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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 109 Answer Let the particle with velocity v be at accelerationnorangularacceleration.Thismeans point P at some instant t. We want to calculate (1) the total force, i.e. the vector sum of the the angular momentum of the particle about an forces, on the rigid body is zero; arbitrary point O. n F1+ F2+ ... Fn= ∑ Fi= 0 (6.30a) i=1 If the total force on the body is zero, then the total linear momentum of the body does not change with time. Eq. (6.30a) gives the condition for the translational equilibrium of the body. (2) The total torque, i.e. the vector sum of the torques on the rigid body is zero, n τ + τ + ..+ τn = ∑ τi= 0 (6.30b) 1 2 i=1 Fig 6.19 The angular momentum is l = r × mv. Its If the total torque on the rigid body is zero, the total angular momentum of the body does magnitude is mvr sin θ, where θ is the angle not change with time. Eq. (6.30 b) gives the between r and v as shown in Fig. 6.19. Although the particle changes position with time, the line condition for the rotationa l equilibrium of the body. of direction of v remains the same and hence One may raise a question, whether the OM = r sin θ. is a constant. Further, the direction of l is perpendicular rotational equilibrium condition [Eq. 6.30(b)] to the plane of r and v. It is into the page of the remains valid, if the origin with respect to which the torques are taken is shifted. One can show figure.This direction does not change with time. Thus, l remains the same in magnitude and that if the translational equilibrium condition direction and is therefore conserved. Is there [Eq. 6.30(a)] holds for a rigid body, then such a shift of origin does not matter, i.e. the rotational any external torque on the particle? ⊳ equilibrium condition is independent of the 6.8 EQUILIBRIUM OF A RIGID BODY location of the origin about which the torques aretaken.Example6.7givesaproofofthisresult We are now going to concentrate on the motion of rigid bodies rather than on the motion of in a special case of a couple, i.e. two forces acting on a rigid body in translational general systems of particles. equilibrium. The generalisation of this result to We shall recapitulate what effect the external forces have on a rigid body. (Henceforth n forces is left as an exercise. we shall omit the adjective ‘external’ because Eq. (6.30a) and Eq. (6.30b), both, are vector equations. They are equ ivalent to three scalar unless stated otherwise, we shall deal with only external forces and torques.) The forces change equations each. Eq. (6.30a) corresponds to the translational state of the motion of the rigid n n n body, i.e. they change its total linear F = 0 F = 0 F = 0 ∑i=1ix ,∑i=1iy and i=1iz (6.31a) momentum in accordance with Eq. (6.17). But this is not the only effect the forces have. The where F ixF iyd F aiz respectively the x, y and total torque on the body may not vanish. Such z components of the forces F. Siiilarly, Eq. (6.30b) is equivalent to three scalar equations a torque changes the rotational state of motion of the rigid body, i.e. it changes the total angular n n momentum of the body in accordance with ∑ τix= 0,∑ τiy= 0 and (6.31b) Eq. (6.28 b). i= i1 where τ ,τ and τ are respectively the x, y and A rigid body is said to be in mechanical ix iy iz equilibrium, if both its linear momentum and z components of the torque τi. angular momentum are not changing with time, Eq. (6.31a) and (6.31b) give six independent or equivalently, the body has neither linear conditions to be satisfied for mechanical Reprint 2026-27