Lecture 25
Today Review:
• Exam covers Chapters 14-17 plus angular momentum, statics
• Assignment
 For Thursday, read through all of Chapter 18
Physics 207: Lecture 25, Pg 1
Angular Momentum Exercise
 A mass m=0.10 kg is attached to a cord passing through a small
hole in a frictionless, horizontal surface as in the Figure. The
mass is initially orbiting with speed wi = 5 rad / s in a circle of
radius ri = 0.20 m. The cord is then slowly pulled from below,
and the radius decreases to r = 0.10 m.
 What is the final angular velocity ?
 Underlying concept: Conservation of Momentum
ri
wi
Physics 207: Lecture 25, Pg 2
Angular Momentum Exercise
 A mass m=0.10 kg is attached to a cord passing through a small
hole in a frictionless, horizontal surface as in the Figure. The
mass is initially orbiting with speed wi = 5 rad / s in a circle of
radius ri = 0.20 m. The cord is then slowly pulled from below,
and the radius decreases to r = 0.10 m.
 What is the final angular velocity ?
No external torque implies
ri
DL = 0 or Li = Lc
Ii wi = If wf
wi
I for a point mass is mr2 where r is the
distance to the axis of rotation
m ri2wi = m rf2 wf
wf = ri2wi / rf2 = (0.20/0.10)2 5 rad/s = 20 rad/s
Physics 207: Lecture 25, Pg 3
Example: Throwing ball from stool
 A student sits on a stool, initially at rest, but which is free to
rotate. The moment of inertia of the student plus the stool is I.
They throw a heavy ball of mass M with speed v such that its
velocity vector has a perpendicular distance d from the axis of
rotation.

 What is the angular speed wF of the student-stool system
after they throw the ball ?
M
Mv
r
wF
I
Top view: before
d
I
after
Physics 207: Lecture 25, Pg 4
Example: Throwing ball from stool
 What is the angular speed wF of the student-stool system after
they throw the ball ?
 Process: (1) Define system (2) Identify Conditions
(1) System: student, stool and ball (No Ext. torque, L is constant)
(2) Momentum is conserved (check |L| = |r| |p| sin q for sign)
Linit = 0 = Lfinal = - M v d + I wf
M
v
wF
I
Top view: before
d
I
after
Physics 207: Lecture 25, Pg 5
Walking the plank…
 A uniform rectangular beam of length L= 5 m and mass M=40 kg is
supported, but not attached, to two posts which are length D=3 m apart. A
child of mass W=20 kg starts walking along the beam.
20 kg
40 kg
5m
3m
 Assuming infinitely rigid posts, how close can the child get to the right end of
the beam without it falling over? [Hint: The upward force exerted by the beam
on the left cannot be negative – this is the limiting condition on how far the
child can be to the right. Set up the static equilibrium condition with the pivot
about the left end of the beam.]
Physics 207: Lecture 25, Pg 6
Walking the plank…
 A uniform rectangular beam of length L= 5 m and mass M=40 kg is
supported, but not attached, to two posts which are length D=3 m apart. A
child of mass W=20 kg starts walking along the beam.
NL
NR
0.5 m
400 N
200 N
 Sum of forces and torques are zero
 As the child walks from left to right NL goes from positive to “negative”
 If NL < 0 (a pull) then the board will tip.
 Choose blue dot as pivot point.
Physics 207: Lecture 25, Pg 7
Walking the plank…
 A uniform rectangular beam of length L= 5 m and mass M=40 kg is
supported, but not attached, to two posts which are length D=3 m apart. A
child of mass W=20 kg starts walking along the beam.
