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Thread: Momentum or Energy?

  1. #1 Momentum or Energy? 
    Space - -1=epii's Avatar
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    Given a disc with a rotational inertia I containing two identical opposite opposing rocket nozzels on opposite sides of the disc(sort of like the Letter "Z"), which are tangent to the outside of the disc, with an amout of propellant (of negligable mass when compared to the mass of the disc) contained inside the disc, and that propellant burns and the gas created flows equally to the two nozzels such that it will spin up the disc to some rotational velocity without causing any translation. Its now desired that we want to spin up the same disk to double the rotational velocity by the addition of more propellant (again of neglible mass when compared to the mass of the disc) while adjusting the combustion chamber and nozzels so that all of the propelant is burned and expelled in the same amount of time as before. Would the propellant have to be doupled- a momentum problem? Or would we n4 times the propellant - an energy problem?


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    That is an interesting question. Yet, we will not do your homework, however, it is note worthy that the ending speed is 2 times the normal (angular velocity) and the ending mass is equal. And in the beginning the rockets also have to proppel the yet to burn mass. Think about it.


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  4. #3  
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    I did think about it, which is why I stated the weight of the propellant was negligable. Its not a home work problem actually, just a question about energy and momentum...
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    If the moment of inertial of the disk remains the same, but the final angular velocity is increased, that would mean you have to apply a greater torque at the perimeter of the disk with respect to the rotational axis. Think about how this can be achieved, and how torque, moment of inertia and angular acceleration are related. The answer is then straightforward.
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    The problem isn't any different than the problem of accelerating a rocket in a straight line. You can work it out for straight line acceleration, then just multiply by the radius.

    I think that assuming negligble mass will get you a nonsense answer, because the only way to accelerate the rocket is to give equal and opposite momentum to the exhaust gas. Since the mass is variable, you have to use the rocket equation.
    Tsiolkovsky rocket equation - Wikipedia, the free encyclopedia

    Think about what happens if you fire a series of bullets from a rifle, on a moving platform. On the first shot, almost all the energy goes into the bullet. If this were not true, a rifle would cause more damage to the shooter than the target. On the second shot, more kinetic energy is imparted to the moving rifle than the first shot, and less to the bullet, from the perspective of a person at rest. This is because the muzzle velocity is constant with respect to the rifle, not the ground. The bullet speed is less by the amount of the rifle's speed. So, you can't use conservation of energy to solve your problem.
    Last edited by Harold14370; December 14th, 2012 at 06:59 AM.
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  7. #6  
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    "that would mean you have to apply a greater torque at the perimeter of the disk with respect to the rotational axis. Think about how this can be achieved, and how torque, moment of inertia and angular acceleration are related. The answer is then straightforward."

    Thanks for thinking about it. But your in error in that the torque has to be increased, I could simply fire the rockets longer.
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    Quote Originally Posted by Harold14370 View Post
    The problem isn't any different than the problem of accelerating a rocket in a straight line. You can work it out for straight line acceleration, then just multiply by the radius.

    I think that assuming negligble mass will get you a nonsense answer, because the only way to accelerate the rocket is to give equal and opposite momentum to the exhaust gas. Since the mass is variable, you have to use the rocket equation.
    Tsiolkovsky rocket equation - Wikipedia, the free encyclopedia

    Think about what happens if you fire a series of bullets from a rifle, on a moving platform. On the first shot, almost all the energy goes into the bullet. If this were not true, a rifle would cause more damage to the shooter than the target. On the second shot, more kinetic energy is imparted to the moving rifle than the first shot, and less to the bullet, from the perspective of a person at rest. This is because the muzzle velocity is constant with respect to the rifle, not the ground. The bullet speed is less by the amount of the rifle's speed. So, you can't use conservation of energy to solve your problem.
    You are correct rotational and linear momentum are anlogous. But you are painfully in error when you said "almost all of the energy goes into the bullet" by newton's 3rd law the force on the gun is exactly equal and opposite to the force on the bullet. true? mgunagun=mbulletabullet Since a=change in V/change in time and the change in time is the same for the bullet and the gun then the change in time can be multiplied out so mgunvgun=mbulletvbullet

    but this gets us back to thte point of the original question. the bullets KE is proportional to its mass and the square of its velocity

