Classical Mechanics LECTURE 21: SYSTEMS OF PARTICLES AND MOMENT OF INERTIA Prof. N. Harnew University of Oxford HT 2017 1 OUTLINE : 21. SYSTEMS OF PARTICLES AND MOMENT OF INERTIA 21.1 NII for system of particles - translation motion 21.1.1 Kinetic energy and the CM 21.2 NII for system of particles - rotational motion 21.2.1 Angular momentum and the CM 21.3 Introduction to Moment of Inertia 21.3.1 Extend the example : J not parallel to ! 21.3.2 Moment of inertia : mass not distributed in a plane 21.3.3 Generalize for rigid bodies 2 21.1 NII for system of particles - translation motion Reminder from MT lectures: d2 ext int I (r ) = F + F Force on particle i: mi dt2 i i i N 2 N N X d X ext X int PN ext I m (r ) = F + F = F i dt2 i i i i i i i i | {z } | {z } | {z } all masses external forces internal forces = zero PN mi ri I rCM = i M PN where M = i mi PN mi r_ i I vCM =r _ CM = i M PN ! PCM = i mi r_i = M vCM ext PN mi ¨ri Fi I aCM = ¨rCM = i M = M 3 21.1.1 Kinetic energy and the CM 1 PN 2 0 I Lab kinetic energy : T = 2 i mi vi ; vi = vi + vCM 0 where vi is velocity of particle i in the CM 1 P 02 1 P 2 P 0 I T = 2 i mi vi + 2 i mi vCM + i mi vi · vCM P 0 P 0 i mi vi I But m v · v = · M v i i i CM M CM | {z } = 0 0 1 2 I T = T + 2 MvCM Same expression as was derived in MT 4 21.2 NII for system of particles - rotational motion I Angular momentum of particle i about O : J = r × p i i i dJi I Torque of i about O: τ = = r × p_ +r _ × p = r × F i dt i i i i i i | {z } = 0 I Total ang. mom. of system J = PN J = PN r × p i i i i i I Internal forces: Pint int int pair τ (i;j) = ri × Fij + rj × Fji int = (ri − rj ) × Fij int = 0 since (ri − rj ) parallel to Fij Hence total torque PN ext dJ τ = i (ri × Fi ) = dt If external torque = 0; J is const. 5 21.2.1 Angular momentum and the CM 0 0 I Lab to CM : ri = ri + rCM ; vi = vi + vCM 0 0 where ri , vi are position & velocity of particle i wrt the CM I Total ang. mom. of system PN PN 0 0 J = i Ji = i mi (ri + rCM ) × (vi + vCM ) P 0 0 P 0 P 0 P = i mi (ri × vi ) + i mi (ri × vCM ) + i mi (rCM × vi ) + i mi (rCM × vCM ) P 0 X 0 P 0 X 0 But i mi (ri × vCM ) = [ mi ri ] ×vCM ; i mi (rCM × vi ) = rCM × [ mi vi ] i i | {z } | {z } = 0 in CM = 0 in CM 0 J = J + rCM × M vCM I Hence |{z} | {z } J wrt CM J of CM translation 6 What we have learned so far I Newton’s laws relate to rotating systems in the same way that the laws relate to transitional motion. I For any system of particles, the rate of change of internal angular momentum about an origin is equal to the total torque of the external forces about the origin. I The total angular momentum about an origin is the sum of the total angular momentum about the CM plus the angular momentum of the translation of the CM. 7 21.3 Introduction to Moment of Inertia I Take the simplest example of 2 particles rotating in circular motion about a common axis of rotation with angular velocity ! = ! z^ I Definition of ! for circular motion : r_ = ! × r I Total angular momentum of the system of particles about O J = r1 × (m1v1) + r2 × (m2v2) I v1 = ! × r1 v2 = ! × r2 I Since ri ? vi 2 2 J = (m1 r1 + m2 r2 ) ! I J = I! (J is parallel to !) 2 2 I Moment of Inertia I = m1 r1 + m2 r2 P 2 Or more generally I = i [mi ri ] 8 21.3.1 Extend the example : J not parallel to ! Now consider the same system but with rotation tilted wrt rotation axis by an angle φ. Again ! = ! z^ I Total angular momentum of the particles about O : J = r1 × (m1v1) + r2 × (m2v2) 0 I NB. J now points along the z axis I J vector is ? to line of m1 and m2 and defines the principal axis of the MoI mass distribution (see later) d1 I Since v1 = d1 ! ; r1 = sin φ d2 v2 = d2 ! ; r2 = sin φ 2 2 (m1 d1 +m2 d2 ) ! I Then jJj = sin φ ! jJj sin φ !^ = Iz ! . Hence Jz = Iz ! where Iz is calculated about the z^ (i.e. !^) axis. 9 21.3.2 Moment of inertia : mass not distributed in a plane Now take a system of particles rotating in circular motion about a common axis of rotation, all with angular velocity ! (where vi = ! × ri ). I Total angular momentum of the system of particles about O (which is on the axis of rotation) PN PN J = i ri × (mi vi ) = i mi ri vi u^i (not necessarily parallel to ! axis) I As before, resolve the angular momentum about the axis of rotation PN Jz z^ = i mi ri vi sin φi !^ I vi = ! × ri ; vi = !ri sin φi Also sin φ = di ; v = ! d i ri i i PN 2 n Jz !^ = i mi di ! ! Jz = Iz ! (Iz : MoI about rot axis) 10 21.3.3 Generalize for rigid bodies A rigid body may be considered as a collection of infinitesimal point particles whose relative distance does not change during motion. P R I i mi ! dm; where dm = ρdV and ρ is the volume density PN 2 R 2 I Iz = i mi di ! V d ρ dV I This integral gives the moment of inertia about axis of rotation (z axis) 11.