Rotational Motion

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Torque and Angular Momentum — The Core Quantities:

Torque (also called moment of force) is the rotational equivalent of force. It is defined as τ = r × F = rF sinθ, where r is the perpendicular distance from the axis of rotation to the point where force is applied, and θ is the angle between r and F. The SI unit of torque is newton-metre (N·m), which has the same dimensions as joule but they are fundamentally different physical quantities — torque is a vector, energy is a scalar.

Angular momentum of a particle about a point is L = r × p = r × mv. For a rigid body rotating about a fixed axis, L = Iω, where I is the moment of inertia about that axis and ω is the angular velocity. This is the rotational analogue of linear momentum p = mv.

Key Equations for Rigid Body Rotation:

• Torque and angular acceleration: τ = Iα (this is Newton’s second law for rotation)
• Conservation of angular momentum: when τ_ext = 0, L = Iω = constant
• Rotational kinetic energy: KE_rot = ½Iω²

⚡ JEE Tip: For problems involving pure rolling (no slipping), apply both translational and rotational equations: v = ωr, a_cm = αr, and use the constraint that the point of contact has zero velocity relative to the surface. JEE Advanced 2022 Qn 35 involved a sphere rolling down an inclined plane — the acceleration is a = (5/7)g sinθ, derived by combining τ = Iα with F = ma.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Moment of Inertia — Detailed Calculation:

Moment of inertia I = Σmr² for a system of particles. For continuous bodies, I = ∫r² dm. It depends on (1) the mass of the body, (2) the distribution of mass relative to the axis, and (3) the position and orientation of the axis.

Moment of inertia of common shapes (about symmetry axis through centre):

Body	I (about axis)	Radius of gyration k
Thin ring	MR²	k = R
Uniform disc	MR²/2	k = R/√2
Solid cylinder	MR²/2	k = R/√2
Hollow sphere	2MR²/3	k = R√(2/3)
Solid sphere	2MR²/5	k = R√(2/5)
Thin rod (about centre)	ML²/12	k = L/√12
Thin rod (about end)	ML²/3	k = L/√3
Rectangular plate (about centre, axis in plane)	M(L² + W²)/12	—

Theorems on Moment of Inertia:

Parallel axis theorem: I = I_cm + Md², where d is the distance between the two parallel axes. This always increases I unless d = 0. Use this for any off-centre axis rotation.

Perpendicular axis theorem: For planar bodies (flat objects with negligible thickness), I_z = I_x + I_y, where z is the axis perpendicular to the plane, and x and y are axes in the plane passing through the centre of mass. This only applies to 2D bodies like rings, discs, rectangles, and triangular plates.

Torque and Angular Acceleration:

For a rigid body under multiple torques, Στ = Iα. This is the rotational analogue of ΣF = ma. In JEE problems, always resolve torques about the same axis — choose the point where unknown forces pass to eliminate them from the equation.

Angular Impulse: The rotational analogue of linear impulse. Angular impulse = τΔt = ΔL. When a force acts for a short time (impulse), it changes angular momentum. Problems involving bats hitting balls, or doors slamming shut, use this principle.

⚡ JEE Tip: When solving pulley-block systems with rotation, draw separate free body diagrams for each translational body AND the rotating pulley. Write ΣF = ma for each body and Στ = Iα for the pulley, then use the constraint relation (string length constant → a = αr for fixed string over pulley).

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation of Torque = Iα from First Principles:

Consider a rigid body of mass M rotating about a fixed axis with angular acceleration α. Take a small element of mass dm at distance r from the axis. The tangential force on this element is F_t = dm × r × α (from Newton’s second law in tangential direction). The torque about the axis due to this force is dτ = r × F_t = r × (dm × r × α) = dm × r² × α. Integrating: τ = ∫ dm r² α = α ∫ dm r² = Iα. This derivation shows that torque = Iα holds for any rigid body rotating about a fixed axis.

Kinetic Energy of Rolling Bodies:

For a body rolling without slipping, total kinetic energy is the sum of translational KE of the centre of mass and rotational KE about the centre of mass:

KE_total = ½Mv_cm² + ½I_cmω²

Using the rolling constraint v_cm = ωR (for pure rolling), KE_total = ½Mv_cm²(1 + k²/R²), where k is the radius of gyration about the centre of mass.

For common rolling bodies:

• Solid sphere (k² = 2R²/5): KE_total = ½Mv²(1 + 2/5) = 7/10Mv²
• Hollow sphere (k² = 2R²/3): KE_total = ½Mv²(1 + 2/3) = 5/6Mv²
• Solid cylinder (k² = R²/2): KE_total = ½Mv²(1 + 1/2) = 3/4Mv²
• Ring (k² = R²): KE_total = ½Mv²(1 + 1) = Mv²

This hierarchy of energies explains why a solid sphere accelerates faster than a hollow sphere rolling down the same incline — more of its kinetic energy goes into translation.

Acceleration of Rolling Bodies Down an Incline:

For a body of mass M and radius R rolling down a plane inclined at angle θ: Mgsinθ - f = Ma (translation), and fR = Iα = I(a/R) (rotation about COM). Solving: a = gsinθ / (1 + I/(MR²)).

• Solid sphere: a = (5/7)g sinθ ≈ 0.714 g sinθ
• Hollow sphere: a = (3/5)g sinθ ≈ 0.6 g sinθ
• Solid cylinder: a = (2/3)g sinθ ≈ 0.667 g sinθ
• Ring: a = (1/2)g sinθ = 0.5 g sinθ

Note: Without rolling (sliding without friction), a = g sinθ. The rolling condition always reduces acceleration because some gravitational potential energy goes into rotational KE.

Precession of a Gyroscope:

When a torque is applied perpendicular to the angular momentum vector of a spinning wheel, the angular momentum vector itself rotates — this is precession. The precessional angular velocity Ω = τ/L = Mgd/(Iω). Gyroscopes, tops, and bicycle wheels exhibit precession. This is a JEE Advanced topic that requires understanding that torque changes the direction (not magnitude) of angular momentum.

⚡ JEE Advanced Pattern: Rotational motion combined with energy conservation and angular momentum conservation appears in nearly every JEE Advanced paper. Common problem setups include: (1) a sphere/cylinder rolling down an incline, (2) a turntable with a mass dropped onto it (conservation of angular momentum, energy dissipated), (3) a rod pivoted at one end struck by a particle (impulse-momentum), and (4) connected bodies over pulleys where the pulley has mass. In JEE Advanced 2023 Paper 1, Qn 42 was a combined translation-rotation problem with a yo-yo moving vertical — the torque of string tension about the centre caused both translational and rotational acceleration simultaneously.