Experiment 09: Angular momentumExperiment 09: Angular momentum

GoalsInvestigate 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.

ApparatusConnect 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.

4

Calibrate tachometer-generatorSpin motor up to full speed, let it coast. Measure and plot voltages for 0.25 s period. Sample Rate: 5000 Hz, and Sensitivity: Low.

Average the output voltage over the same 10 periods.

Time 10 periods to measure ω.

Then calculate ω for 1 V output.

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:Sensitivity: Low Sample rate 500 Hz.Delayed start: NoneAuto 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:

up

up down

( )R

mr g rI

αα α

−=

−downf RIτ α=

Fast collision

Find ω 1 (before) and ω 2 (after), estimate δt for collision.

Calculate

Falls below 0.5V1 sec200 HzLow

Auto StopDelayed StartSample RateSensitivity

2 212 ( )W o iI m r rω= +

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

8.01tNov 15, 2004

Kepler’s Laws1. 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

( ) 1 21,2 2

ˆm mr Gr

= −F 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

( )2

1,2 2ˆF d

dtµ=

rr r

1 2

1 2

m mm m

µ =+

Solution of One Body Problem

• Solving the problem means finding the distance from the origin and angle as functions of time

• Equivalently, finding the distance from the center as a function of angle

• Solution:

( )r t( )tθ

( )r θ

0

1 cosrr

ε θ=

−

Constants of the Motion• Velocity

• Angular Momentum

• Energy

ˆˆdr drdt dt

θ= +v r θ

2 1 212

GmmE vr

µ= −

2tangential

dL rv rdtθµ µ= =

2 22 dr dv r

dt dtθ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 21 21

2Gmmdr dE r

dt dt rθµ

⎡ ⎤⎛ ⎞ ⎛ ⎞= + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

2 21 2

2

12 2

Gmmdr LEdt r r

µµ

⎛ ⎞= + −⎜ ⎟⎝ ⎠

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 QuestionSuppose the potential energy of two particles (reduced mass µ) is given by

where r is the relative distance between the particles. The force between the particles is

1. attractive and has magnitude

2. repulsive and has magnitude

3. attractive and has magnitude

4. repulsive and has magnitude

2

2

1( )2LU rrµ

=

2

3

LFrµ

=

2

3

LFrµ

=

2

2LFrµ

=

2

2LFrµ

=

One Dimensional Description

• Energy

• Kinetic Energy

• Effective Potential Energy

• Repulsive Force

• Gravitational force

2 21 2

2

1 12 2 effective

Gm mdr LE K Udt r r

µµ

⎛ ⎞= + − = +⎜ ⎟⎝ ⎠

212

drKdt

µ ⎛ ⎞= ⎜ ⎟⎝ ⎠

21 2

22effectiveGm mLU

r rµ= −

2 2

2 32centrifugald L LFdr r rµ µ

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

1 22

gravitationalgravitational

dU GmmFdr r

= − = −

Energy Diagram2

1 222effectiveGm mLU

r rµ= −

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 QuestionThe radius and sign of the energy for the lowest energy orbit (circular

orbit) is given by

1. energy is positive,

2. energy is positive,

3. energy is negative,

4. Energy is negative,

2

01 2

LrGm mµ

=

2

01 22

LrGm mµ

=

2

01 22

LrGm mµ

=

2

01 2

LrGm mµ

=

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

• Energy of circular orbit

( ) ( )0

21 2

0 22effective r r

GmmE U

Lµ

== = −

21 2

3 20 effectivedU GmmLdr r rµ

= = − +

2

01 2

LrGm mµ

=

PRS Question

If the earth slows down due to tidal forces will the moon’s angular momentum

1. increase2. decrease3. 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. increase2. decrease3. cannot tell from the information given

Orbit Equation• Solution:

where the two constants are

• radius of circular orbit

• eccentricity

0

1 cosrr

ε θ=

−

2

01 2

LrGm mµ

=

( )

122

21 2

21 ELGmm

εµ

⎛ ⎞= +⎜ ⎟

⎜ ⎟⎝ ⎠

Energy and Angular Momentum

• Energy:

where E0 is the energy of the ‘ground state’

• Angular momentum

where r0 is the radius of the ‘ground state’

( )1

2 20 1E E ε= −

( ) ( )0

21 2

0 22effective r r

Gm mE U

Lµ

== = −

( )1/ 20 1 2L r Gm mµ=

Properties of Ellipse:

( ) 001maximumrr r θ

ε= = =

−

( ) 0

1minimumrr r θ π

ε= = =

+

( ) 0 0 0 1 22

1 12 2 1 1 1 2maximum minimum

r r r Gm ma r rEε ε ε

⎛ ⎞= + = + = = −⎜ ⎟− + −⎝ ⎠

Semi-Major axis

location of the center of the ellipse

Semi-Minor axis

Area

00 21maximum

rx r a aε εε

= − = =−

( )2 2 1/ 2 1/ 20 0b a x a r= − =

3/ 2 1/ 20A ab a rπ π= =

Kepler’s Laws: Equal Area• Area swept out in time ∆t

• Equal Area Law:

( )12 2

rA rr rt t t

θθ ∆∆ ∆ ∆⎛ ⎞= +⎜ ⎟∆ ∆ ∆⎝ ⎠

212

dA drdt dt

θ= 2

d Ldt rθ

µ=

2dA L constantdt µ

= =

Kepler’s Laws: Period• Area

• Integral of Equal Area Law

• Period

• Period squared proportional to cube of the major axis but depends on both masses

3/ 2 1/ 20A ab a rπ π= =

0

2 T

orbit

dA dtLµ

=∫ ∫

3/ 2 1/ 2022 a rT A

L Lµπµ

= =

( )2 2 3 2 3

2 2 302

1 2 1 2

4 4 4a aT a rL Gm m G m mµ π µ ππ= = =

+

Two Body Problem Revisited• Elliptic Case: Each

mass orbits around center of mass with

( )2 1 21 1 2 21 1 1

1 2 1 2 2cm

mm mm m m m m

µ−+′ = − = − = =+ +

r rr rr r R r r

22m

µ′ = −r r