Linear Motion of a System of Particles the Linear Momentum of a System Subject to No Net Ex- Ternal Force Is Conserved. F ˙

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Linear Motion of a System of Particles The linear momentum of a system subject to no net external force is conserved.

Fext P˙

P MR˙ Fext MR¨

Centre of Mass ∑N m r 1 N

¡ i i

i 1

R ∑miri ¢

∑N m M ¡

i ¡ 1 i i 1

P MR˙ Fext MR¨ £ ρ ri R i

Kinetic Energy

1 £ T MR˙ 2 T 2 CM 2

Angular Motion of a System of Particles

¤ Angular equation of motion for each particle i is

d ¦

¥ §©¨ r ¥ F r p i i dt i i

¤ Total angular momentum of the system and total torque are:

N N

τ ¥

L ∑ ri ¥ pi and ∑ ri Fi ¡ i ¡ 1 i 1

¤ Split total force into external and internal parts, same for total torque:

N N £

τ ext ¥

∑ ri ¥ Fi ∑ri ∑Fi j

¡ ¡ ¡

i 1 i 1 j i £

τext τint ¨ 3

¤ Use Newton’s third law (like linear motion case):

£ £ £

τint §

r1 ¥ F12 F13 F1N

£ £ £ £ £ §

r2 ¥ F21 F23 F2N

¦ ¦

§©¨ r1 r2 §¥ F12 other pairs

¤ Assume internal forces act along the lines joining

particle pairs, hence ri r j §¥ Fi j 0: i.e.,

τint

0 for central internal forces ¨

¤ Obtain:

N d N

ext

¥ ¢ ∑ r ¥ F ∑ r p

i i dt i i ¡ i ¡ 1 i 1 i.e.,

τext ˙

L ¨

¤ Result applies when using coordinates in inertial frame (where Newton’s laws apply).

¤ Used both Newton’s third law and the condition that the forces between particles were central. 4

Angular Motion About the Centre of Mass

¤ Total angular momentum using the centre of mass (CM) coordinates:

N N

¦ ¦

£ £

ρ ˙ ρ˙

§¥ §

L ∑ri ¥ mir˙i ∑ R i mi R i ¡ i ¡ 1 i 1

N N £

˙ ρ˙ ¥

∑R ¥ miR ∑R mi i ¡ i ¡ 1 i 1

N N £

£ ρ ˙ ρ ρ˙

¥ ¨

∑ i ¥ miR ∑ i mi i ¡

i ¡ 1 i 1 ¤ ρ Second and third terms on the RHS vanish (∑mi i

ρ 0 and ∑mi ˙ i 0 from CM deﬁnition):

N £

˙ ρ ρ˙

¥ ¢ L R ¥ MR ∑ i mi i

i ¡ 1

write as, £

˙ ¨ L R ¥ MR LCM

1. First term from CM motion relative to origin (orbital angular momentum): frame dipendent.

2. Second term from angular motion relative to CM (e.g., spinning planet orbiting around the Sun): same in all (inertial) frames and intrin- sic (or spin) angular momentum. 5

¤ Take time derivative of the last equation:

dLCM dL

¨ τext ext

¥ R ¥ MR R F dt dt N N

ext ext

∑ri ¥ Fi ∑ R Fi ¡ i ¡ 1 i 1

N ¦

ext ∑ ri R §¥ Fi

i ¡ 1

ρ ext τext ¨ ∑ i ¥ Fi CM

i ¡ 1

¤ Find:

τext ˙ τext ˙ L and CM LCM ¨

¤ Can take moments either about origin of inertial frame or CM.

¤ Rationale:

The angular momentum of a system subject to no external torque is constant.