Experiment 09:
09: Angular
Angular
Experiment
momentum
momentum
Goals
‰
‰
‰
Investigate conservation of angular momentum
and kinetic energy in rotational collisions.
Measure and calculate moments of inertia.
Measure and calculate non-conservative work in
an inelastic collision.
Apparatus
‰
‰
‰
‰
‰
Connect output of
phototransistor to
channel A of 750.
Connect output of
tachometer
generator to channel
B of 750.
Connect power
supply.
Red button is
pressed: Power is
applied to motor.
Red button is
released: Rotor
coasts: Read output
voltage using
DataStudio.
‰
Use black sticker or tape on white plastic
rotor for generator calibration.
Calibrate tachometer-generator
‰
‰
‰
Spin motor up to full speed, let it coast. Measure and plot voltages
for 0.25 s period. Sample Rate: 5000 Hz, and Sensitivity: Low.
Time 10 periods to measure ω.
Then calculate ω for 1 V
output.
‰
Average the output
voltage over the same 10
periods.
4
Measure rotor IR
‰
‰
‰
‰
Start DataStudio and let the weight
drop.
Plot only the generator
voltage for rest of
experiment.
Use a 55 gm weight to
accelerate the rotor.
Settings:
¾ Sensitivity: Low
¾ Sample rate 500 Hz.
¾ Delayed start: None
¾ Auto Stop: 4 seconds
Understand graph output to
measure IR
‰
Generator voltage while measuring IR. What is happening:
1. Along line A-B ?
2. At point B ?
3. Along line B-C ?
‰
How do you use this graph to find IR ?
Measure IR results
‰
Measure and record αup and αdown.
‰
For your report, calculate IR:
τ f = I Rαdown
IR =
mr ( g − rαup )
αup − αdown
Fast collision
Sensitivity Sample Rate
Low
200 Hz
Delayed
Start
Auto Stop
1 sec
Falls below 0.5V
Find ω 1 (before) and ω 2 (after), estimate δt for collision.
Calculate
IW = 12 mω (ro 2 + ri 2 )
Slow collision
‰
Find ω 1 and ω 2 , measure δt , fit or measure to find ac.
‰
Keep a copy of your results for the homework problem.
Kepler Problem and
Planetary Motion
8.01t
Nov 15, 2004
Kepler’s Laws
1. The orbits of planets are ellipses; and the sun is at
one focus
2. The radius vector sweeps out equal areas in equal
time
3. The period T is proportional to the radius to the 3/2
power
T~ r3/2
Kepler Problem
• Find the motion of two
bodies under the
influence of a
gravitational force
using Newtonian
mechanics
G
m1m2
F1,2 ( r ) = −G 2 rˆ
r
Reduction of Two Body Problem
• Reduce two body problem to one body of mass
µ moving about a central point under the
influence of gravity with position vector
corresponding to the vector from mass m2 to
mass m1
m1m2
µ=
m1 + m2
G
G
d r
F1,2 ( r ) rˆ = µ 2
dt
2
Solution of One Body Problem
• Solving the problem means finding the
distance from the origin r ( t ) and
angle θ t as functions of time
()
• Equivalently, finding the distance from the
center as a function of angle r (θ )
• Solution:
r0
r=
1 − ε cos θ
Constants of the Motion
G
dr
d
θ
• Velocity v =
rˆ + r
θˆ
dt
dt
• Angular Momentum
• Energy
2
⎛ dr ⎞ ⎛ dθ ⎞
v = ⎜ ⎟ +⎜r
⎟
dt
dt
⎝ ⎠ ⎝
⎠
2
L = µ rvtangential
dθ
= µr
dt
2
1 2 Gm1m2
E = µv −
2
r
2
2
⎡
1 ⎛ dr ⎞ ⎛ dθ ⎞ ⎤ Gm1m2
E = µ ⎢⎜ ⎟ + ⎜ r
⎟ ⎥−
2
⎢⎣⎝ dt ⎠
2
⎝ dt ⎠ ⎥⎦
Gm1m2
L2
1 ⎛ dr ⎞
E = µ⎜ ⎟ +
−
2
r
2 ⎝ dt ⎠ 2 µ r
r
2
Reduction to One Dimensional
Motion
• Reduce the one body problem in two
dimensions to a one body problem moving
only in the r-direction but under the action of a
