[ ] ( ) L = L = r p. We can deal with this more conveniently by writing: so that L becomes: α α α. α α N N N N

Transcription

1 Lecture 5: Angular Momentum and Energy of a System The total angular momentum is given by the sum of the angular momenta of all particles in the system: We can deal with this more conveniently by writing: = 1 = 1 so that L becomes: ([ ] ) L = r + R p = [ ] ( ) L = L = r p = 1 = 1 r = r + R ( r R m r R ) = + + = r m r + r m R + R m r + R m R = 1 = 1 = 1 = 1

2 To simplify this, focus on the middle two terms: r m R + R m r = 1 = 1 The expression m r appears in both terms but we can rewrite this as: d = m r R + m = 1 dt R r = 1 = 1 m = 1 = 1 ( ) r = m r R = m r R = 1 = 1 = MR R M = m 0

3 So the two middle terms in the expression for L disappear, and we re left with: Angular momentum about the center of mass L = r m r + R MR = 1 = 1 = r p + R P The angular momentum about any origin is the sum of angular momentum of the system about the center of mass plus the angular momentum of the center of mass about the origin Angular momentum of the center of mass

4 Change in Angular Momentum The rate of change of L is given by: L r F = i.e., the sum of the torques on each particle Breaking this up into the internal and external forces gives: e L = r F + f = r F + r f e, The second term can be written as: r f = r f + r f, < > = r f + r f < <

5 From ewton s Third Law, we know that f = f, so:, < < The vector r r is on the line connecting particles and If the strong form of ewton s Third Law applies, f also lies along this line, which means that: This is turn means that: r f = r f r f < ( ) = r r f < ( ) = 0 r r f L r F = = = e e e

6 The rate of change of the system s angular momentum is equal to the total external torque on the system Equivalently, we could say that: The sum of all of the internal torques on the system is zero Keep in mind that both of these conclusions are true only if all the internal forces obey the strong form of ewton s Third Law e.g., the derivation doesn t hold for a system with internal magnetic forces

7 Energy of the System We now want to examine how the energy of the system changes as the particles in the system move To do this, consider the work done in moving all the particles from a set of initial positions (call this configuration 1) to a set of final positions (call this configuration ): W 1 But way back when we proved the work-energy theorem for a single particle, we proved that: F = d r 1 1 F dr = d mv

8 Applying this relation to every term in the sum gives: W = d m v = m v m v = T T = T 1 We again write r in terms of the center of mass: r = r + R r = r + R which means the kinetic energy is: 1 1 = = r r T m v m 1 = m + + ( r R ) ( r R )

9 1 1 T = m r r + m R + m R R r 1 1 d = m r r + m V + R m r dt 1 1 = m v + MVC M + R d ( 0) dt Energy due to motion relative to the center of mass Energy due to motion of the center of mass We can always write the kinetic energy as the sum of these two terms this will come in handy for treating rotating objects!

10 Conservation of Energy We return to considering the work done on the system Again, we break up the force into internal and external components: W 1 = e F + f dr 1 e F dr f dr 1, 1 = + ow we assume that all the forces (both internal and external) are conservative though they may be of different forms This isn t that big of a restriction even frictional forces are conservative, if we consider the forces between individual atoms

11 Call the potential associated with the internal forces U and the one associated with the external forces U The potential depends only on the relative positions of particles and ow write each term in the work in terms of these potentials: U 1 e F r f r 1, 1 W = d + d e F dr = ( U ) dr 1 1 ( U, U,1 ) = i.e., the work done by external forces equals the total change in potential energy due to those forces ot too surprising!

12 The work due to internal forces can be written as: int W = f dr = f dr + f dr, 1 < 1 < 1 = f dr + f dr < 1 < 1 Applying ewton s Third Law (weak form is OK!) gives: int ( ) ( ) r ( ) W = f dr f dr = f dr dr < 1 < 1 < 1 = f d r r f dr < 1 < 1 = U d = du < 1 < 1 Last step is true since depends only on relative positions of and U

13 This means that: ( ) W = U U int,1, < We can now identify the total potential energy of the system as: External potential U = U + U And the work done on the system is: W = U U we also showed earlier that: < 1 1 W1 = T T1 Internal potential

14 Therefore, it must be true that: U U = T T 1 1 T + U = T + U 1 1 E = 1 Energy is conserved for any system on which only conservative forces act This result may seem kind of obvious, but going through the derivation allowed us to determine the form of the internal potential ote that for rigid systems (ones in which the particles can t move relative to one another) the internal potential is a constant, which may be taken to be zero E