Rotational Motion and Gravitation — NEET Physics

These two chapters look unrelated but share one big idea: they are the physics of extended bodies and the forces between them. Rotation is the exact mirror of the linear motion you already know — every quantity and every equation has a twin — so you learn the map once and translate. Gravitation is a single inverse-square law that, applied carefully, explains a falling apple, an orbiting satellite and the motion of the planets. Together they contribute 3–4 NEET questions almost every year, and they are among the most formula-predictable marks in the paper. This chapter builds both from first principles — with every key result derived, not just stated, and worked through with the kind of examples you will actually meet in the exam.

Part A — Rotational Motion

1. Rigid bodies and the centre of mass

A rigid body is an idealised object whose particles keep fixed distances from one another — it can move and spin, but not deform. Its motion splits cleanly into two parts:

• Translation of the centre of mass (CM) — the mass-weighted average position, ${\vec{r}}_{\text{cm}}=\frac{\sum m_i{\vec{r}}_i}{\sum m_i}$. The CM moves exactly as if all the mass and all the external force were concentrated there (${\vec{F}}_{\text{ext}}=M{\vec{a}}_{\text{cm}}$).
• Rotation about an axis through (or fixed relative to) the CM.

This is why a thrown spanner's centre of mass traces a smooth parabola while the spanner tumbles about it: the two motions are independent, and we can analyse each with its own set of laws. The rest of Part A is the rotational set.

2. The translation ↔ rotation analogy

The single most powerful idea here: every rotational quantity mirrors a linear one. Learn the map once, and every rotational formula follows from its linear twin.

Because the algebra is identical, the equations of motion carry over unchanged for constant angular acceleration $\alpha$:

$\omega ={\omega }_0+\alpha t,\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2,{\omega }^2={\omega }_0^2+2\alpha \theta$

A point at radius $r$ on a rotating body has a tangential acceleration $a_{\text{tan}}=r\alpha$ (changing its speed) and, always, a centripetal acceleration $a_c=v^2/r=r{\omega }^2$ pointing toward the axis.

Worked example 2.1. A wheel spinning at ${\omega }_0=10$ rad/s is braked at $\alpha =-2$ rad/s². When does it stop, and through what angle? Stop time from $\omega ={\omega }_0+\alpha t$: $0=10-2t\Rightarrow t=5$ s. Angle from ${\omega }^2={\omega }_0^2+2\alpha \theta$: $0=100-4\theta \Rightarrow \theta =25$ rad $\approx 4$ revolutions.

Worked example 2.2. A point on the rim of that wheel is at $r=0.5$ m. What is its centripetal acceleration at the start? $a_c=r{\omega }_0^2=0.5\times 10^2=50$ m/s² — directed toward the axis. Its tangential acceleration is $r\alpha =0.5\times 2=1$ m/s² (decelerating).

3. Torque, couples and rotational equilibrium

A force produces rotation only if it acts off the axis. The turning effect is the torque:

$\tau =rF\sin \theta =r_{\perp }F$

• $r$ is the distance from axis to point of application; $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.
• $r_{\perp }=r\sin \theta$ is the moment arm — the perpendicular distance from the axis to the line of the force.
• As a vector $\vec{\tau }=\vec{r}\times \vec{F}$; direction along the axis by the right-hand rule. Units N·m.

Torque is zero when the force passes through the axis ($\theta =0°$ or $180°$) — which is why pushing a door at its hinge does nothing while the same push at the far edge swings it open.

A couple is a pair of equal, opposite, parallel forces whose lines are separated by a distance $d$. The net force is zero, but the net torque is $\tau =Fd$ — a pure rotation with no translation (turning a steering wheel or a tap).

Rotational equilibrium requires the net torque to be zero, $\sum \tau =0$. For complete equilibrium a body needs both:

$\sum \vec{F}=0\text{(no translation)}\text{and}\sum \tau =0\text{(no rotation)}$

Worked example 3.1 (spanner). 20 N acts at 15 cm from a bolt, at 30° to the spanner. Torque? $\tau =rF\sin \theta =0.15\times 20\times \sin 30°=0.15\times 20\times 0.5=1.5$ N·m. Perpendicular ($90°$) it would be $3$ N·m — double, which is why you push at right angles.

Worked example 3.2 (see-saw). A 40 kg child sits 1.5 m left of a pivot. Where must a 60 kg adult sit to balance? ($g$ cancels.) Balance means equal and opposite torques: $40\times 1.5=60\times x\Rightarrow x=\frac{60}{60}=1.0$ m to the right. The heavier person sits closer — the moment arm compensates for the larger force.

4. Moment of inertia — rotational "mass"

Moment of inertia $I$ measures how hard it is to change a body's rotation. Unlike mass, it depends on how the mass is distributed about the axis:

$I={\sum }_i{m}_i{r}_i^2$

Distance is squared, so mass far from the axis dominates — the reason a skater, a flywheel and a diver all reshape themselves to control spin.