repulsive force and a gravitational force
PRS Question
Suppose the potential energy of two particles (reduced mass µ) is given by
1 L2
U (r ) =
2 µr 2
where r is the relative distance between the particles. The force
between the particles is
1. attractive and has magnitude
L2
F=
2µ r
2. repulsive and has magnitude
L2
F=
2µ r
3. attractive and has magnitude
L2
F= 3
µr
4. repulsive and has magnitude
L2
F= 3
µr
One Dimensional Description
• Energy
2
1 ⎛ dr ⎞ 1 L2 Gm1m2
E = µ⎜ ⎟ +
−
= K + U effective
2
r
2 ⎝ dt ⎠ 2 µ r
1 ⎛ dr ⎞
K = µ⎜ ⎟
2 ⎝ dt ⎠
• Kinetic Energy
• Effective Potential Energy
• Repulsive Force
U effective
Fcentrifugal
2
Gm1m2
L2
=
−
2
2µ r
r
d ⎛ L2 ⎞ L2
=− ⎜
= 3
2 ⎟
dr ⎝ 2µ r ⎠ µ r
• Gravitational force
Fgravitational = −
dU gravitational
dr
=−
Gm1m2
r2
Energy Diagram
U effective
Gm1m2
L2
=
−
2µ r 2
r
Case 4: Circular Orbit E = Eo
Case 3: Elliptic Orbit Eo < E < 0
Case 2: Parabolic Orbit E = 0
Case 1: Hyberbolic Orbit E > 0
PRS Question
The radius and sign of the energy for the lowest energy orbit (circular
orbit) is given by
1. energy is positive,
L2
r0 =
µ Gm1m2
2. energy is positive,
L2
r0 =
2 µ Gm1m2
3. energy is negative,
L2
r0 =
µ Gm1m2
4. Energy is negative,
L2
r0 =
2 µ Gm1m2
PRS Answer: Circular Orbit
• The lowest energy state corresponds to a
circular orbit where the radius can be
found by finding the minimum of effective
potential energy
0=
dU effective
dr
L2 Gm1m2
=− 3 +
r2
µr
L2
r0 =
µ Gm1m2
• Energy of circular orbit
E0 = (U effective )
r = r0
=−
µ ( Gm1m2 )
2 L2
2
PRS Question
If the earth slows down due to tidal forces
will the moon’s angular momentum
1. increase
2. decrease
3. cannot tell from the information given
PRS Question
If the earth slows down due to tidal forces
will the radius of the moon’s orbit
1. increase
2. decrease
3. cannot tell from the information given
Orbit Equation
r0
r=
1 − ε cos θ
• Solution:
where the two constants are
• radius of circular orbit
• eccentricity
L2
r0 =
µ Gm1m2
⎛
⎞
2 EL2
ε = ⎜1 +
⎟
2
⎜ µ ( Gm m ) ⎟
1 2
⎝
⎠
1
2
Energy and Angular Momentum
• Energy:
E = E0 (1 − ε
1
2 2
)
where E0 is the energy of the ‘ground state’
E0 = (U effective )
r = r0
=−
µ ( Gm1m2 )
2
2 L2
• Angular momentum
L = ( r0 µ Gm1m2 )
1/ 2
where r0 is the radius of the ‘ground state’
Properties of Ellipse:
rminimum = r (θ = π ) =
r0
1+ ε
rmaximum = r (θ = 0 ) =
r0
1− ε
Semi-Major axis
a=
r0
1
1 ⎛ r0
r
r
+
=
+
( maximum minimum ) ⎜
2
2 ⎝ 1− ε 1+ ε
r0
Gm1m2
⎞
=
=
−
⎟
2
2E
⎠ 1− ε
location of the center of the ellipse
ε r0
= εa
x0 = rmaximum − a =
2
1− ε
Semi-Minor axis
Area
b=
(a
2
− x0 2 ) = a1/ 2 r01/ 2
A = π ab = π a 3/ 2 r01/ 2
Kepler’s Laws: Equal Area
• Area swept out in time ∆t
∆A 1 ⎛ ∆θ
= ⎜r
∆t 2 ⎝ ∆t
r ∆θ ) ∆r
(
⎞
⎟r +
2 ∆t
⎠
dA 1 2 dθ
= r
dt 2 dt
dθ
L
= 2
dt µ r
• Equal Area Law:
dA L
=
= constant
dt 2 µ
Kepler’s Laws: Period
• Area
A = π ab = π a 3/ 2 r01/ 2
2µ
dA = ∫ dt
∫
L
orbit
0
T
• Integral of Equal Area Law
2 µπ a 3/ 2 r01/ 2
2µ
T=
A=
L
L
• Period
• Period squared proportional to cube of the
major axis but depends on both masses
4µ 2 2 3
4π 2 µ a 3
4π 2 a 3
=
T = 2 π a r0 =
L
Gm1m2 G ( m1 + m2 )
2
Two Body Problem Revisited
• Elliptic Case: Each
mass orbits around
center of mass with
G G
G
G
G
G G
G m1r1 + m2r2 m2 ( r1 − r2 ) µ G
r1′ = r1 − R cm = r1 −
=
=
r
m1 + m2
m1 + m2
m2
G
µ G
r2′ = −
r
m2