**Unformatted text preview: **PHYS 172: Modern Mechanics Lecture 11 – The Energy Principle Fall 2011 1 Read 6.8 – 6.14 TODAY
• Multiparticle Systems and Potential Energy
• Relationship of Force and Potential Energy
• Energy Graphs 2 ANNOUNCEMENTS
• October Break – 10/10, 10/11!
• No lectures that week, no labs.
• You do have recitation!
• Exam 2 is Tuesday, Oct. 18, 8-10 PM! 3 Last Time: Single Particle System
Energy principle (single particle system): !Esingle particle system = W + Q
where energy is Esingle particle system = ! mc 2 =
and work is Won particle = 0 for now mc 2
1" v / c
2 2 !
!
= Fnet on particle i! r
4 How do we generalize these results to multiparticle systems? Example: Energy in 2-Particle System
!
F2,surr
system !
f1,2
1 !
F1,surr !
f 2,1 !Esystem = W + Q
2 =0 Let s find the energy change
of each particle: !
!!
!
! E1 = f1,2 i! r1 + F1,surr i! r1 = W1,internal + W1,surr
!
!!
!
! E2 = f 2,1 i! r2 + F2,surr i! r2 = W2,internal + W2,surr We re counting the work done by internal forces.
5 Example: Energy in 2-Particle System
!
F2,surr !
f1,2 1
!
F1,surr !
f 2,1 2 Thus ! (E1 + E 2 ) = Wint + Wsurr + Q
where Wint = W1,internal + W2,internal
and Wsurr = W1,surr + W2,surr Put system on left side, surroundings on right side: ! (E1 + E 2 ) " Wint = Wsurr + Q
Now define the change in potential energy as ΔU ≡ – Wint : ! (E1 + E 2 ) + !U = Wsurr + Q 6 Potential Energy (in 2-Particle System)
!
F2,surr !
f1,2 !
f 2,1 2 1
!
F1,surr !
!!
!
!U = "Wint = " f1,2 i! r1 " f 2,1 i! r2
The potential energy U represents a sum of
interaction energies between all pairs of
particles inside the system. NOTE: U is defined to take into account both terms above. !
!
Question: If f1,2 + f 2,1 = 0 , why isn t ΔU = 0 here?
Answer: Usually, !
! r1 " !.r2 ΔU is related to a system changing shape. 7 Energy of a Multiparticle System
sum of single
sum of interaction
Energy of system =
+
particle energies
energies of all pairs Esys = (m1c 2 + m2 c 2 + …)+ (K1 + K 2 + …) + (U12 + …) We now write !Esys = Wsurr + Q Now W is about external forces only
(internal forces show up in U). 8 Energy of a Multiparticle System !Esys = Wsurr + Q
Wsurr is about external forces only Add a mass to the spring.
How much work does the
gravitational force do on the mass? What if the mass oscillates before
coming to equilibrium?
9 Connection: Force and Potential Energy
!
!
!!
!!
!
!!
!
= "W!tU = "1Wi!U1 " f"22,1ii! r12" f1,,2 i! r1 " f 2,1 i! r2
!r 1 !r
2
in = " f , 2 int = = f ,Wint = " 2 1
f 1 ,2 = − f 2 ,1
!
!! !!! !
!
!Equal and opposite
!
F
= " f ,1 ( ) r " 2, i !
= " f 2,1 i(! r2 "2!ir1 !=2" f!1r1() r2 " ! r1 )
!
2
!!
!!!!
!!
=
i! e
= " f 2,1 i! = " f 2,1 i! r " f 2,1whr re r = r2 ! r1
r
2 ,surr !
f1,2 !
f 2,1 1 The combination is independent
of coordinate system. !
F1,surr !
!U
Thus f 2,1 = " !
!r dU
# fr = "
dr
10 Connection: Force and Potential Energy
!
F2,surr !
!U
hus f 2,1 = " !
2
!r
!
f1,2 !
f 2,1 1 dU
# fr = "
dr
For Gravity: !
F1,surr !U
dm1m2
U
" ! # f ="
1 = Gravitation: Frr = -G
!r
drr 2 ! m1m2
U g = "G
r To see this: 11 Gravitational Potential Energy
Ug=0! m1m2
U g = !G
r
Yes, it really is negative! Note: U g = 0 at r = !
r=0! 12 Example: Planet and Star System: planet+star
System: planet+star
!Esys = Wsurr + Q = 0
!Esys = Wsurr + Q = 0
! " E particles + U system $ = 0 &
# ! "E
+ U%ystem $ = 0 &
s
# 2 particles
%2
! " mstar c + K star + mplanet c + K planet + U system $ = 0
2
# ! " m c2 + K + m
c + K planet + U%ystem $ = 0
star
planet
s
# star
%
Rest energiethese is constant !K star ' 0 &
s constant;
Each ne
these &
Rest eof rgies constant; !K staSo' 0 together
r
m
K planet,f + U system, f = K planet , i + U system,ust be constant
K planet,f + U system, f = K planet , i + Uisystem,i
Thus, K planet + U system = const
Thus, K planet + U system = const
13 Rest energies constant;Planet 0 &
Example: !K star ' and Star
K planet,f + U system, f = K planet , i + U system,i Energy Thus, K planet + U system = const K
K+U
r
U 14 Application: Escape Speed
What does it take to launch a rocket so it leaves the Earth's gravitational well? Minimal condition for escape: K +U = 0 Assume: kinetic energy of a planet is negligible, v<<c
2
mvesc ⎡
Mm ⎤
Ki + U i =
+ ⎢−G
⎥=0
2
R⎦
⎣
2
mvesc
Mm
=G
2
R vesc 2GM
=
R
15 Application: Escape Speed
What does it take to launch a rocket so it leaves the Earth's gravitational well? Bound state: K +U < 0 Unbound state: K +U ≥ 0
16 Gravitational U Near Earth's Surface
Mm
U = −G
r Are these
the same? They are the same near the Earth's Surface. RE RE + h Mm
∆U = − G
−
RE + h
U = mgh
Mm
−G
RE Mm
1
= −G
RE 1 + h/RE
1
≈ 1 − x + ...
1+x
Taylor Expansion ≈ −G Mm
1 − h/RE − 1
RE
−1
GM
=m
h ≡ mgh = ∆U
2
RE
g= 2
GM/RE 17 WHAT WE DID TODAY
• Multiparticle Systems and Potential Energy
• Relationship of Force and Potential Energy
• Energy Graphs 18 ...

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