NL
NR
0.5 m
400 N
200 N
 S t = 0 = -NL x 3 + 400 x 0.5 + NR x 0 - d 200
0= 0
+ 200
+0
- d 200
d = 1 meter from pivot point which is 2.0 meter from right side
distance to end (2 – 1) meters or 1 meter
Physics 207: Lecture 25, Pg 8
Velocity and Acceleration
Position:
x(t) = A cos(wt + )
Velocity:
v(t) = -wA sin(wt + )
Acceleration: a(t) = -w2A cos(wt + )
xmax = A
vmax = wA
amax = w2A
k
by taking
derivatives,
since:
dx ( t )
v( t ) 
dt
a( t ) 
dv ( t )
dt
m
0
x
Physics 207: Lecture 25, Pg 9
Measuring g
 To measure the magnitude of the acceleration due to gravity, g,
in an unorthodox manner, a student places a ball bearing on the
concave side of a flexible speaker cone . The speaker cone acts
as a simple harmonic oscillator whose amplitude is A and whose
frequency w can be varied. The student can measure both A
and w. The equations of motion for the speaker are
Position:
y(t) = A cos(wt + )
Velocity:
vy(t) = -wA sin(wt + )
Acceleration: ay (t) = -w2A cos(wt + )
 If the ball bearing is in contact with the speaker, then
S Fy = may= - mg+ N so that N = mg + may
N = mg - mw2A cos(wt + )
Maximum: N = mg + mw2A Minimum: mg - mw2A but must be > 0!
Physics 207: Lecture 25, Pg 10
Ideal Fluid
Bernoulli Equation  P1+ ½ r v12 + r g y1 = constant
A 5 cm radius horizontal pipe carries water at 10 m/s into a
10 cm radius. ( rwater =103 kg/m3 )
What is the pressure difference?
v
P1+ ½ r v12 = P2+ ½ r v22
2
DP = ½ r v22 - ½ r v12
y
2
2
2
DP = ½ r (v2 - v1 )
v
DV
1
p
and A1 v1 = A2 v2
2
y
1
p
1
DP = ½ r v22 (1 – (A2/A1 ) 2 )
= 0.5 x 1000 kg/m x 100 m2/s2 (1- (25/100)2 )
= 47000 Pa
Physics 207: Lecture 25, Pg 11
A water fountain
 A fountain, at sea level, consists of a 10 cm
radius pipe with a 5 cm radius nozzle. The
water sprays up to a height of 20 m.
What is the velocity of the water as it
leaves the nozzle?
What volume of the water per second as it
leaves the nozzle?
What is the velocity of the water in the
pipe?
What is the pressure in the pipe?
How many watts must the water pump
supply?
Physics 207: Lecture 25, Pg 12
A water fountain
 A fountain, at sea level, consists of a 10 cm radius pipe with a






5 cm radius nozzle. The water sprays up to a height of 20 m.
What is the velocity of the water as it leaves the nozzle?
Simple Picture: ½mv2=mgh  v=(2gh)½ = (2x10x20)½ = 20 m/s
What volume of the water per second as it leaves the nozzle?
Q = A vn = 0.0025 x 20 x 3.14 = 0.155 m3/s
What is the velocity of the water in the pipe?
An vn = Ap vp  vp = Q /4 = 5 m/s
What is the pressure in the pipe?
1atm + ½ r vn2 = 1 atm + DP + ½ r vp2  1.9 x 105 N/m2
How many watts must the water pump supply?
Power = Q r g h = 0.0155 m3/s x 103 kg/m3 x 9.8 m/s2 x 20 m
= 3x104 W
Physics 207: Lecture 25, Pg 13
Fluids Buoyancy
 A metal cylinder, 0.5 m in radius and 4.0 m
high is lowered, as shown, from a massles
rope into a vat of oil and water. The
tension, T, in the rope goes to zero when
the cylinder is half in the oil and half in the
water. The densities of the oil is 0.9
gm/cm3 and the water is 1.0 gm/cm3
 What is the average density of the
cylinder?
 What was the tension in the rope when the
cylinder was submerged in the oil?
Physics 207: Lecture 25, Pg 14
Fluids Buoyancy
 r = 0.5 m, h= 4.0 m
 roil = 0.9 gm/cm3 rwater = 1.0 gm/cm3
 What is the average density of the cylinder?
