $L = mvr$ | $r$ = ⊥ distance from axis |
| Rigid body about fixed axis | $L = I\omega$ | Most common form |
| System of particles | $\vec{L}_{total} = \sum \vec{r}_i\times m_i\vec{v}_i$ | Vector sum |
| Conservation (if $\tau_{ext}=0$) | $\boxed{I_1\omega_1 = I_2\omega_2}$ | $L$ constant |
| Unit | kg·m²/s = J·s | — |
With $L$ conserved, $\omega \propto 1/I$: pull mass inward → $I$ drops → $\omega$ rises (figure skater). Conservation of $L$ does not mean conservation of $KE$ — pulling arms in raises $KE$ (the skater does work).
Standard conservation setups:
| Setup | Equation |
|---|---|
| Bullet embeds in rotating disc | $I_{disc}\omega_i + mvr = (I_{disc} + mr^2)\omega_f$ |
| Two discs couple coaxially | $\omega_f = \dfrac{I_1\omega_1}{I_1 + I_2}$ (KE not conserved) |
| Planet / satellite (tangential) | $mv_1 r_1 = mv_2 r_2$ (Kepler’s 2nd law) |
Rotational Kinetic Energy
$$\boxed{KE_{rot} = \tfrac{1}{2}I\omega^2}$$Rolling Motion
Pure rolling (no slipping) condition:
$$\boxed{v_{cm} = R\omega} \qquad a_{cm} = R\alpha$$Velocity of points on a purely rolling body: top $= 2v_{cm}$ (fastest), centre $= v_{cm}$, contact $= 0$ (instantaneously at rest).
Energy in Rolling
$$KE_{total} = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I_{cm}\omega^2$$$$\boxed{KE_{total} = \tfrac{1}{2}Mv_{cm}^2\left(1 + \frac{k^2}{R^2}\right) = \tfrac{1}{2}Mv_{cm}^2\left(1 + \frac{I_{cm}}{MR^2}\right)}$$Energy split: $\dfrac{KE_{rot}}{KE_{trans}} = \dfrac{I_{cm}}{MR^2}$
On an Incline (pure rolling)
| Quantity | Formula |
|---|---|
| Acceleration | $\boxed{a = \dfrac{g\sin\theta}{1 + I_{cm}/MR^2}}$ |
| Velocity at bottom | $\boxed{v = \sqrt{\dfrac{2gh}{1 + I_{cm}/MR^2}}}$ |
| Time down incline (length $L$) | $\boxed{t = \sqrt{\dfrac{2L(1 + I_{cm}/MR^2)}{g\sin\theta}}}$ |
| Friction force | $f = \dfrac{(I_{cm}/MR^2)\,Mg\sin\theta}{1 + I_{cm}/MR^2}$ |
| Min. friction for rolling | $\boxed{\mu_s \ge \dfrac{\tan\theta}{1 + MR^2/I_{cm}}}$ |
Race Comparison ($g\sin\theta$ reference)
| Object | $I/MR^2$ | Acceleration | $v$ at bottom | $KE_{rot}/KE_{total}$ |
|---|---|---|---|---|
| Sliding block (no friction) | 0 | $g\sin\theta$ | — | 0% |
| Solid sphere | 2/5 | $\dfrac{5g\sin\theta}{7}$ | $\sqrt{\tfrac{10gh}{7}}$ | 29% |
| Disc / solid cylinder | 1/2 | $\dfrac{2g\sin\theta}{3}$ | $\sqrt{\tfrac{4gh}{3}}$ | 33% |
| Hollow sphere | 2/3 | $\dfrac{3g\sin\theta}{5}$ | $\sqrt{\tfrac{6gh}{5}}$ | 40% |
| Ring / hollow cylinder | 1 | $\dfrac{g\sin\theta}{2}$ | $\sqrt{gh}$ | 50% |
Acceleration is independent of mass and radius — only the shape factor $I/MR^2$ matters. Smaller ratio → larger acceleration → wins the race. Order fastest to slowest: solid sphere > disc > hollow sphere > ring.
Slipping → Rolling (solid cylinder given $v_0$, no spin, friction $\mu$)
$$t = \frac{v_0}{3\mu g}, \qquad s = \frac{5v_0^2}{18\mu g}, \qquad v_{roll} = \frac{2v_0}{3}$$One-Glance Map
graph TD
A[Rotational Motion] --> B[Centre of Mass]
A --> C[Moment of Inertia]
C --> D[Axis Theorems]
A --> E[Torque: tau = I alpha]
A --> F[Angular Momentum: L = I omega]
E --> F
C --> G[Rolling: v = R omega]
F --> G