Midterm November 8
Classical Mechanics
Lecture 8
Today's Examples:
a) Potential Energy
b) Mechanical Energy
Mechanics Lecture 8, Slide 1
Summary Unit 8: Potential & Mechanical Energy
Mechanics Lecture 8, Slide 2
Energy Conservation Problems in general
For systems with only conservative forces acting
Emechanical  0
Emechanical is a constant
Emechanical  Ki  Ui  K f  U f  K (t )  U (t )
Mechanics Lecture 8, Slide 3
Energy Conservation Problems in general
 conservation of mechanical energy
can be used to “easily” solve
problems.
hL
Emechanical  K  U 
1
mv (h) 2  U (h) gravity
2
 Define coordinates: where is
U=0?
h0
 Identify important
configurations i.e where
potential is minimized U=0.
h  0  U (h  0) gravity  0
h0
1
m v(h  0) 2  m gL
2
 Problem usually states the
configurations of interest!
 Identify important configurations,
i.e starting point where mass is
motionless K=0
Emechanical  K  U  U (h) gravity
U (h) gravity  m gh  m gL
Emechanical  m gL (for conservative forces)
ALWAYS!
Mechanics Lecture 8, Slide 4
Pendulum Problem
Using Work Formalism
h   L  0
Using Conservation of Mechanical energy
1 2
m v  m gh m gL
2
Wnet  Wgravity  Wtension
Emechanical  K  U 
Wgravity   m g( L)  0
 
Wtension   T  dl 0
1 2
m vf  m gL
2
 v f  2 gL
Wnet  m gL
K  Wnet  m gL 
 v f  2 gL
1
m(v 2f  v02 )
2
h0
Conserve Energy from
initial to final position
Mechanics Lecture 8, Slide 5
Pendulum Problem

Fc  2mgˆj

T  3mgˆj


W  mg  mgˆj
Don’t forget centripetal acceleration
and force…required to2 maintain
circular path. Fc  m v
L
At bottom of path:
v  2 gL
Fc  m
2 gL
 2m g
L
  
Fc  T  W  Tˆj  mgˆj  Fc ˆj
Tension is …”what it has to be!”
2 gL
 T mg
L
2 gL
T  mg m
L
T  m g  2m g  3m g
m
Mechanics Lecture 8, Slide 6
Pendulum Problem
T  3m g
Tmax  3mmax g
mmax
Tmax

3g
Mechanics Lecture 8, Slide 7
Pendulum Problem
Emechanical  K  U 
1 2
m v  m gh  m gL
2
1
L'  L
5
1 2
2
m v  m g( L)  m gL
2
5
1 2 3
m v  m gL
2
5
6
v
gL
5
Kinetic energy of mass prior
to string hitting peg is
conserved.
Set h=0 to be at bottom
equilibrium position
Mechanics Lecture 8, Slide 8
Pendulum Problem


W  mg  mgˆj

Fc  6mgˆj
Radius for centripetal
acceleration has been
shortened to L/5 !
Fc  m
v
v2
1 
 L
5 
6
gL
5

T  5mgˆj

  


Fc   Fc j  T  W  Tj  m gj


 Fc j   T  m g j
 Fc   T  m g
6
gL
v
5
Fc  m
m
 6m g
1 
1 
 L
 L
5 
5 
2
6m g  T  m g
T  6m g  m g  5m g
Mechanics Lecture 8, Slide 9
Loop the Loop
v2
Fc  m
L
To stay on loop, the normal
force, N, must be greater
than zero.
vmin  gR

