System of Particles and Rotational Motion

🎯 Key Points

• v=ωr, a_t=αr, a_c=ω²r; rotational kinematics mirror linear kinematics with θ,ω,α replacing x,v,a
• I = Σmr²; Parallel axis: I=Icm+Md²; Perpendicular axis (planar bodies only): Iz=Ix+Iy
• τ = r×F = Iα; L = Iω; τ=dL/dt; L is conserved when net external torque = 0
• Rolling without slipping: KE_total = ½mv²(1+I/mR²); acceleration down incline a=g sinθ/(1+I/mR²)
• KE_rot/KE_total ratios: solid sphere=2/7, disk=1/3, ring=1/2 - smaller I/mR² means faster rolling down a slope

Angular Kinematic Quantities

• Angular displacement: θ (radians)
• Angular velocity: ω = dθ/dt (rad/s)
• Angular acceleration: α = dω/dt (rad/s²)
• Linear-angular relations: v = ωr, a_tangential = αr, a_centripetal = ω²r
• Rotational equations: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ

Moment of Inertia (I)

• I = Σm_i·r_i² (sum of mass × distance² from axis)
• Uniform rod about centre: I = ML²/12; about end: I = ML²/3
• Solid cylinder/disk about axis: I = MR²/2
• Hollow cylinder about axis: I = MR²
• Solid sphere about diameter: I = 2MR²/5
• Hollow sphere about diameter: I = 2MR²/3
• Ring about axis: I = MR²; about diameter: I = MR²/2

Theorems of MI

• Parallel Axis: I = I_cm + Md² (axis parallel to CM axis, distance d away)
• Perpendicular Axis: I_z = I_x + I_y (only for flat planar bodies)

Torque and Angular Momentum

• Torque: τ = r × F = I·α (rotational analogue of force)
• Angular momentum: L = Iω = r × p
• τ = dL/dt (Newton's 2nd law for rotation)
• Conservation of L: if τ_net = 0, then L = constant (e.g., ice skater pulling arms in)

Torque τ = r × F. The moment (lever) arm r_⊥ is the perpendicular distance from the axis to the force's line of action. Image: Guy vandegrift, CC BY-SA 4.0, via Wikimedia Commons.

Rotational Kinetic Energy

• KE_rot = ½Iω²
• Rolling without slipping: KE_total = ½mv² + ½Iω² = ½mv²(1 + I/mR²)
• For solid sphere: KE_rot/KE_total = 2/7; for disk: 1/3; for ring: 1/2
• Acceleration down incline: a = g·sinθ/(1 + I/mR²)

Centre of Mass

• Centre of mass (CM) is the point where the entire mass of a system can be taken to be concentrated for describing its translational motion
• For a system of particles: x_cm = (Σm_i·x_i)/(Σm_i), and similarly for y_cm, z_cm
• For continuous bodies: x_cm = (∫x·dm)/(∫dm)
• The CM of a uniform, symmetric body lies at its geometric centre (centre of a ring, disk, sphere, rod)
• Two particles: CM divides the joining line in the inverse ratio of masses (closer to the heavier mass)

Centre of Mass of Common Bodies

Motion of the Centre of Mass

• Velocity of CM: v_cm = (Σm_i·v_i)/(Σm_i); total momentum P = M·v_cm
• The CM moves as if all external forces acted on the total mass concentrated there: F_ext = M·a_cm
• If net external force is zero, v_cm stays constant (internal forces, like in an explosion, cannot change the CM's motion)
• In projectile motion, if a shell explodes mid-flight, the CM of the fragments continues on the original parabolic path

Equilibrium of a Rigid Body

• A rigid body is in mechanical equilibrium when both conditions hold simultaneously:
• Translational equilibrium: net external force is zero (ΣF = 0)
• Rotational equilibrium: net external torque about any axis is zero (Στ = 0)
• Couple: two equal and opposite parallel forces not along the same line; produces pure rotation, torque = force × perpendicular distance between them
• Centre of gravity is the point where the total gravitational torque on the body is zero; it coincides with the CM in a uniform gravitational field

🚀 JEE Advanced Edge

Race down an incline: Since a = g sinθ/(1+I/mR²), a body with SMALLER I/mR² accelerates faster - so for objects of the same shape category released together: solid sphere > solid cylinder/disk > hollow sphere > hollow cylinder/ring, regardless of their mass or radius (those cancel out of the ratio).

Instantaneous axis of rotation (rolling): For a body rolling without slipping, the point of contact with the ground is momentarily at rest - treating it as a fixed pivot lets you directly compute torque/angular momentum about that point without needing to separately track translational + rotational motion.

Worked problem: A solid sphere and a hollow sphere of the same mass and radius are released from rest at the top of an incline. Find the ratio of times taken to reach the bottom. Approach: a_solid = g sinθ/(1+2/5) = (5/7)g sinθ; a_hollow = g sinθ/(1+2/3) = (3/5)g sinθ. Since distance s=½at² for both (same s), t∝1/√a, so t_solid/t_hollow = √(a_hollow/a_solid) = √((3/5)/(5/7)) = √(21/25) ≈ 0.917 - the solid sphere reaches the bottom faster.

Conservation of Angular Momentum

• When the net external torque on a system is zero, its total angular momentum remains constant: L = I ω = constant.
• If the moment of inertia decreases, the angular velocity increases proportionally, and vice versa.
• A spinning skater speeds up on pulling the arms inward, reducing the moment of inertia while conserving angular momentum.
• A diver curls up to increase spin rate and stretches out to slow down before entering the water.
• This conservation law is the rotational analogue of the conservation of linear momentum.

Rolling Motion

• Rolling combines translation of the centre of mass with rotation about the centre of mass.
• For rolling without slipping, the contact condition is v(cm) = R ω, where R is the radius.
• The velocity of the lowest point is zero, the centre moves with v, and the topmost point moves with 2v.
• The total kinetic energy is KE = (1/2)m v² + (1/2)I ω² = (1/2)m v² (1 + k²/R²), where k is the radius of gyration.
• On rolling down an incline, a body with a smaller k²/R² (like a solid sphere) reaches the bottom faster than one with a larger value (like a ring).

Radius of Gyration

• The radius of gyration k is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia: I = m k^2.
• It is expressed as k = sqrt(I/m) and has the dimension of length.
• For a thin ring about its central axis, k = R; for a solid disc, k = R/sqrt(2).
• The radius of gyration depends on the position and orientation of the axis of rotation, not on the mass alone.
• It provides a convenient way to compare the mass distribution of different bodies about a chosen axis.

Rotational Analogue of Newton's Second Law

• For a rigid body rotating about a fixed axis, the net torque equals the product of moment of inertia and angular acceleration: τ = I α.
• This is the rotational counterpart of F = ma, with torque replacing force, moment of inertia replacing mass, and angular acceleration replacing linear acceleration.
• The work done by a torque in rotating through an angle θ is W = τ θ.
• The power delivered by a torque is P = τ ω.
• These relations let rotational problems be solved with the same techniques as translational dynamics.

Comparison of Translational and Rotational Motion

• Each translational quantity has a rotational counterpart: displacement x with angle θ, velocity v with angular velocity ω, and acceleration a with angular acceleration α.
• Mass m corresponds to moment of inertia I, force F to torque τ, and linear momentum p to angular momentum L.
• Newton's law F = ma corresponds to τ = I α, and p = mv corresponds to L = I ω.
• Translational kinetic energy (1/2)m v² corresponds to rotational kinetic energy (1/2)I ω^2.
• The equations of rotational kinematics (ω = ω0 + α t and so on) mirror the equations for uniformly accelerated linear motion.