    Does it take 2 or 4 times the energy to double a cars velocity?
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  9. #8  
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    by the way, you are also wrong in this case that neglecting the mass of the propellent would give you a nonsensical answer. You would be right if you were launching a rocket into space and a significant portion of its mass were consumed as fuel. Here we have a disc who's consumable mass is much much smaller then the mass of the disc and the mass of the consumed fuel is located near the center of the disc so its contribution to the rotational inertia as even less of a factor.
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  10. #9 true? 
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    Too soon
    Last edited by -1=epii; December 14th, 2012 at 03:46 PM.
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  11. #10  
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    Quote Originally Posted by -1=epii View Post
    I did think about it, which is why I stated the weight of the propellant was negligable. Its not a home work problem actually, just a question about energy and momentum...
    The idea about not doing someones homeworkk also has to do with the fact that simply solving a problem and giving the results goes only a small way in helping the person understand the phyysics. When I was tutoring physics you'd be amazed at how many people came to me at the last minute before a test and wanted me to work out homework problems for them. It'd take me about the same it would take anyone skilled in that section to solve the problem. The job of a tutor is to show the student the principles of physics in actions and how to use them to get from A to B in problem. That is where true learning comes from. We won't do your homework for you. But if you state a problem and show us at attempt by you to solve the problem then we'll be glad to help you.


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    Let's work out an example. Say the mass of the bullet is 1, the mass of the gun is 1000, and the muzzle velocity is 100. This means that by conservation of momentum, the recoil velocity is 100*(1/1000)=0.1. Now we calculate the kinetic energy, which is .05*M*v^2. For the bullet it is 0.5*1*100*100=5000. For the gun it is 0.5*1000*0.1*0.1=5. The bullet gets 1000 times as much kinetic energy as the gun.

    In your disc example, you will have to take into account the energy required to accelerate the propellant from the center of the disc to the spinning perimeter.
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    Quote Originally Posted by -1=epii View Post
    Thanks for thinking about it. But your in error in that the torque has to be increased, I could simply fire the rockets longer.
    You are contradicting your own exercise. In your OP you stated :

    while adjusting the combustion chamber and nozzels so that all of the propelant is burned and expelled in the same amount of time as before.
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  14. #13  
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    Quote Originally Posted by pmb View Post
    Quote Originally Posted by -1=epii View Post
    I did think about it, which is why I stated the weight of the propellant was negligable. Its not a home work problem actually, just a question about energy and momentum...
    The idea about not doing someones homeworkk also has to do with the fact that simply solving a problem and giving the results goes only a small way in helping the person understand the phyysics. When I was tutoring physics you'd be amazed at how many people came to me at the last minute before a test and wanted me to work out homework problems for them. It'd take me about the same it would take anyone skilled in that section to solve the problem. The job of a tutor is to show the student the principles of physics in actions and how to use them to get from A to B in problem. That is where true learning comes from. We won't do your homework for you. But if you state a problem and show us at attempt by you to solve the problem then we'll be glad to help you.


    Pete
    Thanks for your pedantic self-righteousness but, this isn’t homework, as I said before. It’s been more than 30 years since was a research assistant who sometimes tutored; homework is sort of a thing of the past for me. It’s simply an argument I had hope to settle. I am certain I am correct, but so is the person I am arguing with. I worked out the answer both ways: it is either 2 times the propellant or 4 times the propellant to achieve twice the spin rate. But none of you have the guts to go on the record as to which it is – come on put your money down on 2 or 4
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  15. #14  
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    Quote Originally Posted by -1=epii View Post
    But none of you have the guts to go on the record as to which it is come on put your money down on 2 or 4
    So then how did the two of your arrive at the figures "2" and "4" ?
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  16. #15  
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    wow,

    ok - thought that was clear -

    Energy is proportional to the square of velocity - if the velocity is doubled then the energy is 2-squared

