Unit 5: Rotational Motion

Rotational Motion is a high-yield NEET unit — typically 2–3 MCQs per year. The questions break down into: (i) finding centre of mass of standard composite bodies, (ii) using moment of inertia of standard shapes (with parallel-axis), (iii) angular-momentum conservation problems (ice-skater type), and (iv) rolling-without-slipping on inclines.

Concept Map

• Centre of mass of discrete and continuous bodies
• Translation of CM: $\vec F_\text{ext} = M \vec a_\text{CM}$
• Angular kinematics (analogue of linear)
• Torque, angular momentum
• Moment of inertia (theorems, standard shapes)
• Newton's second law for rotation: $\tau = I\alpha$
• Conservation of angular momentum
• Rolling without slipping
• Combined translation + rotation (KE)

Topic 1: Centre of Mass

Sub-topic A: System of Particles

For point masses $m_i$ at positions $\vec r_i$:

For two particles: $\vec r_\text{CM} = (m_1 \vec r_1 + m_2 \vec r_2)/(m_1 + m_2)$. The CM lies on the line joining them, closer to the heavier one.

Sub-topic B: Continuous Bodies

Sub-topic C: CM of Composite / Cavity Bodies

For a disc of radius $R$ with a circular hole of radius $r$ at distance $d$ from centre:

(taken from main centre, hole's side negative).

Sub-topic D: Motion of CM

In absence of external forces, $\vec v_\text{CM} =$ constant. Internal forces (explosion, collision) cannot alter $\vec v_\text{CM}$.

Topic 2: Angular Kinematics

Relations: $v = r\omega$, $a_t = r\alpha$ (tangential), $a_c = r\omega^2 = v^2/r$ (centripetal).

Topic 3: Torque and Angular Momentum

Sub-topic A: Torque

Units: N·m. SI dimension $[M L^2 T^{-2}]$ (same as energy, but torque is a vector — not energy).

Sub-topic B: Angular Momentum

For a rigid body rotating about a fixed axis: $L = I\omega$.

Sub-topic C: Newton's Second Law for Rotation

For fixed-axis rotation, $\tau = I\alpha$.

Sub-topic D: Conservation of Angular Momentum

If $\vec \tau_\text{ext} = 0$, $\vec L =$ constant.

Classic example: an ice-skater pulls in her arms, reducing $I$, so $\omega$ increases to keep $L = I\omega$ constant. Her KE actually increases (she does work against centrifugal effect).

Topic 4: Moment of Inertia

Sub-topic A: Definition

with $r$ the perpendicular distance from the axis.

Sub-topic B: Standard Moments of Inertia (about symmetry axis)

Sub-topic C: Parallel-Axis Theorem

where $d$ is distance between the parallel axis and CM-axis. Example: rod about end $= \tfrac{1}{12} M L^2 + M (L/2)^2 = \tfrac{1}{3} M L^2$.

Sub-topic D: Perpendicular-Axis Theorem (planar bodies only)

For a flat (lamina) body in the $xy$-plane:

Example: for a disc, $I_z = \tfrac{1}{2} M R^2$ (perpendicular). By symmetry $I_x = I_y$, so $I_x = I_y = \tfrac{1}{4} M R^2$.

Sub-topic E: Radius of Gyration

For a solid sphere $k = R\sqrt{2/5}$; for a ring $k = R$.

Topic 5: Rotational Kinetic Energy and Work

Sub-topic A: KE of Rotation

Sub-topic B: Combined Translation + Rotation

For a body rolling or moving with CM velocity $v$:

Sub-topic C: Work-Energy in Rotation

Power: $P = \tau \omega$.

Topic 6: Rolling Without Slipping

Sub-topic A: Constraint

A body rolling without slipping has $v_\text{CM} = R\omega$. The point of contact is momentarily at rest.

Sub-topic B: KE in Pure Rolling

For specific bodies the factor $(1 + k^2/R^2)$ is:

Sub-topic C: Rolling Down an Incline

For a body rolling down an incline of angle $\theta$ (without slipping):

For a solid sphere this is $(5/7) g\sin\theta$; for a disc $(2/3) g\sin\theta$; for a ring $(1/2) g\sin\theta$.

Speed at bottom of incline of height $h$:

Order of finish (fastest to slowest down an incline): solid sphere $>$ disc $>$ hollow sphere $>$ ring.

Friction needed (to enforce rolling):

For pure rolling without sliding: $\mu_s \ge \tan\theta / (1 + M R^2/I)$.

Sub-topic D: Slipping vs Rolling on a Belt

If the contact point has velocity, kinetic friction acts there. Friction reduces relative motion until rolling condition is reached.

NEET Pattern MCQ Tips

• CM problems: composite bodies and bodies with cavities — use the negative-mass trick.
• MI recall: NEET asks for $I$ of standard shapes about specific axes; memorise both the symmetry axis and the diameter cases.
• Parallel-axis numericals: shift from CM-axis.
• Conservation of L: ice-skater style; what happens to $\omega$, $L$, KE.
• Rolling: "which reaches bottom first" — the body with the smallest $k^2/R^2$.
• Assertion-Reason: "Internal forces cannot change CM velocity" (true).

Common Confusions and Traps

• A body in pure rolling has zero velocity at the contact point — so kinetic friction does no work on it.
• $I$ depends on the axis — never quote a single $I$ for a body.
• Parallel-axis theorem requires the parallel axis through the CM as the reference.
• Perpendicular-axis theorem applies only to planar (lamina) bodies.
• For a sphere about its diameter use $\tfrac{2}{5} MR^2$ (solid) or $\tfrac{2}{3} MR^2$ (hollow shell) — easy to swap.
• Torque has SI units of N·m, but never call it joule — it is a different quantity from energy.
• In conservation of $L$ when arms are pulled in, KE increases (work done by internal muscular force).

