Download Physics 207: Lecture 2 Notes

Lecture 15
Goals:
• Chapter 11
 Employ conservative and non-conservative forces
 Use the concept of power (i.e., energy per time)
• Chapter 12
 Extend the particle model to rigid-bodies
 Understand the equilibrium of an extended object.
 Understand rotation about a fixed axis.
 Employ “conservation of angular momentum” concept
Assignment:
 HW7 due March 25th
 For Thursday: Read Chapter 12, Sections 7-11
do not concern yourself with the integration process in
regards to “center of mass” or “moment of inertia”
Physics 207: Lecture 15, Pg 1
More Work: “2-D” Example
(constant force)
 An angled force, F = 10 N, pushes a box across a
frictionless floor for a distance x = 5 m and y = 0 m
Start
F
Finish
q = -45°
Fx
x
 (Net)
Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm = 35 J
 Notice that work reflects energy transfer
Physics 207: Lecture 15, Pg 2
Work in 3D….(assigning U to be external to the system)
 x, y and z with constant F:

Fy ( y f  yi )  Fy y 
1
2
mv xf
2
1
2
mv yf
2
Fx ( x f  xi )  Fx x 
Fz ( z f  zi )  Fz z 
1
2

Fx x  Fy y  Fz z  
with v
2
2
 vx

2
vy

1
2
2
mv zf
2
mv f


1
2
1
2
mv xi
2
1
2
mv yi
2
1
2
mv zi
2
2
mvi
 K
2
vz
Physics 207: Lecture 15, Pg 4
A tool: Scalar Product (or Dot Product)
A · B ≡ |A| |B| cos(q)
 Useful for performing projections.
A  î = Ax
A
îî=1
îj=0
q Ay
î Ax
 Calculation can be made in terms of
components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos q
You choose the way that works best for you!
Physics 207: Lecture 15, Pg 5
Scalar Product (or Dot Product)
Compare:
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Redefine A  F (force), B  r (displacement)
Notice:
F  r = (Fx )(x) + (Fy )(z ) + (Fz )(z)
So here
F  r = W
More generally a Force acting over a Distance does Work
Physics 207: Lecture 15, Pg 6
Definition of Work, The basics
Ingredients: Force ( F ), displacement (  r )
Work, W, of a constant force F
acts through a displacement  r :
W = F · r
F
q
r
(Work is a scalar)
“Scalar or Dot Product”
 
F  dr
If we know the angle the force makes with the
path, the dot product gives us F cos q and r
If the path is curved

r
f
and

W

dW

 
F  dr
at each point

 F  dr
ri
Physics 207: Lecture 15, Pg 7
Exercise
Work in the presence of friction and non-contact forces
 A box is pulled up a rough (m > 0) incline by a rope-pulley-
weight arrangement as shown below.
 How many forces (including non-contact ones) are
doing work on the box ?
 Of these which are positive and which are negative?
 State the system (here, just the box)
 Use a Free Body Diagram
 Compare force and path
v
A. 2
B. 3
C. 4
D. 5
Physics 207: Lecture 15, Pg 10
Exercise
Work in the presence of friction and non-contact forces
 A box is pulled up a rough (m > 0) incline by a rope-pulley-
weight arrangement as shown below.
 How many forces are doing work on the box ?
 And which are positive (T) and which are negative (f, mg)?
(For mg only the component along the surface is relevant)
 Use a Free Body Diagram
(A) 2
v
T
(B) 3 is correct
N
(C) 4
mg
f
(D) 5
Physics 207: Lecture 15, Pg 11
Work and Varying Forces (1D)
Area = Fx x
F is increasing
Here W = F · r
becomes dW = F dx
xf
 Consider a varying force F(x)
Fx
W
  F ( x ) dx
x
x
xi
Finish
Start
F
F
q = 0°
x
Work has units of energy and is a scalar!
Physics 207: Lecture 15, Pg 12
•
Example: Hooke’s Law Spring (xi equilibrium)
How much will the spring compress (i.e. x = xf - xi) to bring
the box to a stop (i.e., v = 0 ) if the object is moving initially at a
constant velocity (vi) on frictionless surface as shown below
and xi is the equilibrium position of the spring?
xf
Wbox
ti vi
xi
m
Wbox
xf
  - k ( x  x ) dx
i
xi
spring at an equilibrium position
x
f
Wbox  - k ( x  xi ) |
xi
x
1
2
V=0
t
  F ( x ) dx
F
m
spring compressed
Wbox  - 2 k ( x f
1
2
 xi )2  12 k 02  K
- k x  m0  mv
1
2
2
1
2
2
1
2
Physics 207: Lecture 15, Pg 13
2
i
Work signs
ti vi
Notice that the spring force is
opposite the displacement
m
spring at an equilibrium position
x
For the mass m, work is negative
V=0
t
F
For the spring, work is positive
m
spring compressed
They are opposite, and equal (spring is conservative)
Physics 207: Lecture 15, Pg 15
Conservative Forces & Potential Energy
 For any conservative force F we can define a potential energy
function U in the following way:
W =
 F ·dr = - U
The work done by a conservative force is equal and opposite to
the change in the potential energy function. r
U
f
 This can be written as:
U = Uf - Ui = - W = -
f
rf
rF • dr
i
ri
Ui
Physics 207: Lecture 15, Pg 16
Conservative Forces and Potential Energy
 So we can also describe work and changes in
potential energy (for conservative forces)
U = - W
 Recalling (if 1D)
W = Fx x
 Combining these two,
U = - Fx x
 Letting small quantities go to infinitesimals,
dU = - Fx dx
 Or,
Fx = -dU / dx
Physics 207: Lecture 15, Pg 17
Exercise
Work Done by Gravity
 An frictionless track is at an angle of 30° with respect to the
horizontal. A cart (mass 1 kg) is released from rest. It slides
1 meter downwards along the track bounces and then
slides upwards to its original position.
 How much total work is done by gravity on the cart when it
reaches its original position? (g = 10 m/s2)
30°
(A) 5 J
(B) 10 J
(C) 20 J
h = 1 m sin 30°
= 0.5 m
(D) 0 J
Physics 207: Lecture 15, Pg 18
Non-conservative Forces :
 If the work done does not depend on the path taken, the
force involved is said to be conservative.
 If the work done does depend on the path taken, the force
involved is said to be non-conservative.
 An example of a non-conservative force is friction:
 Pushing a box across the floor, the amount of work that is
done by friction depends on the path taken.
and work done is proportional to the length of the path !
Physics 207: Lecture 15, Pg 23
A Non-Conservative Force, Friction
 Looking down on an air-hockey table with no air
flowing (m > 0).
 Now compare two paths in which the puck starts
out with the same speed (Ki path 1 = Ki path 2) .
Path 2
Path 1
Physics 207: Lecture 15, Pg 24
A Non-Conservative Force
Path 2
Path 1
Since path2 distance >path1 distance the puck will be traveling
slower at the end of path 2.
Work done by a non-conservative force irreversibly removes
energy out of the “system”.
Here WNC = Efinal - Einitial < 0  and reflects Ethermal
Physics 207: Lecture 15, Pg 25
Work & Power:
 Two cars go up a hill, a Corvette and a ordinary Chevy