v2 ˆ
Fc  m
j
R


W  mg  mgˆj

N   Nˆj

v2 ˆ  
Fc  m
j  N  W   Nˆj  m gˆj
R
v2 ˆ
m
j  ( N  m g) ˆj
R
v2
N  m  mg 0
R
2
vmin
m
 mg
R
Mechanics Lecture 8, Slide 10
1 2
m vmin  m g(2 R)  m gh
2
1
m gR m g(2 R)  m gh
2
1
5
 h  R  2R  R
2
2
 Mass must start higher
than top of loop
Mechanics Lecture 8, Slide 11
1 2
m v  m g(0)  m gh
2
5
v  2 gh  2 g R  5 gR
2
1 2
m v  m g( R )  m gh
2
1 2
5
m v  m g( R )  m gR
2
2
3
v  2 gh  2 g R  3 gR
2
Mechanics Lecture 8, Slide 12
v  3 gR
K  U spring 
1 2 1
m v  k (x) 2  0
2
2
m v2
m
x 
v
k
k
x 
3m gR
k
Mechanics Lecture 8, Slide 13
1 2
5
m v  m g(2 R)  m gh  m g R
2
2
1 2 1
m v  m gR
2
2
vstart  gR
Mechanics Lecture 8, Slide 14
Wnormal
 
  N  dl  0
Mechanics Lecture 8, Slide 15
Mechanics Lecture 8, Slide 16
Gravity and Springs…Oh my!
K  U  K  U gravity  U spring  mgh Emechanical
h  d  x sin 
1
1
Emechanical  m0 2  0  kx 2  mg d  x sin 
2
2
1 2
kx  mg d  x sin   0
2
Conservation of Energy
Coordinate System
Initial configuration
Final configuration
Solve Quadratic Equation for x!
1 2
kx  m gxsin   m gdsin   0
2
1
a  k ; b  m g sin  ; c  m gdsin 
2
 b  b 2  4ac
x
2a
x
m g sin  
m gsin  2  2km gdsin 
k
Mechanics Lecture 8, Slide 17
Pendulum 2
Given speed and tension at
certain point in pendulum
trajectory can use
conservation of mechanical
energy to solve for L and m….
Emechanical  K  U 
h0
1 2
m v  m gh  m gL
2
1 2
m vf  m gL
2
1 2
m vf v 2
f
2
L

mg
2g
Emechanical  K  U 
1
m v(h) 2  U (h) gravity
2
U (h) gravity  m gh
v2
T  mg m
L
T  m g  2m g  3m g
T
m
3g
Mechanics Lecture 8, Slide 18
Pendulum 2
T  3m g
Tmax  3mmax g
Mechanics Lecture 8, Slide 19
Pendulum 2
Emechanical  K  U 
1 2
m v  m gh  m gL
2
1
L'  L
5
1 2
1
m v  m g( L)  m gL
2
5
1 2 4
m v  m gL
2
5
8
v
gL
5
Mechanics Lecture 8, Slide 20
Pendulum 2

Fc  8mgiˆ

T  8mgiˆ


W  mg  mgˆj
v2
T m
1 
 L
5 
8
gL
5
8
gL
5
T m
1 
 L
5 
T  8m g
v
Mechanics Lecture 8, Slide 21
Classical Mechanics
Lecture 9
Today's Concepts:
a) Energy and Friction
b) Potential energy & force
Mechanics Lecture 9, Slide 22
Main Points
Mechanics Lecture 8, Slide 23
Main Points
Mechanics Lecture 8, Slide 24
Macroscopic Work done by Friction
Mechanics Lecture 9, Slide 25
Macroscopic Work done by Friction
Mechanics Lecture 9, Slide 26
Macroscopic Work:
This is not a new idea – it’s the same “work” you are used to.
 