When T = 0 Fbuoyancy = Wcylinder
Fbuoyancy = roil g ½ Vcyl.+ rwater g ½ Vcyl.
Wcylinder = rcyl g Vcyl.
rcyl g Vcyl. = roil g ½ Vcyl.+ rwater g ½ Vcyl.
rcyl = ½ roill.+ ½ rwater
What was the tension in the rope when the
cylinder was submerged in the oil?
Use a Free Body Diagram !
Physics 207: Lecture 25, Pg 15
Fluids Buoyancy
 r = 0.5 m, h= 4.0 m
Vcyl. = p r2 h
 roil = 0.9 gm/cm3 rwater = 1.0 gm/cm3
 What is the average density of the cylinder?
When T = 0 Fbuoyancy = Wcylinder
Fbuoyancy = roil g ½ Vcyl.+ rwater g ½ Vcyl.
Wcylinder = rcyl g Vcyl.
rcyl g Vcyl. = roil g ½ Vcyl.+ rwater g ½ Vcyl.
rcyl = ½ roill.+ ½ rwater = 0.95 gm/cm3
What was the tension in the rope when the
cylinder was submerged in the oil?
Use a Free Body Diagram!
S Fz = 0 = T - Wcylinder +
Fbuoyancy
T = Wcyl - Fbuoy = g ( rcyl .- roil ) Vcyl
T = 9.8 x 0.05 x 103 x p x 0.52 x 4 .0 = 1500 NPhysics 207: Lecture 25, Pg 16
Fluids Buoyancy
 r = 0.5 m, h= 4.0 m
Vcyl. = p r2 h
 roil = 0.9 gm/cm3 rwater = 1.0 gm/cm3
 What is the average density of the cylinder?
When T = 0 Fbuoyancy = Wcylinder
.
rcyl = ½ roill.+ ½ rwater = 0.95 gm/cm3
What was the tension in the rope when the
cylinder was submerged in the oil?
Use a Free Body Diagram!
S Fz = 0 = T - Wcylinder +
Fbuoyancy
T = Wcyl - Fbuoy = g ( rcyl .- roil ) Vcyl
T = 9.8 x 0.05 x 103 x p x 0.52 x 4 .0 = 1500 N
Physics 207: Lecture 25, Pg 17
A new trick
 Two trapeze artists, of mass 100 kg
and 50 kg respectively are testing a
new trick and want to get the timing
right. They both start at the same
time using ropes of 10 m in length
and, at the turnaround point the
smaller grabs hold of the larger artist
and together they swing back to the
starting platform. A model of the
stunt is shown at right.
 How long will this stunt require if the
angle is small ?
Physics 207: Lecture 25, Pg 18
A new trick
 How long will this stunt require?
Period of a pendulum is just
w = (g/L)½
T = 2p (L/g)½
Time before ½ period
Time after ½ period
So, t = T = 2p(L/g)½ = 2p sec
Key points: Period is one full swing
and independent of mass
(this is SHM but very different than a
spring. SHM requires only a linear
restoring force.)
Physics 207: Lecture 25, Pg 19
Example
 A Hooke’s Law spring, k=200 N/m, is on a horizontal
frictionless surface is stretched 2.0 m from its equilibrium
position. A 1.0 kg mass is initially attached to the spring
however, at a displacement of 1.0 m a 2.0 kg lump of clay is
dropped onto the mass. The clay sticks.
M
What is the new amplitude?
k
m
M
m
k
-2
0(≡Xeq) 2
Physics 207: Lecture 25, Pg 20
Example
 A Hooke’s Law spring, k=200 N/m, is on a horizontal
frictionless surface is stretched 2.0 m from its equilibrium
position. A 1.0 kg mass is initially attached to the spring
however, at a displacement of 1.0 m a 2.0 kg lump of clay is
dropped onto the mass.
M
What is the new amplitude?
Sequence: SHM, collision, SHM
k
m
½ k A02 = const.