Standard bodies (memorise these)

The radius of gyration $k$ is defined by $I=M{k}^2$, so $k=\sqrt{I/M}$ — the single distance at which all the mass could sit to give the same $I$.

The two theorems

Parallel axis theorem — for any axis parallel to one through the CM, a distance $d$ away:

$I=I_{\text{cm}}+M{d}^2$

Perpendicular axis theorem — for a flat (planar) body only, the moment about the axis perpendicular to the plane equals the sum of the two in-plane axes:

$I_z=I_x+I_y$

Worked example 4.1 (parallel axis). Derive the rod's end value from its centre value. $I_{\text{end}}=I_{\text{cm}}+M{d}^2=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2=\frac{1}{12}M{L}^2+\frac{3}{12}M{L}^2=\frac{1}{3}M{L}^2.$ ✓ (matches the table).

Worked example 4.2 (perpendicular axis). A disc has $I=\frac{1}{2}M{R}^2$ about its central axis. Find $I$ about a diameter. By symmetry the two in-plane diameters are equal, $I_x=I_y=I_d$. Perpendicular axis: $I_z=I_x+I_y\Rightarrow \frac{1}{2}M{R}^2=2I_d\Rightarrow I_d=\frac{1}{4}M{R}^2.$

5. Rotational dynamics, angular momentum and angular impulse

Newton's second law, rotational form:

${\tau }_{\text{net}}=I\alpha$

Its momentum cousin is angular momentum $L=I\omega$, and the torque-over-time form is the angular impulse:

$\tau \Delta t=\Delta L=I\Delta \omega$

Conservation of angular momentum — when no external torque acts, $L$ is constant, so a drop in $I$ forces a rise in $\omega$:

$I_1{\omega }_1=I_2{\omega }_2$

• A skater pulling her arms in cuts $I$ and spins faster; a diver tucks to somersault, then opens to enter cleanly; a collapsing star spins up as its radius shrinks.

Worked example 5.1 (angular impulse). A torque of 5 N·m acts for 4 s on a body of $I=2$ kg·m², initially at rest. Final angular velocity? $\tau \Delta t=I\Delta \omega \Rightarrow 5\times 4=2\times \omega \Rightarrow \omega =10$ rad/s.

Worked example 5.2 (conservation). A disc of moment of inertia $I$ rotating at $\omega$ has an identical stationary disc dropped coaxially onto it; they then rotate together. Common angular speed, and what happens to the energy? $L$ conserved: $I\omega =(I+I){\omega }^{\prime }\Rightarrow {\omega }^{\prime }=\omega /2$. Initial $KE=\frac{1}{2}I{\omega }^2$; final $KE=\frac{1}{2}(2I)(\omega /2{)}^2=\frac{1}{4}I{\omega }^2$ — exactly half. The rest is dissipated as heat by friction between the discs. Angular momentum is conserved; kinetic energy is not (this coupling is the rotational analogue of a perfectly inelastic collision).

6. Rolling motion — translation and rotation together

A rolling wheel moves forward and spins. Rolling without slipping locks them:

$v_{\text{cm}}=\omega R$

Total kinetic energy is the sum of both parts:

$KE=\frac{1}{2}m{v}_{\text{cm}}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}_{\text{cm}}^2\left(1+\frac{I}{m{R}^2}\right)$

The factor $\frac{I}{m{R}^2}$ is the shape signature — $\frac{2}{5}$ (solid sphere), $\frac{1}{2}$ (disc), $1$ (ring). It controls everything about how a body rolls.

Acceleration down an incline (angle $\theta$, rolling without slipping):

$a=\frac{g\sin \theta }{1+\frac{I}{m{R}^2}}$

The smaller the shape signature, the larger the acceleration:

$\text{solid sphere }(\frac{5}{7}g\sin \theta )>\text{disc }(\frac{2}{3}g\sin \theta )>\text{ring }(\frac{1}{2}g\sin \theta )$

A solid sphere always wins the race down; a ring always loses — and mass and radius are irrelevant, only the shape. A perennial NEET question.

Worked example 6.1 (energy split). What fraction of a rolling solid sphere's KE is rotational? $K{E}_{\text{rot}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}(\frac{2}{5}m{R}^2){\omega }^2=\frac{1}{5}m{R}^2{\omega }^2$; with $v=\omega R$, $K{E}_{\text{trans}}=\frac{1}{2}m{R}^2{\omega }^2$. Total $=\frac{7}{10}m{R}^2{\omega }^2$, so rotational fraction $=\frac{1/5}{7/10}=\frac{2}{7}$ (and translational $\frac{5}{7}$).

