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Conservation of angular momentum (hard)

  1. Apr 1, 2004 #1
    The rear view of a space capsule that is rotating about its longitudinal axis at 6rev/min. The occupants want to stop this rotation. They have small jets counted tangentially at a distance R=3m from the axis, as indicated, and can eject 10g/s of gas from jet with a nozzle velocity of 800m/s. For how long must they turn on these jets to stop the rotation? The moment of inertia of the ship about its axis assumed to be constant is 4000kg.m^2.

    They must keep turning it until it stops rotating! hehe..
    I had no problem with rotation though, but I am really weak in the conservation consept, and it doesn't seem to get it. I'm not looking for the answer as much as I want to learn the concept. So if you have time please explain it to me, I would really appreciate it!
  2. jcsd
  3. Apr 1, 2004 #2


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    Just as linear momentum is mv so rotational momentum is Iω. You know both of those for the space capsule so you can calculate its momentum. If you want to stop the rotation, that momentum has to go somewhere! How many grams of gas at 3m from the axis and 800 m/s will have that angular momentum? Okay, how many second will it take to eject that much gas?
  4. Apr 1, 2004 #3
    [tex]L = I\omega[/tex]

    You're given I and you're given [itex]\omega[/itex], so you know how much momentum the capsule has initially.


    [tex] L = pr = mvr[/tex]
    in the special case that the linear momentum is acting exactly perpendicular to the axis of rotation.

    The gas is shooting out at 800m/s from a nozzle 3m off the radius perfectly perpendicular to the axis. That gives us 2 of the three variables on the right-hand side of the second equation for L. The left-hand side is given by the first equation. So we have 1 variable and one equation, solve for mass.

    Now that you know how much mass must be shot out, all you have to do is calculate how long it will take for that to happen. You know that mass is being shot out at 10g/s, so just divide the mass by .01kg/s to get the time.

  5. Apr 2, 2004 #4


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    Linear momentum (like the gas) has a formula like this

    [tex]F \Delta t = m \Delta v[/tex]

    Angular momentum has a formula like this

    [tex]T \Delta t = I \Delta \omega[/tex]

    Torque (T) is really force x distance where the distance is the radius, so sub that

    [tex]Fr \Delta t = I \Delta \omega[/tex]

    F delta t is also equal to m delta v, so you can substitute that in as well

    [tex]rm \Delta v = I \Delta \omega[/tex]

    Now just fill in values. The variable you are looking for is m. You are given a rate of mass, you need the total mass, and you want to find time.
    Last edited: Apr 2, 2004
  6. Apr 2, 2004 #5
    you mean
    L = m * 800m/s * 3m
    and L = I*w= 4000kg.m^2 * 0.6286 rad/sec = 2513.27 kgm^2/s

    so 2513.27 = m * 800m/s * 3m
    and m = 2513.27 /(800m/s*3m)= 2513.27 / 2400 = 1.04 kg (I.m not sure of the rounding again)

    t = 1.04kg/0.01kg/s = 104s ?
  7. Apr 2, 2004 #6
    As Shawn said:
    [tex]rm \Delta v = I \Delta \omega[/tex]
    You need to find the amount of mass that needs to be released in order to stop the capsule, so:
    [tex]m = \frac{I \Delta \omega }{\Delta vr}[/tex]
    [tex]\Delta v[/tex] is the initial velocity minus the final velocity, and since the final velocity is 0 it's just the initial velocity. To find it, use:
    [tex]V = \omega r[/tex]
    And to find omega:
    [tex]\omega = 2\pi f[/tex]
    The frequency, f, is 6rev/min or 0.1Hz. So the final equation is:
    [tex]m = \frac{I 2\pi f}{2\pi fr^2} = \frac{I}{r^2}[/tex]
    Now, the mass is ejected at 10 grams per second. Once you know how much mass needs to be ejected, you can find the time.
    Last edited: Apr 2, 2004
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