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Conservation of Angular Momentum of Train on Disk

  1. Aug 4, 2017 #1
    1. The problem statement, all variables and given/known data
    A horizontal plywood disk with mass 6.90 kg and diameter 1.14 m pivots on frictionless bearings about a vertical axis through its center. You attach a circular model-railroad track of negligible mass and average diameter 1.04 m to the disk. A 1.40 −kg , battery-driven model train rests on the tracks. To demonstrate conservation of angular momentum, you switch on the train's engine. The train moves counterclockwise, soon attaining a constant speed of 0.790 m/s relative to the tracks. What is the magnitude and direction of the angular velocity?

    2. Relevant equations
    L=rp
    L=Iω
    v=ωr

    3. The attempt at a solution

    As the problem states, this is definitely a conservation of angular momentum problem. The train is a particle with angular momentum, so it's angular momentum is given by L=rp.

    Lt=rtps=Lt=rtmsvt

    Since the train is part of the system, when it gained this angular momentum, to disk must have also gained an equal amount of angular momentum in the opposite direction.

    Ld=Idωd=0.5mdrd2ωd
    Ld+Lt=0

    Setting them equal to each other gets:

    rtmtvt=0.5mdrd2ωd

    Solving for ωd gets 0.513 rad/s (as the magnitude), but this isn't the correct answer. What am I missing?
     
  2. jcsd
  3. Aug 4, 2017 #2

    jbriggs444

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    When someone writes down an equation like this, my first reaction is almost always:

    What is ##L_t##? What is ##r_t##? What is ##p_s##, what is ##m_s## and what is ##v_t##.

    Document your variable names.

    I would guess that ##L_t## is the angular momentum of the train and ##r_t## is the radius of the tracks. And I would guess that ##v_t## is the velocity of the train relative to the disc. But then what are ##p_s## and ##m_s##? One would have expected ##p_t## and ##m_t## as the linear momentum of the [short] train, and its mass respectively.

    There is a key gotcha here. By not carefully documenting the variable names, it is left unclear what ##v_t## is relative to. Is that supposed to be a velocity relative to the lab frame or a velocity relative to the tracks? The given in the problem is a velocity relative to the tracks. The relevant velocity you need is velocity relative to the non-rotating lab frame.

    What is ##L_d##, what is ##l_d##, what is ##\omega_d##, what is ##m_d##, what is ##r_d##?

    Guessing... ##L_d## is the angular momentum of the disk, ##I_d## is its moment of inertia, ##\omega_d## is the angular rotation rate of the disk [against a non-rotating reference], ##m_d## is the mask of the disk and ##r_d## is its radius.

    As for what went wrong... what about that gotcha I mentioned above?

    Edit: One last minor nitpick...
    There is a sign error there. ##L_d+L_t=0## yields ##L_d=-L_t##. You should not be "setting them equal". You should be "substituting into" the latter equation.

    [Documenting variable names is something that was drilled into our heads when learning computer programming in the '70's. It is a good practice in many other disciplines as well. Including physics].
     
    Last edited: Aug 4, 2017
  4. Aug 4, 2017 #3
    Thanks for your reply! Relative velocity has always been a weak point for me.

    I seem to have made a few typos. The subscripts "s" are supposed to be "t" for the train/track. The subscripts "d" are for "disk."

    So if the train is moving with a linear velocity of 0.790 m/s relative to the tracks, its linear velocity with respect to the Earth would be 0.790 m/s + vtracks,rel,earth. The linear velocity of the tracks and the disk (relative to the Earth) would be the same, since the tracks are attached to the disk. Using the velocity of the train with respect to the earth, its angular momentum equation becomes:

    Ltrain=rtracksmtrain(vtrain+vdisc)

    The angular velocity of the disc relative to the Earth would still be:

    Ldisc=0.5mdiscr2discωdisc

    I used the relationship v=rω to get angular velocity instead of linear velocity in the first equation:

    Ltrain=rtracksmtrain(vtrain+rdiscωdisc).

    Then I used the conservation of angular momentum to solve for ω, getting 0.815 rad/s, but this was incorrect. Was my setup wrong somewhere?
     
  5. Aug 4, 2017 #4

    jbriggs444

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    Ok. Then we're at least agreed on what the variable names mean.

    Without working through the arithmetic myself, it seems to me that you have a sign error. The velocity of the disk is in the opposite direction to the velocity of the train. If you are using a sign convention where the velocity of the disk is positive, that means that you have to subtract it, not add.
     
  6. Aug 4, 2017 #5
    Yes I forgot to account for the opposite direction.

    Using the equation:

    Ltrain=rtracksmtrain(vtrain-vdisc)

    I got the answer to be 0.374, which the website was nice enough to let me know that is was close, but not quite. I'm assuming my set-up is correct.

    If you could check through my arithmetic that would be great!

    Ltrain=Ldisc

    rtracksmtrain(vtrain-vdisc)=0.5mdiscr2discωdisc

    rtracksmtrainvtrain-rdiscωdiscrtracksmtrain=0.5mdiscr2discωdisc

    rtracksmtrainvtrain=rdiscωdiscrtracksmtrain+0.5mdiscr2discωdisc

    rtracksmtrainvtraindisc(0.5mdiscrdisc2+rdiscrtracksmtrain)

    I then divided by (0.5mdiscrdisc2+rdiscrtracksmtrain) to get my answer. I've triple checked to make sure I didn't make a typo in my calculator. Is it a coincidence my answer is close :/ ?
     
  7. Aug 4, 2017 #6

    jbriggs444

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    Dang. I hate checking arithmetic. Let's see if I can follow your formulas first.
    You've set ##L_{train}## equal to ##L_{disc}##. So we're using a sign convention where both can be positive. That's fine.
    You've substituted in your formulas for the angular momentum of both into the previous equality. That's good. And you're keeping everything symbolic for now. That's double-plus good.
    Wait a minute here. You have the velocity of the train (lab frame) being calculated as the velocity of the train (disk frame) minus the velocity of the disk. That much is good. But the relevant disk velocity is the velocity of the disk where the tracks are. You wrote down a formula for the velocity of the disk at its rim.

    *whew*. I don't have to check your arithmetic. It's the formulas that need work.
     
    Last edited: Aug 4, 2017
  8. Aug 4, 2017 #7
    Mhm....

    I'm not sure I follow. I think you're saying the left side of the equation is on the right track, but my disc velocity is incorrect.

    rtracksmtrainvtrain-rdiscωdiscrtracksmtrain=0.5mdiscr2discωdisc

    After some additional thought, it seems that I made an incorrect assumption that the linear velocities of the rims of the disk and train were equal, which is not the case since the radii is different. I'd need to calculate the linear velocity of the disk/train tracks at the radius of the train tracks, and use that instead of the velocity of the rim of the disk.

    So my equation should now be:

    rtracksmtrain(vtrain-vtracks)=Ltrain

    I would then set this equal to the angular momentum of the disk.

    I have to go to class now so I can't attempt a solution, but I thought I'd post this in case there's something wrong with my assessment here.
     
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