Quick Revision Card

• $\vec r_\text{CM} = \sum m_i \vec r_i / \sum m_i$.
• CM of semicircular wire: $2R/\pi$ from diameter.
• CM of solid hemisphere: $3R/8$ from base.
• $\tau = I\alpha$; $L = I\omega$; $K_\text{rot} = \tfrac{1}{2} I\omega^2$.
• Disc $I = \tfrac{1}{2} MR^2$; Ring $I = MR^2$; Solid sphere $I = \tfrac{2}{5} MR^2$; Hollow sphere $I = \tfrac{2}{3} MR^2$.
• Parallel-axis: $I = I_\text{CM} + Md^2$; Perpendicular-axis (lamina): $I_z = I_x + I_y$.
• Rolling: $K = \tfrac{1}{2} Mv^2 (1 + k^2/R^2)$.
• Rolling down incline: $a = g\sin\theta/(1 + k^2/R^2)$.
• Sphere wins the race down the incline.

Worked NEET Examples

Example 1: CM of a Two-Particle System

Particles 2 kg at (0, 0) and 4 kg at (3, 6). CM: $x_\text{CM} = (2 \cdot 0 + 4 \cdot 3)/6 = 2$, $y_\text{CM} = (0 + 24)/6 = 4$. So CM at (2, 4).

Example 2: CM of a Half-Disc

A uniform disc of radius $R$ with center at origin, cut to leave only the upper half. The CM is at $(0, 4R/(3\pi))$ above center.

Example 3: Ring Rolling on Floor

Ring of mass $M$, radius $R$, rolling without slipping at $v$ on horizontal floor. Total KE: $K = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}(MR^2)(v/R)^2 = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}Mv^2 = Mv^2$. So rolling KE of ring = $Mv^2$ (twice translational).

Example 4: Angular Momentum of a Rotating Earth

Earth's MI (treated as solid sphere): $I = (2/5)MR^2 = (2/5)(6 \times 10^{24})(6.4 \times 10^6)^2 \approx 9.8 \times 10^{37}$ kg m². Angular speed $\omega = 2\pi/86400 \approx 7.3 \times 10^{-5}$ rad/s. So $L = I\omega \approx 7.2 \times 10^{33}$ kg m²/s.

Example 5: Ice-Skater

An ice-skater spins at $\omega_1$ with arms extended (MI $I_1$). She pulls in to $I_2 = I_1/4$. Then $\omega_2 = 4\omega_1$. Rotational KE goes from $\tfrac{1}{2}I_1\omega_1^2$ to $\tfrac{1}{2}(I_1/4)(4\omega_1)^2 = 4 \times \tfrac{1}{2}I_1\omega_1^2$ — KE quadruples. The skater does work against centripetal force in pulling arms in.

Derivations Summary

Moment of Inertia of a Rod (Through Center)

A uniform rod of length $L$ and mass $M$. Linear mass density $\mu = M/L$. Take element $dm = \mu\,dx$ at distance $x$ from center.

MI of a Ring About Central Axis

All mass is at distance $R$ from axis: $I = MR^2$.

About a diameter, by perpendicular-axis theorem: $I_z = I_x + I_y$. By symmetry $I_x = I_y$, so $I_x = I_z/2 = MR^2/2$.

MI of a Disc About Center

Use rings of width $dr$: each ring has $dm = 2\pi r\,dr \cdot M/(\pi R^2)$, contributing $dI = r^2\,dm$.

Solid Sphere About Diameter

Result: $I = (2/5)MR^2$. Derive via volume integration or by treating as a stack of discs.

Hollow Sphere About Diameter

Result: $I = (2/3)MR^2$. Larger than solid sphere — mass is farther from axis.

Combined Translation + Rotation Problems

Rolling Disc Down an Incline

Disc of mass $M$, radius $R$, on an incline of angle $\theta$. Equation along incline: $Mg\sin\theta - f = Ma$, where $f$ is friction (up the incline). Torque about CM: $fR = I\alpha = (MR^2/2)\alpha = (MR^2/2)(a/R) = MRa/2$. So $f = Ma/2$. Combining:

Friction $f = Mg\sin\theta/3$, must be $\le \mu_s Mg\cos\theta$, so $\mu_s \ge \tan\theta/3$.

Spool Pulled by String

A spool of yarn lies on a table. A horizontal string emerges from underneath the spool. If pulled, the spool rolls toward the puller — counterintuitive! The torque from the string about the contact point (instantaneous pivot) is in the direction of pull.

Rolling vs Sliding

A ball is given speed $v_0$ on a rough surface with no initial spin. Friction acts backward on the body but creates torque about CM, spinning up the ball. CM slows down: $v(t) = v_0 - \mu g t$. Angular speed grows: $\omega(t) = \mu g t/R \cdot (5/2)$ for a solid sphere (since $\mu mg \cdot R = I\alpha = (2/5)mR^2 \alpha$, so $\alpha = 5\mu g/(2R)$).

Rolling condition $v = R\omega$ reached when

giving $t = 2v_0/(7\mu g)$. At this time $v = (5/7)v_0$, KE has dropped from $(1/2)mv_0^2$ to $(5/7) \cdot \tfrac{1}{2}m v_0^2 \cdot$ (factor for translational + rotational).

Standard Combined-Motion Results (Race Down Incline)

Formula Sheet