½ k A02 = ½ mv2 + ½ k (A0/2)2
M
k
¾ k A02 = m v2  v = ( ¾ k A02 / m )½
m
½
v = (0.75*200*4 / 1 ) = 24.5 m/s
-2
0(≡Xeq) 2
Conservation of x-momentum:
mv= (m+M) V  V = mv/(m+M)
V = 24.5/3 m/s = 8.2 m/s
Physics 207: Lecture 25, Pg 21
Example
 A Hooke’s Law spring, k=200 N/m, is on a horizontal
frictionless surface is stretched 2.0 m from its equilibrium
position. A 1.0 kg mass is initially attached to the spring
however, at a displacement of 1.0 m a 2.0 kg lump of clay is
dropped onto the mass. The clay sticks.
M
What is the new amplitude?
Sequence: SHM, collision, SHM
k
m
V = 24.5/3 m/s = 8.2 m/s
½ k Af2 = const.
M
½ k Af2 = ½ (m+M)V2 + ½ k (Ai)2
k
m
A 2 = [(m+M)V2 /k + (A )2 ]½
f
i
Af2 = [3 x 8.22 /200 + (1)2 ]½
Af2 = [1 + 1]½  Af2 = 1.4 m
Key point: K+U is constant in SHM
-2
0(≡Xeq) 2
Physics 207: Lecture 25, Pg 22
Fluids Buoyancy & SHM
 A metal cylinder, 0.5 m in radius and 4.0 m
high is lowered, as shown, from a rope into
a vat of oil and water. The tension, T, in
the rope goes to zero when the cylinder is
half in the oil and half in the water. The
densities of the oil is 0.9 gm/cm3 and the
water is 1.0 gm/cm3
 Refer to earlier example
 Now the metal cylinder is lifted slightly
from its equilibrium position. What is the
relationship between the displacement and
the rope’s tension?
 If the rope is cut and the drum undergoes
SHM, what is the period of the oscillation if
undamped?
Physics 207: Lecture 25, Pg 23
Fluids Buoyancy & SHM
 Refer to earlier example
 Now the metal cylinder is lifted Dy from its
equilibrium position. What is the relationship
between the displacement and the rope’s tension?
0 = T + Fbuoyancy – Wcylinder
T = - Fbuoyancy + Wcylinder
T =-[rog(h/2+Dy) Ac+rwgAc(h/2-Dy)] + Wcyl
T =-[ghAc(ro +rw)/2 + Dy gAc(ro -rw)]+ Wcyl
T =-[Wcyl + Dy gAc(ro -rw)]+ Wcyl
T = [ g Ac (rw -ro) ] Dy
If the is rope cut, net force is towards equilibrium
position with a proportionality constant
g Ac (rw -ro) [& with g=10 m/s2]
If F = - k Dy then k = g Ac (ro -rw) = p/4 x103 N/m
Physics 207: Lecture 25, Pg 24
Fluids Buoyancy & SHM
 A metal cylinder, 0.5 m in radius and 4.0 m high is
lowered, as shown, from a rope into a vat of oil
and water. The tension, T, in the rope goes to
zero when the cylinder is half in the oil and half in
the water. The densities of the oil is 0.9 gm/cm3
and the water is 1.0 gm/cm3
 If the rope is cut and the drum undergoes SHM,
what is the period of the oscillation if undamped?
F = ma = - k Dy and with SHM …. w = (k/m)½
where k is a “spring” constant and m is the
inertial mass (resistance to motion), the cylinder
So w = (1000p /4 mcyl)½
= (1000p / 4rcylVcyl)½ = ( 0.25/0.95 )½
= 0.51 rad/sec
T = 3.2 sec
Physics 207: Lecture 25, Pg 25
Underdamped SHM
if
x(t )  A exp - 2btm cos(wt   )
(
wo  b / 2m
)
If the period is 2.0 sec and, after four cycles, the amplitude
drops by 75%, what is the time constant t if t ≡ 2m / b ?