W   F dl
b
Applied to big (i.e. macroscopic) objects
rather than point particles (picky detail)
a
We call it “macroscopic” to distinguish it from “microscopic”.
Mechanics Lecture 9, Slide 27
Mechanics Lecture 9, Slide 28
Mechanics Lecture 9, Slide 29
f
f
Mechanics Lecture 9, Slide 30
“Heat” is just the kinetic energy of the atoms!
Mechanics Lecture 9, Slide 31
Incline with Friction:Newton’s law and kinematics
Mechanics Lecture 8, Slide 32
Incline with Friction: Work-Kinetic Energy
Mechanics Lecture 8, Slide 33
Checkpoint
A.
B.
C.
D.
A block of mass m, initially held at rest on a frictionless ramp a vertical
distance H above the floor, slides down the ramp and onto a floor
where friction causes it to stop a distance r from the bottom of the
ramp. The coefficient of kinetic friction between the box and the floor
is mk. What is the macroscopic work done on the block by friction
during this process?
B) –mgH
A) mgH
C) mk mgD
D) 0
0%
0%
0%
0%
m
H
D
Mechanics Lecture 9,
8, Slide 34
Work by Friction :Wfriction<0
K  Wtot  Wgravity  Wfriction
Motionless box released on
frictionless incline which
then slides on horizontal
surface with friction to a
stop.  K  0
N1
0  Wgravity  Wfriction
0  mgH Wfriction
must be negative
m
N2
H
mg
m mg
mg
Mechanics Lecture 9, Slide 35
CheckPoint
What is the macroscopic work done on
the block by friction during this process?
B) –mgH
A) mgH
C) mk mgD
D) 0
B) All of the potential energy goes to kinetic as it slides
down the ramp, then the friction does negative work to
slow the box to stop
C) Since the floor has friction, the work done by the block
by friction is the normal force times the coefficient of
kinetic friction times the distance.
m
H
D
Mechanics Lecture 9, Slide 36
Checkpoint
A.
B.
C.
D.
A block of mass m, initially held at rest on a frictionless ramp a vertical
distance H above the floor, slides down the ramp and onto a floor
where friction causes it to stop a distance D from the bottom of the
ramp. The coefficient of kinetic friction between the box and the floor
is mk. What is the total macroscopic work done on the block by all forces
during this process?
B) –mgH
A) mgH
C) mk mgD
D) 0
m
0%
0%
0%
0%
K  Wtot
H
D
Mechanics Lecture 9,
8, Slide 37
CheckPoint
What is the total macroscopic work done on
the block by all forces during this process?
B) –mgH
A) mgH
C) mk mgD
D) 0
B) work= change in potential energy
C) The only work being done on the object is by the friction
force.
D) total work is equal to the change in kinetic energy. since
the box starts and ends at rest, the change in kinetic energy
is zero.
m
K  Wtot
H
D
Mechanics Lecture 9, Slide 38
Mechanics Lecture 8, Slide 39
Force from Potential Energy:1D
start here
Mechanics Lecture 8, Slide 40
Force from Potential Energy
Mechanics Lecture 8, Slide 41
Potential Energy vs. Force
dU ( x)
F ( x)  
dx
Mechanics Lecture 9, Slide 42
Potential Energy vs. Force
1 2
U ( x)   kx
2
dU ( x)
F ( x)  
dx
 kx
Mechanics Lecture 9, Slide 43
Potential Energy vs. Force
Mechanics Lecture 9, Slide 44
Potential Energy vs. Force
Mechanics Lecture 9, Slide 45
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
D.
Where is the force on the object zero?
A) (a)
B) (b)
C) (c)
D) (d)
25%
25%
25%
25%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 46
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
D.
Where is the force on the object in the +x direction?
A) To the left of (b)
B) To the right of (b)
C) Nowhere
25%
25%
25%
25%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 47
Clicker Question
A.
B.
C.
Suppose the potential energy of some object U as a function of x
looks like the plot shown below.
Where is the force on the object biggest in the –x direction?
A) (a)
B) (b)
C) (c)
D) (d)
25%
25%
25%
D.
25%
U(x)
x
(a)
(b)
(c)
(d)
dU ( x)
F ( x)  
dx
Mechanics Lecture 8, Slide 48
Equilibrium
Mechanics Lecture 8, Slide 49
Equilibrium points
Mechanics Lecture 8, Slide 50
Equilibrium points
Mechanics Lecture 8, Slide 51
Equilibrium points
Mechanics Lecture 8, Slide 52