Worked example 6.2 (final speed). A solid sphere rolls from rest down a height $h$. Its speed at the bottom (energy method)? $mgh=\frac{1}{2}m{v}^2(1+\frac{2}{5})=\frac{7}{10}m{v}^2\Rightarrow v=\sqrt{\frac{10gh}{7}}.$ Less than the $\sqrt{2gh}$ of a sliding block, because some energy goes into spin.

Part B — Gravitation

7. Newton's law of gravitation

Every pair of masses attracts along the line joining them:

$F=\frac{G{m}_1{m}_2}{r^2},G=6.67\times 10^{-11}{\text{N⋅m}}^2/{\text{kg}}^2$

• Inverse-square: double the separation → force to a quarter; triple → a ninth.
• $G$ (universal constant) is not $g$ (acceleration due to gravity). At a planet's surface they link as:

$g=\frac{GM}{R^2}$

Worked example 7.1. Estimate $g$ from Earth data ($M=6\times 10^{24}$ kg, $R=6.4\times 10^6$ m). $g=\frac{(6.67\times 10^{-11})(6\times 10^{24})}{(6.4\times 10^6{)}^2}=\frac{4.0\times 10^{14}}{4.1\times 10^{13}}\approx 9.8$ m/s². ✓

8. Gravitational field and potential

Three related but distinct quantities NEET loves to separate:

• Gravitational field $E=\frac{F}{m}=\frac{GM}{r^2}$ — force per unit mass (N/kg); a vector pointing toward $M$. (Numerically equal to $g$.)
• Gravitational potential $V=-\frac{GM}{r}$ — potential energy per unit mass (J/kg); a scalar, always negative, zero at infinity.
• Gravitational potential energy $U=mV=-\frac{GMm}{r}$ — the energy of the pair.

They are linked by $E=-\frac{dV}{dr}$: the field is the (negative) slope of the potential. Field is a vector about direction of force; potential is a scalar about energy — do not confuse the two.

Worked example 8.1. Gravitational potential at Earth's surface? $V=-\frac{GM}{R}=-gR=-(9.8)(6.4\times 10^6)\approx -6.3\times 10^7$ J/kg. The large negative value reflects how deeply bound anything at the surface is.

9. Variation of g

The surface value $g=9.8$ m/s² changes with altitude, depth and latitude.

With altitude $h$:

$g_h=g{\left(\frac{R}{R+h}\right)}^2\approx g\left(1-\frac{2h}{R}\right)(h\ll R)$

At $h=R$: $g_h=g(R/2R{)}^2=g/4$.

With depth $d$ (uniform Earth):

$g_d=g\left(1-\frac{d}{R}\right)$

$g$ decreases linearly downward, reaching zero at the centre ($d=R$). At $d=R/2$, $g_d=g/2$.

With latitude: Earth's rotation and its equatorial bulge make $g$ smallest at the equator, largest at the poles.

Contrast to fix in memory. Up: $g$ falls as $1/(R+h{)}^2$ (fast). Down: $g$ falls linearly to zero at the centre (slower near the surface). Students routinely swap these.

Worked example 9.1. At what depth is $g$ half its surface value? $g/2=g(1-d/R)\Rightarrow d=R/2$ — halfway to the centre.

10. Satellites: orbit, energy, binding energy and weightlessness

Orbital velocity — set gravity equal to the centripetal requirement, $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$:

$v_{\text{orbit}}=\sqrt{\frac{GM}{r}}(\approx 7.9\text{ km/s for a low orbit})$

Total mechanical energy of a satellite:

$E=\frac{1}{2}m{v}_{\text{orbit}}^2-\frac{GMm}{r}=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$

The negative sign is the signature of a bound orbit; its magnitude, $\frac{GMm}{2r}$, is the binding energy — the energy you'd add to just free the satellite to infinity.

Weightlessness in orbit is not the absence of gravity (gravity provides the centripetal force!). The satellite and its occupant fall together with the same acceleration, so the astronaut presses on nothing and nothing presses back — apparent weight is zero, exactly like a freely falling lift.

Worked example 10.1 (period). For a circular orbit, using $v=2\pi r/T$ and $v=\sqrt{GM/r}$, show $T=2\pi \sqrt{r^3/GM}$. $\frac{2\pi r}{T}=\sqrt{\frac{GM}{r}}\Rightarrow T=2\pi r\sqrt{\frac{r}{GM}}=2\pi \sqrt{\frac{r^3}{GM}}$ — which is exactly Kepler's third law in disguise.

Worked example 10.2 (energy comparison). Which needs more energy: raising a satellite to a higher orbit, or the same body's orbital KE? As $r$ grows, $E=-GMm/2r$ rises toward zero (more energy), while orbital speed $\sqrt{GM/r}$ drops. So higher orbits have more total energy but move slower — a counter-intuitive favourite.