1.2
1
Four cycles implies 8 sec
0.8
0.6
So
0.4
A
0.2
0.25 A0 = A0 exp(-8b / 2m)
0
-0.2
ln(1/4)= -8 1/t
-0.4
-0.6
-0.8
t = -8/ ln(1/4) = 5.8 sec
-1
wt
Physics 207: Lecture 25, Pg 26
Anti-global warming or the nuclear winter scenario
 Assume P/A =
P = 1340 W/m2 from the sun is incident on
a thick dust cloud above the Earth and this energy is
absorbed, equilibrated and then reradiated towards space
where the Earth’s surface is in thermal equilibrium with
cloud. Let e (the emissivity) be unity for all wavelengths of
light.
 What is the Earth’s temperature?
P=
 A T4=  (4p r2) T4 = P p r2  T = [P / (4 x  )]¼
  = 5.710-8 W/m2 K4
 T = 277 K (A little on the chilly side.)
Physics 207: Lecture 25, Pg 27
Time for a swim
 At solar equinox, how much could you reasonably expect for the





top 1 meter of water in a lake at 45 deg. latitude to warm on a
calm but bright sunny day assuming that the sun’s flux (P/A) is
1340 W/m2 if all the sun’s energy were absorbed in that layer ?
At 45 deg. , sin 45 deg or 0.7071 x 1340 or 950 W/m2
But only 6 hours of full sun
So Q available is 6 x 60 x 60 seconds with 950 J/s/m2
Q = 2 x 107 J in one square meter
Q = m c DT with m 103 kg/m3 and c = 4.2 x 103 J/kg-K
 about 5 K per day.
Physics 207: Lecture 25, Pg 28
Q, W and DETH

0.20 moles of an ideal monoatomic
gas are cycled clockwise through
the pV sequence shown at right
(ADCBA). (R = 8.3 J/K mol)
 What are the temperatures of the
two isotherms?
 Fill out the table below with the
appropriate values.
(a) pV = nRT T = pV/nR
TA = 16600 x 0.10 / (0.20 x 8.3)
TA = 1000 K
TB = 500 K
(b)
DETH=W + Q
Ideal Gas at constant V
Q = n CV DT
Physics 207: Lecture 25, Pg 29
Ch. 12
|L|=mvrperpendicular
Physics 207: Lecture 25, Pg 30
Ch. 12
Physics 207: Lecture 25, Pg 31
Hooke’s Law Springs and a Restoring Force
 Key fact: w = (k / m)½ is general result where k reflects a
constant of the linear restoring force and m is the inertial response
(e.g., the “physical pendulum” where w = (k / I)½
Physics 207: Lecture 25, Pg 32
Simple Harmonic Motion
Maximum
potential
energy
Maximum
kinetic
energy
Physics 207: Lecture 25, Pg 33
Resonance and
damping
 Energy transfer is
optimal when the
driving force varies at
the resonant
frequency.
 Types of motion
 Undamped
 Underdamped
 Critically damped
 Overdamped
Physics 207: Lecture 25, Pg 34
Fluid Flow
Physics 207: Lecture 25, Pg 35
Density and pressure
Physics 207: Lecture 25, Pg 36
Response to forces
Physics 207: Lecture 25, Pg 37
States of Matter and Phase Diagrams
Physics 207: Lecture 25, Pg 38
Ideal gas equation of state
Physics 207: Lecture 25, Pg 39
pV diagrams
Physics 207: Lecture 25, Pg 40
Thermodynamics
Physics 207: Lecture 25, Pg 41
Work, Pressure, Volume, Heat
T can change!
In steady-state T=constant and so heat in equals heat out
Physics 207: Lecture 25, Pg 42
Gas Processes
Physics 207: Lecture 25, Pg 43
Lecture 25
• Exam covers Chapters 14-17 plus angular momentum, statics
• Assignment
 For Thursday, read through all of Chapter 18
Physics 207: Lecture 25, Pg 44