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Rocket sled

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A rocket sled moves along a horizontal plane, and is retarded by a friction force ##f_{friction} = \mu W##, where ##\mu## is the kinetic friction constant and ##W## is the weight of the sled.
    The sled's initial mass is ##M_0##, and its rocket engine expels mass at a constant rate ##\displaystyle \frac{dM}{dt} = \gamma##; the expelled mass has constant speed ##v_0## relative to the rocket.
    The rocket sled starts from rest and the engine stops when half the sled's total mass is gone. Find an expression for the maximum speed.

    2. Relevant equations
    Rocket equation

    3. The attempt at a solution

    We start by using the rocket equation, and defining that movement to the right is positive and movement to the left is negative, and that the sled is moving to the right:

    ##\displaystyle \frac{dP}{dt} = M\frac{dv}{dt} - v_0 \frac{dM}{dt}##

    then since ##M = M_0 + \gamma t## and ##\displaystyle \frac{dM}{dt} = \gamma##

    ##\displaystyle \frac{dP}{dt} = (M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma##

    Then since there is external friction force acting on the system, we have that ##\displaystyle \frac{dP}{dt} = (M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma = - \mu (m_0 + \gamma t) g##

    then

    ##\displaystyle \frac{dv}{dt} = \frac{v_0 \gamma - \mu (M_0 + \gamma t) g}{M_) + \gamma t} = \frac{v_0 \gamma}{M_0 + \gamma t} - \mu g##

    Solving for velocity, we find that

    ##\displaystyle v = v_0 \log (1 + \frac{\gamma t}{M_0}) - \mu g t##

    Now the final velocity of the rocket sled is when the mass expelled is half of the initial mass: ##M_0 + \gamma t_f = \frac{1}{2} M_0 \implies t_f = \frac{-m_0}{2 \gamma}##. Plugging this in to our formula for velocity, we get ##\displaystyle v_f = \mu g \frac{M_0}{2 \gamma} - v_0 \log (2)##

    Is this the correct expression for the maximum velocity?
     
  2. jcsd
  3. Sep 25, 2016 #2

    kuruman

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    Your starting equation does not look right. dp/dt is the generalized form of ma that takes into account the changing mass. The right side should be the sum of all the forces in the horizontal direction. Also, what is v0 doing there and where did the negative sign come from?
     
  4. Sep 25, 2016 #3
    I do have the sum of the horizontal forces in the horizontal direction. That is what ##(M_0 + \gamma t)\frac{dv}{dt} - v_0 \gamma = - \mu (m_0 + \gamma t) g##, where the LHS is the rate at which the momentum changes and the RHS is the sum of the forces, which is only the friction force acting to the left since the rocket is moving to the right. Also, ##v_0## is the relative velocity of the expelled mass to the rocket sled (I'm not sure why it is called ##v_0##, but that is how is stated in my book. You could just pretend that it is ##u## or something, as it has nothing to do with the initial velocity of the rocket sled. Also, the negative sign where? The negative sign in the beginning equation comes from the rocket equation.
     
  5. Sep 25, 2016 #4

    kuruman

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    I see, v0 confused me a bit. The way you use γ is also confusing. I would use a positive γ and then write M(t) = M0 - γt. The final equation would be easier to read and interpret then. Also, you are looking for the maximum speed. You calculated the final speed. Is that the maximum? It would be if v(t) were monotonically increasing. Is it? It's hard to tell with γ being implicitly negative.
     
  6. Sep 25, 2016 #5
    When I use negatives I get ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t##. This doesn't seem like a monotonically increasing function, but it seems like it would have to be because until the engines are cut off the force from the sled is greater than the friction, which is why it is accelerating in the first place
     
  7. Sep 25, 2016 #6

    kuruman

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    The function is what it is and does not have to be anything else. How can you tell if it has an extremum, a maximum or a minimum?
     
  8. Sep 25, 2016 #7
    Well if the function is monotonically increasing, then the maximum would have to be when the engines turn off, because at that point friction kicks in immediately. If the function is not monotonically increasing, then we could do the first derivative test to determine if there is an extremum.
     
  9. Sep 25, 2016 #8

    kuruman

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    Right, do the first derivative rule and see what you get. If there is an extremum, you need to determine if it's a maximum or a minimum.
    Friction "kicked in" as soon as the sled started to move. What you want to say is that if the function is monotonically increasing, the maximum value is reached at the moment when the rocket engine shuts down because after that the velocity will be monotonically decreasing.
     
  10. Sep 25, 2016 #9
    So I got ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## as an extremum. When I plug this in to the derivative of the acceleration, I get ##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##, which is negative because ##v_0 < 0## (right?).
     
  11. Sep 26, 2016 #10

    kuruman

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    It depends. Look at the derivation of the rocket equation that you used. If the relative velocity of the expelled gases is written as v + v0, it is negative; if it is written as v - v0, it is positive. I have seen both forms used.
     
  12. Sep 26, 2016 #11
    In the derivation, it is written as ##v + v_0##, so it would be negative
     
  13. Sep 26, 2016 #12

    kuruman

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    OK, I agree.
    You mean the derivative of the velocity, not acceleration. Anyway, do you see what you need to do now?
     
  14. Sep 26, 2016 #13
    Do I plug ##t = \frac{M_0}{\gamma} + \frac{v_0}{\mu g}## into ##\displaystyle v = v_0 \log (\frac{M_0}{M_0 - \gamma t}) - \mu g t## to find the minimum velocity?
     
  15. Sep 26, 2016 #14

    kuruman

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    You could do that, but how will you be able to tell whether the velocity is a minimum or a maximum at that time? Also, how do you know that the rocket fuel will last until that time?
     
  16. Sep 26, 2016 #15
    Didn't I just find at that time the velocity is at a minimum? Also, could I just take the difference between the time when the engines shut off and the time at which be have an extemum to determine which one is after the other?
     
  17. Sep 26, 2016 #16

    kuruman

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    You found the time when the velocity is an extremum, this means either a maximum or a minimum.
    You can try that, but without numbers for the variables (except g), how can you tell which is the shorter time? All you can hope for is a condition involving an inequality.

    Think this through.
    1. If the fuel runs out first, no extremum is reached, so it doesn't matter if the extremum is a maximum or a minimum.
    2. If the extremum is reached, and it is a maximum, then that maximum is the maximum speed.
    3. If the extremum is reached, and it is a minimum, then the final speed is the maximum speed.
     
  18. Sep 26, 2016 #17
    Isn't the extremum a minimum since we we plug it into the the derivative of the acceleration we get a value that is always negative, namely
    ##v_0 (\frac{\gamma}{M_0 - \gamma t})^2##?
     
  19. Sep 26, 2016 #18

    kuruman

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    f(x) = x2 has a minimum at x = 0. Its second derivative at x = 0 is positive.
     
  20. Sep 26, 2016 #19
    But I thought that I was using the second derivative test to test the sign of the extremum...
     
  21. Sep 27, 2016 #20
    One more thing. Why would the author give me the information that the rocket's engines turn off when it is half its initial mass if this is not the exact maximum time?
     
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