11. Escape velocity

The minimum speed to break free entirely. Set total energy to zero, $\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$:

$v_{\text{escape}}=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}=\sqrt{2}{v}_{\text{orbit}}\approx 11.2\text{km/s (Earth)}$

Three tested facts: it is $\sqrt{2}$ times the surface orbital speed; it is independent of the escaping body's mass; and it is $\approx 11.2$ km/s for Earth.

Worked example 11.1. Why is escape velocity mass-independent? The mass $m$ cancels on both sides of $\frac{1}{2}m{v}_e^2=GMm/R$, leaving $v_e=\sqrt{2GM/R}$. A pebble and a spacecraft need the same 11.2 km/s; a rocket merely needs more energy to give more mass that speed.

12. Kepler's three laws

1. Law of orbits. Each planet moves in an ellipse with the Sun at one focus (not the centre).
2. Law of areas. The Sun–planet line sweeps equal areas in equal times — this is conservation of angular momentum, so a planet is fastest at perihelion (nearest) and slowest at aphelion (farthest).
3. Law of periods. $T^2\propto r^3$ (strictly, the semi-major axis $a^3$).

Deriving law 3 for a circular orbit: equate gravity to centripetal force with $v=2\pi r/T$:

$\frac{GM}{r^2}=\frac{4{\pi }^{2}r}{T^2}\Rightarrow T^2=\frac{4{\pi }^2}{GM}{r}^3\Rightarrow T^2\propto r^3.$

Because it is a 3/2 power, quadrupling the radius multiplies the period by $4^{3/2}=8$, not 4 — the classic trap.

Geostationary vs polar satellites: a geostationary satellite has $T=24$ h, orbits ~36,000 km above the equator west-to-east, so it appears fixed (TV/communication). A polar satellite orbits low over the poles, scanning fresh ground each pass (weather, imaging, spying).

Worked example 12.1. Two satellites orbit at $r$ and $4r$. Ratio of periods? $(T_2/T_1{)}^2=(4r/r{)}^3=64\Rightarrow T_2/T_1=8$.

13. Common traps NEET sets here

• Wrong-axis moment of inertia. $I$ is quoted for a stated axis; use the theorems to convert (disc: $\frac{1}{2}M{R}^2$ central, $\frac{1}{4}M{R}^2$ diameter).
• Escape vs orbital velocity. They differ by exactly $\sqrt{2}$; escape is from the surface, orbital is for a radius.
• Altitude vs depth for $g$. Up = inverse-square (fast); down = linear to zero at the centre. Never swap.
• Kepler's 3/2 power. $r\to 4r$ gives $T\to 8T$, not $4T$.
• Rolling race. Only $I/m{R}^2$ decides — mass and radius are irrelevant. Sphere first, ring last.
• Angular momentum conserved, KE not in drop-on-a-spinning-disc problems.
• Field vs potential. Field is a vector (N/kg); potential is a scalar (J/kg). Both are about the source mass, not a test mass.
• "Weightless means no gravity." Wrong — orbiting bodies are weightless because gravity makes them free-fall together.

14. Memory aids

• "Sphere Disc Ring, fast to slow" — the rolling-race order (increasing $I/m{R}^2$: $\frac{2}{5},\frac{1}{2},1$).
• "Up squared, down straight" — $g$ with altitude (inverse-square) vs depth (linear).
• "Escape is root-two orbit" — $v_e=\sqrt{2}{v}_o$; Earth 11.2 vs 7.9 km/s.
• "Bound is negative" — a satellite's total energy $-GMm/2r$ is negative; zero means escape.
• "Torque = force × arm" — always the perpendicular distance to the line of force.

15. Exam protocol

1. Translate rotation into linear terms via the analogy table; $\tau =I\alpha$ works exactly like $F=ma$.
2. Torque $=rF\sin \theta$; a force through the axis gives zero torque; a couple gives pure rotation.
3. Memorise the standard moments of inertia; move axes with the parallel/perpendicular theorems.
4. In changing-shape spin problems, conserve angular momentum ($I_1{\omega }_1=I_2{\omega }_2$), not KE.
5. For rolling, the shape factor $I/m{R}^2$ sets the KE split and the incline race — solid sphere fastest, ring slowest.
6. Gravitation is inverse-square; keep $G$ and $g$ distinct, and field (vector) vs potential (scalar) distinct.
7. $g$ fall-off: up inverse-square, down linear to zero at the centre.
8. $v_{\text{escape}}=\sqrt{2}{v}_{\text{orbit}}\approx 11.2$ km/s, mass-independent; a bound satellite has $E=-GMm/2r$; weightlessness = free fall.
9. Kepler: $T^2\propto r^3$, so use the 3/2 power in ratios.