Lecture 11
Goals:
Chapter 8:
Employ rotational motion models with friction or in free fall
Chapter 9: Momentum & Impulse
 Understand what momentum is and how it relates to forces
 Employ momentum conservation principles
 In problems with 1D and 2D Collisions
 In problems having an impulse (Force vs. time)
Assignment:


Read through Chapter 10, 1st four section
MP HW6, due Wednesday 3/3
Physics 207: Lecture 11, Pg 1
Zero Gravity Ride

One last reprisal of the
Free Body Diagram

Remember
1 Normal Force is  to the surface
2 Friction is parallel to the contact surface
3 Radial (aka, centripetal) acceleration requires a net force
Physics 207: Lecture 11, Pg 2
Zero Gravity Ride
A rider in a horizontal “0 gravity ride” finds
herself stuck with her back to the wall.
Which diagram correctly shows the forces
acting on her?
Physics 207: Lecture 11, Pg 3
Banked Curves
In the previous car scenario, we drew the following free body
diagram for a race car going around a curve at constant
speed on a flat track.
Because the acceleration is radial (i.e., velocity changes in
direction only) we need to modify our view of friction.
n
Ff
mg
So, what differs on a banked curve?
Physics 207: Lecture 11, Pg 4
Banked Curves (high speed)
1 Draw a Free Body Diagram for a banked curve.
2 Use a rotated x-y coordinates
3 Resolve into components parallel and
N
perpendicular to bank
mar
Ff
q
y
mg
Physics 207: Lecture 11, Pg 5
x
Banked Curves, high speed
y
4 Apply Newton’s 1st and 2nd Laws
x
N
mar
mar sin q
q
mar cos q
Ff
S Fx = -mar cos q = - Ff - mg sin q
S Fy =
mar sin q = 0 - mg cos q + N
q
mg cos q
mg sin q
Friction model  Ff ≤ m N (maximum speed when equal)
Physics 207: Lecture 11, Pg 7
Banked Curves, low speed
y
4 Apply Newton’s 1st and 2nd Laws
x
N
mar
mar sin q
q
mar cos q
S Fx = -mar cos q = + Ff - mg sin q
S Fy =
Ff
mar sin q = 0 - mg cos q + N
q
mg cos q
mg sin q
Friction model  Ff ≤ m N (minimum speed when equal but
not less than zero!)
Physics 207: Lecture 11, Pg 8
vmax
Banked Curves, constant speed
= (gr)½ [ (m + tan q) / (1 - m tan q)] ½
vmin = (gr)½ [ (tan q - m) / (1 + m tan q)] ½
Dry pavement
“Typical” values of r = 30 m, g = 9.8 m/s2, m = 0.8, q = 20°
vmax = 20 m/s (45 mph)
vmin =
0 m/s (as long as m > 0.36 )
Wet Ice
“Typical values” of r = 30 m, g = 9.8 m/s2, m = 0.1, q =20°
vmax = 12 m/s (25 mph)
vmin =
9 m/s
(Ideal speed is when frictional force goes to zero)
Physics 207: Lecture 11, Pg 9
Navigating a hill
Knight concept exercise: A car is rolling over the top of a hill
at speed v. At this instant,
A. n > w.
B. n = w.
C. n < w.
D. We can’t tell about n without
knowing v.
At what speed does the car lose
contact?
This occurs when the normal force goes to zero or, equivalently, when
all the weight is used to achieve circular motion.
Fc = mg = m v2 /r  v = (gr)½ (just like an object in orbit)
Note this approach can also be used to estimate the maximum walking
speed.
Physics 207: Lecture 11, Pg 11
Orbiting satellites vT = (gr)½
Physics 207: Lecture 11, Pg 12
Locomotion: how fast can a biped walk?
Physics 207: Lecture 11, Pg 13
How fast can a biped walk?
What about weight?
(a) A heavier person of equal
height and proportions
can walk faster than a
lighter person
(b) A lighter person of equal
height and proportions
can walk faster than a
heavier person
(c) To first order, size doesn’t
matter
Physics 207: Lecture 11, Pg 14
How fast can a biped walk?
What about height?
(a) A taller person of equal
weight and proportions
can walk faster than a
shorter person
(b) A shorter person of equal
weight and proportions
can walk faster than a
taller person
(c) To first order, height
doesn’t matter
Physics 207: Lecture 11, Pg 15
How fast can a biped walk?
What can we say about the walker’s
acceleration if there is UCM (a
smooth walker) ?
Acceleration is radial !
So where does it, ar, come from?
(i.e., what external forces act on
the walker?)
1. Weight of walker, downwards
2. Friction with the ground, sideways
Physics 207: Lecture 11, Pg 16
Impulse & Linear Momentum

