Difference between revisions of "Forest UCM Ch3 Rockets"

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Consider a rocket of mass <math>m</math> moving at a speed <math>v</math> ejecting rocket fuel at a speed <math>u</math>.
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=Rocket problem 1=
  
  
[[File:Forest_UCM_Ch3_Rockets_Fig.png]][[File:Forest_UCM_Ch3_Rockets_Fig.xfig.txt]]
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Remember the hobo problem?  This is how rockets are propelled by expending fuel (mass).
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Lets first think about the inverse rocket where mass is added instead of expelled.
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Consider a railroad car with frictionless wheels moving at speed <math>v</math> on level ground.
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A hobo handing from a tree drops down onto the railroad car as it is moving under the tree.
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What is the final velocity of a railroad car after the  hobo jumps on.
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;Conservation of momentum
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:<math>Mv = (M+m) v_f</math>
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:<math>v_f = \frac{M}{M+m} v</math>
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The railroad car slows down after the hobo jumps on.
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 +
Rockets work in a similar fashion.  They speed up after the fuel "jumps" off.
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=Rocket Problem 2=
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Consider a rocket of mass <math>m</math> moving at a speed <math>v</math> ejecting rocket fuel for propulsion.
 +
 
 +
 
 +
[[File:Forest_UCM_Ch3_Rockets_Fig.png | 400 px]][[File:Forest_UCM_Ch3_Rockets_Fig.xfig.txt]]
  
  
 
<math>m =</math> mass of Fuel + Rocket
 
<math>m =</math> mass of Fuel + Rocket
  
<math>v_F =</math> velocity of Fuel relative to the Rocket
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<math>-dm =</math> mass of Fuel ejected over time interval <math>dt</math> ( <math>dm</math> = mass lost by rocket < 0)
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<math>u =v_{FR}</math> velocity of Fuel relative to the Rocket
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<math>v =</math> velocity of rocket relative to the ground before ejecting fuel of mass <math>(-dm)</math>
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<math>v+dv =v_{RG}</math> velocity of the rocket relative to the ground after ejecting fuel
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<math>V_FG =</math> Velocity of the fuel with respect to the ground
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= Apply Conservation of Momentum=
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:<math>mv = (-dm)V_{FG} + (m-(-dm))(v+dv)</math>
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:<math>dmV_{FG} = mdv + dm (v+dv)</math>
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The velocity of the fuel with respect to the ground<math> (V_{FG})</math> may be written as the vector sum of the rocket's velocity with respect to the ground <math>(V_{RG})</math> and the velocity of the fuel with respect to the rocket <math>(V_{FR})</math> :Galilean transormation
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:<math>\vec{V}_{FG} - \vec{V}_{FR} + \vec{V}_{RG}</math>
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assuming 1-D motion and using the velocity variables defined above
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:<math>V_{FG} = -u + (v+dv)</math>
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substituting
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:<math>dm \left (-u + (v+dv) \right ) = mdv + dm (v+dv)</math>
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:<math>-udm  = mdv </math>
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==solving for the velocity==
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: <math>\int_{v_0}^v  = \int_{m_0}^m -u \frac{dm}{m}</math>
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: <math>v - v_0 = u \ln \left (\frac{m_0}{m} \right )</math>
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: <math>v  = v_0 + u \ln \left (\frac{m_0}{m} \right )</math>
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where <math>m_0</math> is the initial Rocket mass
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==In terms of Thrust==
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Defining
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:<math>\mbox{THRUST} = - \dot{m} u</math>
  
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Then
  
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:<math>m \dot{v} =</math> THRUST = Force on rocket as a result of expelling fuel
  
 
[[Forest_UCM_MnAM#Rockets]]
 
[[Forest_UCM_MnAM#Rockets]]

Latest revision as of 13:59, 13 September 2014

Rocket problem 1

Remember the hobo problem? This is how rockets are propelled by expending fuel (mass).

Lets first think about the inverse rocket where mass is added instead of expelled.

Consider a railroad car with frictionless wheels moving at speed [math]v[/math] on level ground.

A hobo handing from a tree drops down onto the railroad car as it is moving under the tree.

What is the final velocity of a railroad car after the hobo jumps on.

Conservation of momentum
[math]Mv = (M+m) v_f[/math]
[math]v_f = \frac{M}{M+m} v[/math]

The railroad car slows down after the hobo jumps on.

Rockets work in a similar fashion. They speed up after the fuel "jumps" off.

Rocket Problem 2

Consider a rocket of mass [math]m[/math] moving at a speed [math]v[/math] ejecting rocket fuel for propulsion.


Forest UCM Ch3 Rockets Fig.pngFile:Forest UCM Ch3 Rockets Fig.xfig.txt


[math]m =[/math] mass of Fuel + Rocket

[math]-dm =[/math] mass of Fuel ejected over time interval [math]dt[/math] ( [math]dm[/math] = mass lost by rocket < 0)

[math]u =v_{FR}[/math] velocity of Fuel relative to the Rocket

[math]v =[/math] velocity of rocket relative to the ground before ejecting fuel of mass [math](-dm)[/math]

[math]v+dv =v_{RG}[/math] velocity of the rocket relative to the ground after ejecting fuel

[math]V_FG =[/math] Velocity of the fuel with respect to the ground

Apply Conservation of Momentum

[math]mv = (-dm)V_{FG} + (m-(-dm))(v+dv)[/math]
[math]dmV_{FG} = mdv + dm (v+dv)[/math]

The velocity of the fuel with respect to the ground[math] (V_{FG})[/math] may be written as the vector sum of the rocket's velocity with respect to the ground [math](V_{RG})[/math] and the velocity of the fuel with respect to the rocket [math](V_{FR})[/math] :Galilean transormation

[math]\vec{V}_{FG} - \vec{V}_{FR} + \vec{V}_{RG}[/math]

assuming 1-D motion and using the velocity variables defined above

[math]V_{FG} = -u + (v+dv)[/math]

substituting

[math]dm \left (-u + (v+dv) \right ) = mdv + dm (v+dv)[/math]
[math]-udm = mdv [/math]

solving for the velocity

[math]\int_{v_0}^v = \int_{m_0}^m -u \frac{dm}{m}[/math]
[math]v - v_0 = u \ln \left (\frac{m_0}{m} \right )[/math]
[math]v = v_0 + u \ln \left (\frac{m_0}{m} \right )[/math]

where [math]m_0[/math] is the initial Rocket mass

In terms of Thrust

Defining

[math]\mbox{THRUST} = - \dot{m} u[/math]

Then

[math]m \dot{v} =[/math] THRUST = Force on rocket as a result of expelling fuel

Forest_UCM_MnAM#Rockets