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 ω 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 ω = ω zˆ

I Deﬁnition of ω for circular motion : r˙ = ω × r

I Total angular momentum of the system of particles about O

J = r1 × (m1v1) + r2 × (m2v2)

I v1 = ω × r1

v2 = ω × r2

I Since ri ⊥ vi 2 2 J = (m1 r1 + m2 r2 ) ω

I J = Iω (J is parallel to ω) 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 ω Now consider the same system but with rotation tilted wrt rotation axis by an angle φ. Again ω = ω zˆ

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 ω ; r1 = sin φ

d2 v2 = d2 ω ; r2 = sin φ 2 2 (m1 d1 +m2 d2 ) ω I Then |J| = sin φ → |J| sin φ ωˆ = Iz ω . Hence Jz = Iz ω where Iz is calculated about the zˆ (i.e. ωˆ) 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 ω (where vi = ω × 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 uˆi (not necessarily parallel to ω axis)

I As before, resolve the angular momentum about the axis of rotation PN Jz zˆ = i mi ri vi sin φi ωˆ

I vi = ω × ri ; vi = ωri sin φi Also sin φ = di ; v = ω d i ri i i PN 2 n Jz ωˆ = i mi di ω → Jz = Iz ω (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)

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