System of Particles and Rotational Motion

In this chapter: rigid bodies and types of motion, centre of mass and its motion, vector product, angular velocity, torque and angular momentum, equilibrium of a rigid body, principle of moments, centre of gravity, moment of inertia, rotational kinematics and dynamics about a fixed axis, and conservation of angular momentum.

Rigid bodies and types of motion

A rigid body is one in which the distances between all pairs of particles stay fixed. Its motion can be pure translation (every particle has the same velocity at an instant), rotation about a fixed axis (every particle moves in a circle centred on the axis), or a combination of the two, as in a rolling wheel.

Centre of mass

The centre of mass (CM) of a system is the point whose position is the mass-weighted average of the positions of its particles.

For two particles, the CM lies on the line joining them and divides it in the inverse ratio of the masses, so it sits closer to the heavier one. For uniform bodies with symmetry, the CM is at the geometric centre: the middle of a rod, the centre of a ring, disc or sphere, the point where the diagonals of a rectangular plate meet, and the centroid of a triangular plate. The CM may lie outside the material of the body, as for a ring.

Worked example: Masses of 2 kg and 3 kg are placed at x = 0 and x = 5 m. Where is their centre of mass?
Solution: X = (2 × 0 + 3 × 5)/(2 + 3) = 15/5 = 3 m from the 2 kg mass, which is 2 m from the 3 kg mass. The distances are in the ratio 3 : 2, the inverse of the masses.

Motion of the centre of mass

Internal forces cannot change the motion of the CM. If a shell explodes in mid-air, the fragments fly apart, but their CM keeps following the original parabola until a fragment hits the ground. If the net external force is zero, the total momentum and the velocity of the CM stay constant.

Vector product

The vector (cross) product is A × B = AB sin θ n̂, where n̂ is perpendicular to the plane of A and B, given by the right-hand rule: curl the fingers from A to B through the smaller angle, and the thumb points along n̂.

• It is not commutative: B × A = −A × B. It is distributive over addition.
• A × A = 0, so î × î = ĵ × ĵ = k̂ × k̂ = 0.
• î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (cyclic order), and reversing the order gives a minus sign.
• In components, A × B is the determinant with rows (î ĵ k̂), (A_x A_y A_z), (B_x B_y B_z).

Angular velocity, torque and angular momentum

For rotation about a fixed axis, each particle at perpendicular distance r from the axis moves in a circle with speed v = ωr; in vector form v = ω × r. The angular velocity ω points along the axis (right-hand rule) and is the same for every particle. Angular acceleration is α = dω/dt.

Torque

Torque, or the moment of a force, measures the turning effect of a force about a point or axis:

r sin θ is the perpendicular distance of the line of action of the force from the axis, called the lever arm. A force whose line of action passes through the axis produces no torque, however large it is. For this reason a door is pushed at the handle, far from the hinges.

Worked example: Find the torque of a force F = (7î + 3ĵ − 5k̂) N about the origin, acting at a point with position vector r = (î − ĵ + k̂) m.
Solution: τ = r × F = î[(−1)(−5) − (1)(3)] − ĵ[(1)(−5) − (1)(7)] + k̂[(1)(3) − (−1)(7)] = î(5 − 3) − ĵ(−12) + k̂(10) = (2î + 12ĵ + 10k̂) N m.

Angular momentum

Conservation of angular momentum: if the net external torque on a system is zero, its total angular momentum stays constant.

Equilibrium of a rigid body

A rigid body is in mechanical equilibrium when both the net external force and the net external torque are zero. The first condition gives translational equilibrium, the second rotational equilibrium. A couple (two equal and opposite forces with different lines of action) has zero net force but a non-zero torque, so it rotates the body without translating it. Turning a tap or a steering wheel applies a couple.

Principle of moments: for a lever in equilibrium, load × load arm = effort × effort arm. The ratio load/effort is the mechanical advantage. The centre of gravity is the point about which the total gravitational torque is zero. For bodies small compared with the Earth, where g is the same at every point, it coincides with the centre of mass.

Moment of inertia

The kinetic energy of a rotating body is K = ½Iω², where the moment of inertia about the axis is

Moment of inertia plays the role in rotation that mass plays in translation. Unlike mass, it depends on the axis and on how the mass is distributed about it. A flywheel has most of its mass at the rim to give a large I, which keeps the rotation of an engine shaft steady.

Parallel and perpendicular axis theorems

These two theorems are in the JEE Main syllabus and are needed for many NEET questions, so learn them.

• Parallel axis theorem: I = I_cm + Md², where d is the distance between the given axis and a parallel axis through the centre of mass. It works for any body.
• Perpendicular axis theorem: for a flat (planar) body, I_z = I_x + I_y, where x and y lie in the plane and z is perpendicular to it, all three meeting at one point. This is how the disc's diameter value, MR²/4, follows from MR²/2.

Rotation about a fixed axis

For constant angular acceleration, the equations have the same form as in Chapter 2:

Worked example: A cord of negligible mass is wound round the rim of a flywheel of mass 20 kg and radius 20 cm (treat it as a disc). A steady pull of 25 N is applied on the cord. Find the angular acceleration and check that the work done in the first 2 s equals the gain in kinetic energy.
Solution: I = MR²/2 = 20 × (0.2)²/2 = 0.4 kg m². τ = FR = 25 × 0.2 = 5 N m, so α = τ/I = 12.5 rad s⁻². In 2 s from rest, θ = ½ × 12.5 × 4 = 25 rad, so work = τθ = 5 × 25 = 125 J. Also ω = 12.5 × 2 = 25 rad s⁻¹, and KE = ½ × 0.4 × 625 = 125 J. The two agree.

Conservation of angular momentum about a fixed axis

For rotation about a fixed axis, L = Iω. If no external torque acts, I₁ω₁ = I₂ω₂. A skater or dancer spinning with arms outstretched speeds up on pulling them in, because I decreases. A diver tucks in to spin faster during the fall. Kinetic energy is not conserved in these cases; the person does work with their muscles to pull the arms in, and ½Iω² = L²/2I increases.

Common mistakes: (1) Using the distance from the axis to the point of application instead of the lever arm when finding torque. (2) Applying the perpendicular axis theorem to a three-dimensional body such as a sphere or a cylinder; it holds only for planar bodies. (3) Using the parallel axis theorem between two axes when neither passes through the CM. (4) Assuming kinetic energy stays constant when angular momentum is conserved. (5) Getting the order of a cross product wrong; r × F and F × r have opposite signs.

JEE and NEET focus

• Centre of mass of two or three particles and of simple composite shapes, including a shape with a part removed.
• Moments of inertia from the table, with the parallel and perpendicular axis theorems.
• τ = Iα for a pulley or flywheel with a hanging mass or cord.
• Conservation of angular momentum: skater, person on a rotating platform, two discs brought into contact.
• Torque as r × F in component form, and equilibrium of a rod or ladder using the principle of moments.

Practice questions