Transition from forces to conservation laws
Newton’s Laws  Conservation Laws
Conservation Laws  Newton’s Laws
They are different faces of the same physics
NOTE: We have studied “impulse” and “momentum”
but we have not explicitly named them as such
Conservation of momentum is far more general than
conservation of mechanical energy
Physics 207: Lecture 11, Pg 17
Forces vs time (and space, Ch. 10)
Underlying any “new” concept in Chapter 9 is
(1) A net force changes velocity (either magnitude or
direction)
(2) For any action there is an equal and opposite
reaction


If we emphasize Newton’s 3rd Law and emphasize
changes with time then this leads to the
Conservation of Momentum Principle
Physics 207: Lecture 11, Pg 18
Example
There many situations in which the sum of the product
“mass times velocity” is constant over time
 To each product we assign the name, “momentum”
and associate it with a conservation law.
(Units: kg m/s or N s)
 A force applied for a certain period of time can be
graphed and the area under the curve is the “impulse”

F (N)
10
0
Area under curve : “impulse”
With: m Dv = Favg Dt
2
Time (sec)
Physics 207: Lecture 11, Pg 22
Force curves are usually a bit different in the
real world
Physics 207: Lecture 11, Pg 23
Example with Action-Reaction
Now the 10 N force from before is applied by person
A on person B while standing on a frictionless surface
 For the force of A on B there is an equal and opposite
force of B on A
10

F (N)
A on B
0
B on A
-10
0
2
Time (sec)
MA x DVA = Area of top curve
MB x DVB = Area of bottom curve
Area (top) + Area (bottom) = 0
Physics 207: Lecture 11, Pg 24
Example with Action-Reaction
MA DVA + MB DVB = 0
MA [VA(final) - VA(initial)] + MB [VB(final) - VB(initial)] = 0
Rearranging terms
MAVA(final) +MB VB(final) =
MAVA(initial) +MB VB(initial)
which is constant regardless of M or DV
(Remember: frictionless surface)
Physics 207: Lecture 11, Pg 25
Example with Action-Reaction
MAVA(final) +MB VB(final) = MAVA(initial) +MB VB(initial)
which is constant regardless of M or DV
Define MV to be the “momentum” and this
is conserved in a system if and only if the
system is not acted on by a net external
force (choosing the system is key)
Conservation of momentum is a special
case of applying Newton’s Laws
Physics 207: Lecture 11, Pg 26
Applications of Momentum Conservation
Radioactive decay:
238U
Alpha
Decay
234Th
v2
4He
v1
Explosions
Collisions
Physics 207: Lecture 11, Pg 27
Impulse & Linear Momentum
Definition: For a single particle, the momentum p is defined as:

p ≡ mv
(p is a vector since v is a vector)
So px = mvx and so on (y and z directions)

Newton’s 2nd Law:

dv d

 m  (mv)
dt dt

F = ma

 dp
F
dt
This is the most general statement of Newton’s 2nd Law
Physics 207: Lecture 11, Pg 28
Momentum Conservation
FEXT
dP

dt
dP
0
dt
FEXT  0

Momentum conservation (recasts Newton’s 2nd Law when
net external F = 0) is an important principle

It is a vector expression (Px, Py and Pz) .
 And applies to any situation in which there is NO net
external force applied (in terms of the x, y & z axes).
Physics 207: Lecture 11, Pg 29
Momentum Conservation
Many problems can be addressed through
momentum conservation even if other physical
quantities (e.g. mechanical energy) are not
conserved
 Momentum is a vector quantity and we can
independently assess its conservation in the x, y and
z directions
(e.g., net forces in the z direction do not affect the
momentum of the x & y directions)

Physics 207: Lecture 11, Pg 30
Exercise 2
Momentum Conservation

Two balls of equal mass are thrown horizontally with the
same initial velocity. They hit identical stationary boxes
resting on a frictionless horizontal surface.

The ball hitting box 1 bounces elastically back, while the ball
hitting box 2 sticks.
 Which box ends up moving fastest ?
A.
B.
C.
Box 1
Box 2
same
1
2
Physics 207: Lecture 11, Pg 31
Lecture 11
Assignment:
For Monday: Read through Chapter 10, 1st
four sections
 MP HW6 due Wednesday 3/3

Physics 207: Lecture 11, Pg 32