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Lecture 30
• Review (the final is, to a large degree, cumulative)
~50% refers to material in Ch. 1-12
 ~50% refers to material in Ch. 13,14,15
-- Chapter 13: Gravitation
-- Chapter 14: Newtonian Fluids
-- Chapter 15: Oscillatory Motion

• Today we will review chapters 13-15
• 1st is short mention of resonance
• Order Ch. 15 to 13
Physics 201: Lecture 30, Pg 1
Final Exam Details
 Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet
room
 Format:
 Closed book
 Up to 4 8½x1 sheets, hand written only
 Approximately 50% from Chapters 13-15 and 50% 1-12
 Bring a calculator
 Special needs/ conflicts:
All requests for alternative test arrangements should be
made by today (except for medical emergency)
Physics 201: Lecture 30, Pg 2
Driven SHM with Resistance
 Apply a sinusoidal force, F0 cos (wt), and now
consider what A and b do,
steady state amplitude
d 2 x b dx k
F

 x  cos wt
2
dt
m dt m
m
A
b/m small
Not Zero!!!
F0 / m
bw 2
(w  w )  ( )
m
2
2 2
0
b/m middling
b large
w  w0
w
Physics 201: Lecture 30, Pg 3
Dramatic example of resonance
 In 1940, a steady wind set up a torsional vibration in the
Tacoma Narrows Bridge
Physics 201: Lecture 30, Pg 4
Dramatic example of resonance
 Eventually it collapsed
Physics 201: Lecture 30, Pg 5
Mechanical Energy of the Spring-Mass System
x(t) =
A cos( wt + f )
v(t) = -wA sin( wt + f )
a(t) = -w2A cos( wt + f ) & F(t)=ma(t)
Kinetic energy:
K = ½ mv2 = ½ m(wA)2 sin2(wt+f)
Potential energy:
U = ½ k x2 = ½ k A2 cos2(wt + f)
And w2 = k / m or k = m w2
K + U = constant
Physics 201: Lecture 30, Pg 6
SHM
x(t) =
A cos( wt + f )
v(t) = -wA sin( wt + f )
a(t) = -w2A cos( wt + f )
A : amplitude
w : angular frequency
w =2p f =2p/T
f : phase constant
xmax =
A
vmax = wA
amax = w2A
Physics 201: Lecture 30, Pg 7
Recognizing the phase constant
 An oscillation is described by x(t) =A cos(ωt+φ).
Find φ for each of the following figures:
Answers
φ=0
φ = π/2
x(t)= A cos (π)
φ=π
Physics 201: Lecture 30, Pg 8
Common SHMs
w
k
m
w
g
L
w

I
Physics 201: Lecture 30, Pg 9
SHM: Friction with velocity dependent Drag force -bv
b is the drag coefficient; soln is a damped exponential
x(t )  A exp ( b t ) cos(wt  f )
2m
if
wo  b / 2m
1.2
1
k  b 
w


m  2m 
0.8
0.6
0.4
2
A
0.2
0
-0.2
-0.4
 b 
w  w  
 2m 
2
o
-0.6
-0.8
-1
wt
Physics 201: Lecture 30, Pg 10
2
SHM: Friction with velocity dependent Drag force -bv
If the maximum amplitude drop 50% in 10 seconds,
what will the relative drop be in 30 more seconds?
x(t  10 s)  12 x(t ) x(t  30) A exp( 2bm (t  30))

x(t )
x(t  30 s)  cx(t )
A exp(  b t )
2m
1.2
b 20)
x
(
t

10
)
exp
(

x(t  30)
2m

x(t )
x(t )
b 10)]2
[exp
(

x(t  30)
2m

x(t )
2
1
0.8
0.6
0.4
A
0.2
0
-0.2
-0.4
x(t  30) 12 2 1
 
x(t )
2 8
-0.6
-0.8
-1
wt
Physics 201: Lecture 30, Pg 11
Chapter 14 Fluids
 Density ρ = m/V
P1 atm = 1x105 N/m2
Force is normal to container surface
 Pressure with Depth/Height P = P0 + ρgh
 Gauge vs. Absolute pressure
 Pascal’s Principle: Same depth  Same pressure
 Buoyancy, force, B, is always upwards
 B = ρfluid Vfluid displaced g (Archimedes’s Principle)
 Flow
Continuity: Q = v2A2 = v1A1 (volume / time or m3/s)
Bernoulli’s eqn: P1+ ½ ρv12 + ρgh1 = P2+ ½ ρv22 + ρgh2
 Pressure P = F/A
Physics 201: Lecture 30, Pg 12
Example problem
 A piece of iron (ρ=7.9x103 kg/m3) block weighs 1.0 N in air.
How much does the scale read in water?
 Solution:
 In air
T1 = mg = ρiromV g
 In water: B+T2-mg = 0
T2 = mg-B
= mg – ρwaterVg
= mg – ( ρwater /ρiron ) ρiron Vg
= mg (1-ρwater /ρiron )
= 0.87mg = 0.87 N
Physics 201: Lecture 30, Pg 13
Another buoyancy problem
A spherical balloon is filled with air (rair 1.2 kg/m3). The
radius of the balloon is 0.50 m and the wall thickness of
the latex wall is 0.01 m (rlatex 103 kg/m3). The balloon is
anchored to the bottom of stream which is flowing from
left to right at 2.0 m/s. The massless string makes an
angle of 30° from the stream bed.
 What is the magnitude of the drag force
on the balloon?
Key physics: Equilbrium and buoyancy.
SFx=0 & SFy=0

