Announcements
1. Remember -- Tuesday, Oct. 28th, 9:30 AM – Second exam
(covering Chapters 9-14 of HRW) – Bring the following:
a) 1 equation sheet
b) Calculator
c) Pencil
d) Clear head
e) Note: If you have kept up with your HW, you may drop
your lowest exam grade
2. Today --Thursday, Oct. 23th, 4 PM – Physics Colloquium
by Professor Bernd Schüttler, Dept. of Physics, U. Ga – will
discuss the analysis of biological systems in terms of a
physical and mathematical model
3. Today’s lecture – review Chapters 9-14, problem solving
techniques
10/23/2003
PHY 113 -- Lecture 14R
1
Gravitational forces and energy
m
Potential energy :
r
v0
GMm
U (r ) = −
r
Energy needed to escape Earth’s
gravitational field, assuming an
initial velocity v0 :
Ki + U i = K f + U f
1
2
GMm
mv −
=0
RE
2
0
2GM
⇒ v0 =
≈ 25000 km/h
RE
10/23/2003
PHY 113 -- Lecture 14R
2
Energy needed to go from one stable circular orbit to another:
E = K +U
From Newton' s law for cirular orbit :
R1
R2
v2
GMm
−m = − 2
r
r
GMm
⇒ mv =
r
GMm
⇒E=−
2r
E2 − E1 = −
10/23/2003
PHY 113 -- Lecture 14R
2
GMm GMm
+
2 R2
2 R1
3
Energy needed to go from one stable circular orbit to another -Example:
R1
R2
How much energy is needed to
take a satellite of mass m=100kg
from the international space
station (R1=RE+390 km) to its
usual orbit (R2=RE+600 km)?
GMm ⎛ 1 1 ⎞
E2 − E1 = −
⎜ − ⎟
2 ⎝ R2 R1 ⎠
≈ 9 ×10 7 J
10/23/2003
PHY 113 -- Lecture 14R
4
τ = r×F
L = r×p
Fgravity
Gm1m2
=
r̂12
2
r12
Problem solving skills
Math skills
Equation Sheet
Advice:
1. Keep basic concepts and equations at the top of your head.
2. Practice problem solving and math skills
3. Develop an equation sheet that you can consult.
10/23/2003
PHY 113 -- Lecture 14R
5
Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle applies
3. Write down the appropriate equations using the variables
defined in step 1.
4. Check whether you have the correct amount of
information to solve the problem (same number of
knowns and unknowns.
5. Solve the equations.
6. Check whether your answer makes sense (units, order of
magnitude, etc.).
10/23/2003
PHY 113 -- Lecture 14R
6
ŷ
of mass
ri
rj
r COM ≡
∑mr
∑m
i
i
i
i
i
x̂
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PHY 113 -- Lecture 14R
7
Position of the center of mass:
r com ≡
∑mr
∑m
i
i
i
i
i
Velocity of the center of mass:
v com ≡
∑mv
∑m
i
i
i
i
i
Acceleration of the center of mass: a com ≡
∑ma
∑m
i
i
i
i
i
10/23/2003
PHY 113 -- Lecture 14R
8
Physics of composite systems:
dmi v i
dp i
∑i Fi = ∑i miai =∑i dt = ∑i dt
Center-of-mass velocity:
v com
∑m v ∑m v
≡
≡
M
∑m
i
i
i
i
i
i
i
i
Note that:
∑ Fi ≡ Ftotal = M
i
10/23/2003
dv com
dt
PHY 113 -- Lecture 14R
9
A new way to look at Newton’s second law:
F = ma = m
dv d (mv ) dp
=
≡
dt
dt
dt
Define linear momentum p = mv
Consequences:
1. If F = 0
Î
dp
=0
dt
2. For system of particles:
Î p = constant
∑ Fi = ∑
i
If
∑F = 0
i
i
10/23/2003
dp i
⇒∑
=0
dt
i
PHY 113 -- Lecture 14R
i
dp i
dt
⇒ ∑ p i = constant
i
10
Statement of conservation of momentum:
m1v1i = m1v1 f cos θ + m2 v2 f cos φ
0 = m1v1 f sinθ − m2 v2 f sinφ
If mechanical (kinetic) energy is conserved, then:
1
2
m
v
2 1 1i =
10/23/2003
1
2
1 m v2
m
v
+
2 1 1f
2 2 2f
PHY 113 -- Lecture 14R
11
Snapshot of a collision:
Pi
Impulse:
dp
F (t ) =
⇒ dp = F(t )dt
dt
Pf
10/23/2003
t2
t2
t1
t1
∫ dp = ∫ F(t )dt ≡ J
PHY 113 -- Lecture 14R
12
Angular motion
s
angular “displacement” Î θ(t)
dθ
angular “velocity” Î ω(t) =
dt
dω
angular “acceleration” Î
α(t) =
dt
“natural” unit == 1 radian
Relation to linear variables:
sθ = r (θf-θi)
vθ = r ω
10/23/2003
PHY 113 -- Lecture 14R
aθ = r α
13
v1=r1ω
r1
ω
r2
v2=r2ω
Special case of constant angular acceleration: α = α0:
ω(t) = ωi + α0 t
θ(t) = θi + ωi t + ½ α0 t2
( ω(t))2 = ωi2 + 2 α0 (θ(t) - θi )
10/23/2003
PHY 113 -- Lecture 14R
14
Newton’s second law applied to center-of-mass motion
dv i
dv CM
⇒ Ftotal = M
∑ Fi = ∑ mi
dt
dt
i
i
Newton’s second law applied to rotational motion
Fi = mi
dv i
dv
⇒ ri × Fi = ri × mi i
dt
dt
I ≡ ∑ mi d i2
τ i = ri × Fi
i
v i = ω × ri
ri
d (ω × ri )
⇒ τ i = mi ri ×
dt
dimi
Fi
dω
⇒ τ total = I
= Iα (for rotating about principal axis)
dt
10/23/2003
PHY 113 -- Lecture 14R
15
Object rotating with constant angular velocity (α = 0)
ω
R
v=Rω
v=0
Kinetic energy associated with rotation:
K = ∑ 1 2mi vi2 = ∑ 1 2mi ri2 ω 2 ≡
i
i
where : I ≡ ∑ mi ri2
1
2
I
ω
;
2
“moment of inertia”
i
10/23/2003
PHY 113 -- Lecture 14R
16
Kinetic energy associated with rolling without slipping:
I ≡ ∑ mi ri 2
K rot = 12 Iω 2
i
Distance to axis
of rotation
K tot = K com + K rot
Rolling:
If there is no slipping :
vcom = Rω
I ⎞ 2
⎛
⇒ K tot = M ⎜1 +
v
2 ⎟ com
⎝ MR ⎠
1
2
10/23/2003
PHY 113 -- Lecture 14R
17
Torque and angular momentum
Define angular momentum: L ≡ r × p
For composite object: L = Iω
Newton’s law for torque:
dω dL
τ total = I
=
dt
dt
Î If τtotal = 0 then L = constant
In the absence of a net torque on a system,
angular momentum is conserved.
10/23/2003
PHY 113 -- Lecture 14R
18
Center-of-mass
rCM ≡
∑ mi ri
i
∑ mi
i
Torque on an extended object due to gravity (near
surface of the earth) is the same as the torque on a point
mass M located at the center of mass.
ri
10/23/2003
mi
rCM
τ = ∑ ri × {mi g (− j)} = rCM × {Mg (− j)}
i
PHY 113 -- Lecture 14R
19
Notion of equilibrium:
Notion of stability:
θ
∑ Fi = 0
i
i
F=ma Î
r
∑ τi = 0
T- mg cos θ = 0
−mg sin θ = −maθ
T
τ=I α Î r mg sin θ = mr2 α = mraθ
mg(-j)
10/23/2003
Example of stable equilibrium.
PHY 113 -- Lecture 14R
20
Analysis of stability:
∑ Fi = 0
i
10/23/2003
PHY 113 -- Lecture 14R
∑ τi = 0
i
21
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PHY 113 -- Lecture 14R
22
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PHY 113 -- Lecture 14R
23
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PHY 113 -- Lecture 14R
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PHY 113 -- Lecture 14R
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PHY 113 -- Lecture 14R
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