Class 11 Physics · Chapter 6 NotesSystem of Particles and Rotational Motion

A rigid body is an idealised body whose shape does not change, so the distance between any two of its particles stays the same. No real body is perfectly rigid, but in many cases the deformation is small enough to be ignored. A rigid body that is not fixed in any way can have pure translational motion, in which every particle has the same velocity at every instant, or a combination of translation and rotation, as in a cylinder rolling down an inclined plane. When a body is pivoted or fixed along a line, it can only rotate. In rotation about a fixed axis, every particle moves in a circle lying in a plane perpendicular to the axis, with the centre of the circle on the axis. Particles lying on the axis remain stationary.

• Pure translationall particles have the same velocity at any instant.
• Rotation about a fixed axisevery particle has the same angular velocity at any instant.
• Rolling motion is a combination of translation and rotation.

The centre of mass of a system of particles is the point whose position vector is R = (Σ mᵢrᵢ)/M, where M is the total mass. For two particles of equal mass on a line, it lies exactly midway between them. For three particles of equal mass forming a triangle, it coincides with the centroid of the triangle. For a continuous body the sums become integrals, and for a homogeneous body of regular shape the centre of mass lies at its geometric centre because of symmetry. The centre of mass need not lie inside the material of the body.

• X = (Σ mᵢxᵢ)/M, Y = (Σ mᵢyᵢ)/M, Z = (Σ mᵢzᵢ)/M
• For a homogeneous thin rod, ring, disc or sphere, the centre of mass is at the geometric centre.
• The centre of gravity coincides with the centre of mass when g is the same throughout the body.

Differentiating the definition of the centre of mass gives M V = Σ mᵢvᵢ, so the total linear momentum of a system is P = M V. Differentiating again gives M A = F_ext, which means the centre of mass moves as if the whole mass were concentrated there and all external forces acted at that point. Internal forces do not affect this motion. If the total external force is zero, the total linear momentum is constant and the centre of mass moves with constant velocity. This explains why the centre of mass of an exploding projectile continues along the same parabolic path.

• P = M V
• M A = F_ext
• If F_ext = 0, then P = constant and the velocity of the centre of mass is constant.

The vector product of two vectors a and b is a vector of magnitude ab sinθ, directed perpendicular to the plane of a and b according to the right hand rule. It is not commutative: a × b = −b × a. For a rigid body rotating about a fixed axis, the angular velocity ω is a vector along the axis, and the linear velocity of any particle is v = ω × r, with magnitude v = ωr. Angular acceleration is α = dω/dt. For rotation about a fixed axis the direction of ω is fixed and the vector equation reduces to the scalar form α = dω/dt.

• |a × b| = ab sinθ
• v = ω × r and v = ωr
• α = dω/dt

The moment of a force, or torque, about a point is τ = r × F, with magnitude τ = rF sinθ. It vanishes if the force is zero or if its line of action passes through the origin. The angular momentum of a particle about a point is l = r × p, with magnitude l = rp sinθ. Differentiating l with respect to time gives dl/dt = τ, the rotational analogue of F = dp/dt. For a system of particles, the total angular momentum changes at a rate equal to the total external torque, since internal torques cancel in pairs. If the total external torque is zero, the total angular momentum is conserved.

• τ = r × F, τ = rF sinθ
• l = r × p, l = rp sinθ
• dL/dt = τ_ext; if τ_ext = 0, L = constant

A rigid body is in mechanical equilibrium when both its linear momentum and its angular momentum do not change with time. This requires the total external force to be zero and the total external torque to be zero. These two vector conditions give six scalar equations. When all forces are coplanar, only three conditions are needed: the sums of force components along two perpendicular axes in the plane must vanish, and the sum of torques about an axis perpendicular to the plane must vanish. A pair of equal and opposite forces with different lines of action forms a couple, which produces rotation without translation. The principle of moments for a lever follows from rotational equilibrium.

• Σ F = 0 (translational equilibrium)
• Σ τ = 0 (rotational equilibrium)
• For a couple, total force is zero but total torque is not zero.
• Leverload × load arm = effort × effort arm

For a rigid body rotating about a fixed axis, the kinetic energy is K = ½ Iω², where the moment of inertia is I = Σ mᵢrᵢ² and rᵢ is the perpendicular distance of the ith particle from the axis. The moment of inertia plays the role of mass in rotational motion and depends on the mass, its distribution and the axis chosen. For a thin ring about its centre, I = MR²; for a thin rod about its midpoint, I = ML²/12; for a solid sphere about a diameter, I = 2MR²/5. The radius of gyration k is defined by I = Mk². The rotational form of Newton's second law is τ = Iα, and the work done by a torque is W = τθ with power P = τω.

• I = Σ mᵢrᵢ²
• K = ½ Iω²
• τ = Iα
• W = τθ, P = τω
• I = Mk²

For rotation about a fixed axis, the component of angular momentum along the axis is L_z = Iω. For bodies symmetric about the axis of rotation, the perpendicular component L_⊥ is zero and the total angular momentum is L = Iω along the axis. Differentiating gives d(Iω)/dt = τ. If the moment of inertia does not change with time, this reduces to τ = Iα. If the external torque is zero, Iω remains constant. This conservation law explains why a skater or a person on a swivel chair spins faster when the arms are brought closer to the body, since the moment of inertia decreases.

• L_z = Iω
• d(Iω)/dt = τ
• If τ = 0, Iω = constant