    E=1/2 Iω2

    (Efast=1/2 I(2ω)2)/(Eslow=1/2 I(ω)2)

    Efast/Eslow=〖(2ω)〗22

    Efast/Eslow =4

    Efast=4Eslow


    thus it takes 4 times the propellant

    Momentum is proportional to the velocity - if the velocity is doubled the momentum is doubled - the math here is trivial
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    Quote Originally Posted by -1=epii View Post
    wow,

    ok - thought that was clear -

    Energy is proportional to the square of velocity - if the velocity is doubled then the energy is 2-squared

    E=1/2 Iω2

    (Efast=1/2 I(2ω)2)/(Eslow=1/2 I(ω)2)

    Efast/Eslow=〖(2ω)〗22

    Efast/Eslow =4

    Efast=4Eslow


    thus it takes 4 times the propellant

    Momentum is proportional to the velocity - if the velocity is doubled the momentum is doubled - the math here is trivial
    If you get two different answers by working it out two different ways, that should be a clue that there is something wrong with your assumptions. In this case, your assumption of negligible mass of propellant gives you nonsensical results, as I said before. If the mass is negligible, your disc isn't going anywhere.
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  18. #17  
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    ok then whats the answer??
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  19. #18  
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    no one?
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    To a good approximation, you can use conservation of momentum to calculate the additional propellant required. Conservation of energy doesn't work, because most of the energy is going into the propellant initially, with a greater proportion going to the vehicle as it speeds up. You can prove this to yourself by calculating the momentum and energy after firing a second shot, in the example from post 11.
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  21. #20  
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    Yea we have someone willing to put their money down.

    But I am afraid I dont agree. a car traveling at a speed V has a certain energy. The same car traveling at twice the velocity has 4 times the energy. If I have to provide that energy lets say by using gasoline to fuel the engine it will take me 4 times the gasoline to get the car up to twice its velocity
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    You have not paid attention to what I wrote. The energy of the fuel used does not all go into powering the vehicle. Most of it initially goes into the kinetic energy of the exhaust gas. I have shown that in the calculation I did in post 11. This changes as the vehicle accelerates.
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  23. #22  
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    Quote Originally Posted by Harold14370 View Post
    To a good approximation, you can use conservation of momentum to calculate the additional propellant required. Conservation of energy doesn't work, because most of the energy is going into the propellant initially, with a greater proportion going to the vehicle as it speeds up. You can prove this to yourself by calculating the momentum and energy after firing a second shot, in the example from post 11.
    Let's go back to your rocket-disk example. You asked how much mass you would need if you modified the nozzles so that the new mass amount was expelled in the same time . Now I'm going to assume that you meant without changing the exhaust velocities of the rockets (otherwise, all bets are off). So let's imagine that you double your reaction mass. Without changing the exhaust velocity, in order to expel twice the mass an an equal amount of time, you double the thrust. You could do this by modifying your rocket or just doubling the number of rocket nozzles (twice as many rockets will burn the fuel twice as fast allowing you to use up your fuel in the same time.). Twice the Thrust means twice the force acting on the disk, and twice the force means twice the acceleration, and twice the acceleration in an equal amount of time leads to twice the velocity. So twice the velocity means ~twice the mass.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


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  24. #23  
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    And just for giggles, here's some math using the rocket equation given in post #5.

    The equation in the link uses M0 and M1 for the initial and final masses. In this problem , M0 = M+m and M1=M, where M is the mass of the disk and m the reaction mass.

    We are looking for what factor(x) we need to apply to m to get a increased factor of 2 for our delta v.

    We set that up like thus;



    And then solve for x

    We lose Ve (the exhaust velocity) and rearrange:



    Raise e to the power of both sides



    rearrange the exponents on the left



    Which allows us to simplify and lose the other natural log



    Now its just a matter of simple algebra to find x











    As m gets smaller with respect to M, the factor you need to multiple m by to double the speed approaches 2.
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