Physics 201: Lecture 30, Pg 14
Another buoyancy problem
A spherical balloon is filled with air (rair 1.2 kg/m3). The
radius of the balloon is 0.50 m and the wall thickness of the
latex wall is 1.0 cm (rlatex 103 kg/m3). The balloon is
anchored to the bottom of stream which is flowing from left to
right at 2.0 m/s. The massless string makes an angle of 45°
from the stream bed.
 What is the magnitude of the drag force on the balloon?
Key physics: Equilibrium and buoyancy. SFx=0 & SFy=0
SFx=-T cos q + D = 0
SFy=-T sin q + Fb - Wair = 0
Fb
Wair = rair V g= rair (4/3 pr3) g with r=0.49 m
D
Fb= rwater V g= rwater (4/3 pr3) g
T
Variation: What is the maximum wall
Wair
thickness of a lead balloon filled with He?

Physics 201: Lecture 30, Pg 15
Pascal’s Principle
 Is PA = PB ?
Answer: No!
Same level, same pressure, only if same fluid density
B
A
Physics 201: Lecture 30, Pg 16
Power from a river
 Water in a river has a rectangular cross section which
is 50 m wide and 5 m deep. The water is flowing at
1.5 m/s horizontally. A little bit downstream the water
goes over a water fall 50 m high. How much power is
potentially being generated in the fall?
 W = Fd = mgh
 Pavg= W / t = (m/t) gh
 Q = Av and m/t = rwater Q (kg/m3 m3/s)
 Pavg= rwater Av gh
= 103 kg/m3 x 250 m2 x 1.5 m/s x 10 m/s2 x 50 m
= 1.8x108 kg m2/s2/s = 180 MW
Physics 201: Lecture 30, Pg 17
Chapter 13 Gravitation
 Universal gravitation force





 Always attractive
 Proportional to the mass ( m1m2 )
 Inversely proportional to the square of the distance (1/r2)
 Central force: orbits conserve angular momentum
Gravitational potential energy
 Always negative
 Proportional to the mass ( m1m2 )
 Inversely proportional to the distance (1/r)
Circular orbits: Dynamical quantities (v,E,K,U,F) involve radius
 K(r) = - ½ U(r)
Employ conservation of angular momentum in elliptical orbits
No need to derive Kepler’s Laws (know the reasons for them)
Energy transfer when orbit radius changes(e.g. escape velocity)
Physics 201: Lecture 30, Pg 18
Key equations
 Newton’s Universal “Law” of Gravity


m1m2
Fon 2 by 1  G 2 r̂12   Fon1by 2
r
Universal Gravitational Constant G = 6.673 x 10-11 Nm2 / kg2
The force points along the line connecting the two objects.
g surface
GM
 2
RE
On Earth, near sea level, it can
be shown that gsurface ≈ 9.8 m/s2.
 Gravitational potential energy
U (r )  Gm1m2 / r
“Zero” of potential energy defined to be at r = ∞, force  0
 At apogee and perigee:
L  mvr
Physics 201: Lecture 30, Pg 19
Dynamics of Circular Orbits
For a circular orbit:
 Force on m: FG= GMm/r2
 Orbiting speed: v2 = GM/r (independent of m)
 Kinetic energy: K = ½ mv2 = ½ GMm/r
 Potential energy UG= - GMm/r
 Notice UG = -2 K
 Total Mech. Energy:
 E = KE + UG = - ½ GMm/r
Physics 201: Lecture 30, Pg 20
Changing orbit
 A 200 kg satellite is launched into a circular orbit at height
h= 200 km above the Earth’s surface.
What is the minimum energy required to put it into the orbit ?
(ignore Earth’s spin) (ME = 5.97x1024 kg, RE = 6.37x106 m,
G = 6.67x10-11 Nm2/kg2)
 Solution:
Initial: h=0, ri = RE
 Ei = Ki +Ui = 0 + (-GMEm/RE )
= -1.25x1010J
 In orbit: h = 200 km, rf = RE + 200 km
Ef = Kf + Uf = - ½ Uf+ Uf = ½ Uf
= - ½ GMEm/(RE+200 km)
= -6.06x109J
DE = Ef – Ei = 6.4x109 J
Physics 201: Lecture 30, Pg 21
Escaping Earth orbit
 Exercise: suppose an object of mass m is
projected vertically upwards from the
Earth’s surface at an initial speed v, how
high can it go ? (Ignore Earth’s rotation)
Ei  U G ( RE )  mv
1
2
2
i
E f  U G ( RE  h)  0
2 2
E i
Rv
h
2
(2GM  RE vi )
2GM  RE vi2  0
implies infinite height
vEscape
2GM

 11.2 km/s
RE
Physics 201: Lecture 30, Pg 22
We hope everyone does well on Sunday
Have a great summer!
Physics 201: Lecture 30, Pg 23