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Experiment 09: Angularmomentum

Experiment 09: Angularmomentum

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Goals Investigate conservation of angular momentum

and kinetic energy in rotational collisions.

Measure and calculate moments of inertia.

Measure and calculate non-conservative work inan inelastic collision.

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 Apparatus Connect output of

phototransistor tochannel A of 750.

Connect output oftachometergenerator to channelB of 750.

Connect powersupply.

Red button ispressed: Power isapplied to motor.

Red button isreleased: Rotorcoasts: Read outputvoltage using

DataStudio.

Use black sticker or tape on white plastic

rotor for generator calibration.

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4

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.

Average the outputvoltage over the same 10

periods.

Time 10 periods to measure ω.

Then calculate ω for 1 Voutput.

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Measure IR

results

Measure and record αup and αdown.

For your report, calculate IR:

up

up down

( ) R

mr g r   I 

α 

α α 

−=

−

down f R I τ α =

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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 i I m r r ω = +

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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.

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Kepler Problem and

Planetary Motion

8.01t

Nov 15, 2004

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Kepler Problem• Find the motion of two

bodies under theinfluence of a

gravitational force

using Newtonianmechanics

( )  1 2

1,2   2   ˆ

m m

r G r = −F r

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Solution of One Body Problem

• Solving the problem means finding the

distance from the origin andangle as functions of time

• Equivalently, finding the distance from the

center as a function of angle

• Solution:

( )r t ( )t θ 

( )r  θ 

0

1 cos

r r 

ε θ =

−

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Constants of the Motion• Velocity

• Angular Momentum

• Energy

ˆˆdr d 

r 

dt dt  

θ = +v r   θ

2   1 21

2

Gm m E v

r µ = −

2

tangential 

d  L rv r 

dt 

θ µ µ = =

2 2

2   dr d v r 

dt dt  

θ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 2

1 21

2

Gm mdr d 

 E r dt dt r  

θ 

µ 

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

2 2

1 2

2

1

2 2

Gm mdr L E 

dt r r  

µ 

µ 

⎛ ⎞= + −⎜ ⎟⎝ ⎠

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Reduction to One Dimensional

Motion

• Reduce the one body problem in two

dimensions to a one body problem movingonly in the r-direction but under the action of a

repulsive force and a gravitational force

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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

( ) 2

 L

U r  r =

2

3

 L F 

r µ 

=

2

3 F 

r =

2

2 F 

r =

2

2 F  r =

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One Dimensional Description

• Energy

• Kinetic Energy

• Effective Potential Energy

• Repulsive Force

• Gravitational force

2 2

1 2

2

1 1

2 2  effective

Gm mdr L E K U 

dt r r  µ 

µ 

⎛ ⎞= + − = +⎜ ⎟

⎝ ⎠ 21

2

dr  K 

dt µ 

⎛ ⎞=   ⎜ ⎟⎝ ⎠

2

1 222effective

Gm m LU  r r µ = −

2 2

2 32centrifugal 

d L L F 

dr r r  µ 

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

1 2

2

 gravitational 

 gravitational 

dU    Gm m F 

dr r = − = −

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Energy Diagram2

1 2

22effectiveGm m L

U  r r µ = −

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

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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

0

1 2

 Lr 

Gm mµ =

2

0

1 22 Lr 

Gm mµ =

2

0

1 22

 Lr 

Gm mµ =

2

0

1 2

 Lr 

Gm mµ 

=

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PRS Answer: Circular Orbit

• The lowest energy state corresponds to a

circular orbit where the radius can befound by finding the minimum of effective

potential energy

• Energy of circular orbit

( )   ( )0

2

1 20   22

effectiver r 

Gm m E U 

 L

µ 

== = −

2

1 2

3 20

  effectivedU    Gm m L

dr r r  µ = = − +

2

0

1 2

 L

r  Gm mµ =

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PRS QuestionIf the earth slows down due to tidal forces

will the radius of the moon’s orbit

1. increase2. decrease

3. cannot tell from the information given

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Orbit Equation• Solution:

where the two constants are

• radius of circular orbit

• eccentricity

0

1 cos

r r 

ε θ =

−

2

0

1 2

 Lr Gm mµ 

=

( )

122

2

1 2

21

  EL

Gm mε 

µ 

⎛ ⎞= +⎜ ⎟

⎜ ⎟⎝ ⎠

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Energy and Angular Momentum

• Energy:

where E0 is the energy of the ‘ground state’

• Angular momentum

where r 0 is the radius of the ‘ground state’

( )1

2 20   1 E E    ε = −

( )

  ( )0

2

1 2

0  22

effectiver r 

Gm m E U 

 L

µ 

=

= = −

( )1/ 2

0 1 2

 L r Gm mµ =

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Properties of Ellipse:

( )   001

maximumr r r   θ 

ε = = =

−

( )   01

minimum

r r r   θ π 

ε = = =

+

( )   0 0 0   1 221 12 2 1 1 1 2maximum minimumr r r    Gm ma r r 

 E ε ε ε ⎛ ⎞= + = + = = −⎜ ⎟− + −⎝ ⎠

Semi-Major axis

location of the center of the ellipse

Semi-Minor axis

 Area

00   2

1maximum

r  x r a a

ε ε 

ε 

= − = =

−

( )2 2 1/ 2 1/ 20 0b a x a r  = − =

3/ 2 1/ 2

0ab a r  π π = =

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

• Equal Area Law:

( )12 2

r    r r r 

t t t 

θ θ    ∆∆ ∆ ∆⎛ ⎞= +⎜ ⎟∆ ∆ ∆⎝ ⎠

21

2

dA d r 

dt dt  

θ =2

d L

dt r 

θ 

µ =

2

dA Lconstant 

dt    µ = =

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Kepler’s Laws: Period• Area

• Integral of Equal Area Law

• Period

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

3/ 2 1/ 2

0 A ab a r π π = =

0

2  T 

orbit 

dA dt   L

µ =∫ ∫

3/ 2 1/ 2

022   a r T A L L

µπ µ = =

( )

2 2 3 2 32 2 3

02

1 2 1 2

4 4 4a aT a r 

Gm m G m m

µ π µ π  π = = =

+

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Two Body Problem Revisited• Elliptic Case: Each

mass orbits aroundcenter of mass with

( )2 1 21 1 2 21 1 1

1 2 1 2 2

cm

mm m

m m m m m

µ −+′ = − = − = =+ +

r rr rr r R r r

2

2mµ ′ = −r r