Malibu. Both cars have the same mass.
Assuming identical friction, both engines do the same amount
of work to get up the hill.
Are the cars essentially the same ?
NO. The Corvette can get up the hill quicker
It has a more powerful engine.
Physics 207: Lecture 15, Pg 26
Work & Power:
 Power is the rate at which work is done.
 Average Power is,
 Instantaneous Power is,
W
P 
t
dW
P
dt
 If force constant, W= F x = F (v0 t + ½ at2)
and
P = W / t = F (v0 + at)
Physics 207: Lecture 15, Pg 27
Work & Power:
 Power is the rate at which work is done.
Average
Power:
Instantaneous
Power:
W
P
t
dW
P
dt
Units (SI) are
Watts (W):
1 W = 1 J / 1s
Example:
 A person of mass 80.0 kg walks up to 3rd floor (12.0m). If
he/she climbs in 20.0 sec what is the average power used.
 Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W
 P = 470. W
Physics 207: Lecture 15, Pg 28
Exercise
Work & Power
 Starting from rest, a car drives up a hill at constant
acceleration and then suddenly stops at the top.
 The instantaneous power delivered by the engine during this
drive looks like which of the following,
A. Top
time
B. Middle
C. Bottom
time
time
Physics 207: Lecture 15, Pg 29
Chap. 12: Rotational Dynamics
 Up until now rotation has been only in terms of circular motion
with ac = v2 / R and | aT | = d| v | / dt
 Rotation is common in the world around us.
 Many ideas developed for translational motion are transferable.
Physics 207: Lecture 15, Pg 30
Conservation of angular momentum has consequences
How does one describe rotation (magnitude and direction)?
Physics 207: Lecture 15, Pg 31
Rotational Dynamics: A child’s toy, a physics
playground or a student’s nightmare
 A merry-go-round is spinning and we run and jump on
it. What does it do?
 We are standing on the rim and our “friends” spin it
faster. What happens to us?
 We are standing on the rim a walk towards the center.
Does anything change?
Physics 207: Lecture 15, Pg 32
Overview (with comparison to 1-D kinematics)
Angular
Linear
  constant
a  constant
  0  t
q  q0  0 t  t
1
2
v  v0  at
2
x  x0  v0 t  12 at 2
And for a point at a distance R from the rotation axis:
x = R qv =  R
aT =  R
Here aT refers to tangential acceleration
Physics 207: Lecture 15, Pg 35
Lecture 15
Assignment:
 HW7 due March 25th
 For Thursday: Read Chapter 12, Sections 7-11
do not concern yourself with the integration process in
regards to “center of mass” or “moment of inertia”
Physics 207: Lecture 15, Pg 37