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USAPhO 2009 F=MA exam #25, (torque, rotational energy)

  1. Jan 27, 2014 #1
    http://www.aapt.org/physicsteam/2010/upload/2009_F-maSolutions.pdf

    1. The problem statement, all variables and given/known data

    Two discs are mounted on thin, lightweight rods oriented through their centers and normal to the discs.These axles are constrained to be vertical at all times, and the discs can pivot frictionlessly on the rods. The discs have identical thickness and are made of the same material, but have differing radii r1 and r2. The discs are given angular velocities of magnitudes ω1 and ω2, respectively, and brought into contact at their edges. After the discs interact via friction it is found that both discs come exactly to a halt. Which of the following must hold? Ignore effects associated with the vertical rods.

    2. Relevant equations

    Conservation of rotational energy

    Conservation of angular momentum (although I don't think I need this. Momentum is not conserved because there are forces acting).

    Torque = Iα = rF

    3. The attempt at a solution

    Tried it two ways. Once with conservation of rotational energy, once with torque. The answer is ω1r13 = ωr23

    1) If they slowed each other to a halt, their energies must've been =

    I1 ω12 /2 = I2 ω22 /2

    So mass times radius squared times angular velocity squared is equal. Densities are equal, heights are equal, pi cancels out, so (using area of circle)

    r12*r12ω12 = r22*r22ω22

    Which simplifies to one of the answer choices, C: ω1r12 = ω2r22

    2) Here the reasoning is that if they halted each other, the torques must've been equal:

    T1 = T2

    Torque equals moment of inertia times angular acceleration, which equals moment of inertia times angular velocity over time

    I1ω1 / t = I2ω2 / t

    The times are equal; they cancel out. Using the same technique as last time, I get M1 = (r12 / r22) M2

    ((r12 / r22) M2)r12ω1 = M2r22ω22

    Which turns into

    ω1r14 = ω2r24

    Answer choice E.

    But the correct answer is D: ω1r13 = ω2r23

    How? How did both methods fail? Gah!

    Thanks in advance.
     
  2. jcsd
  3. Jan 27, 2014 #2

    lightgrav

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    bad 1) rotational KE is wasted (turned to Thermal E) during friction at the edges; it is not conserved!

    bad 2) the frictional torques are not equal; instead, the frictional Forces are _opposite_.

    use 3) angular momentums are _opposite_, so the total always was zero, even before stopping.
     
  4. Jan 27, 2014 #3
    Darn it. Okay. But wouldn't thermal energy waste effect the conservation of momentum as well?

    And why wouldn't method 2 work? All I did wrong (it seems) is forget a negative sign. That wouldn't effect the rx value, would it?

    And adding the angular momenta:

    I1ω1 + I2ω2 = 0

    I1ω1 = -I2ω2

    Now I get what I got three steps down method 2, but with a negative sign attached. The outcome has the same exponents.
     
  5. Jan 27, 2014 #4

    haruspex

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    No, energy is energy, momentum is momentum. Work energy converting to thermal energy says nothing about momentum.
    The torques will be different. Think about the tangential forces exerted by friction.
    Sounds like you are taking the discs to be coaxial. I read it that they are coplanar. Angular momentum is not conserved either - if you take moments about one spindle then there's an unknown force from the other spindle that has a moment.
     
  6. Jan 28, 2014 #5
    Each disk's angular momentum changes to 0 in time t. Therefore, they each exert a torque t on each other (equal to the change in momentum over the change in time). The change in time is constant, therefore the changes in momentums are equal and opposite, which is what I did in my last post, which led me to

    I1ω1 = -I2ω2

    Which leads to

    ω1r14 = ω2r24
     
  7. Jan 29, 2014 #6

    lightgrav

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    oops, my suggestion #3 was no good (thanks to Haruspex for pointing that out).
    as I said in #2, it is the friction Forces that are opposite (not the friction torques) via Newton's 3rd Law.
     
  8. Jan 29, 2014 #7
    Guys, hear the desperation in my voice. The exam is tomorrow.

    I need some straightforward answers or at least some quality hints. I don't see how my logic is wrong.

    Help. Thanks.
     
  9. Jan 29, 2014 #8
    Just apply τnet = dL/dt to both the disks keeping in mind that the friction acting on disk A is equal in magnitude to friction acting on disk B.

    This will give you the correct answer .
     
  10. Jan 29, 2014 #9
    Okay.

    T1 = T2

    I1ω1/t = I2ω2/t

    The times are equal so they cancel out and you're left with conservation of angular momentum. I established this in my very first post. This leads to ω1r14 = ω2r24.

    I'm not challenging your advice. It's just that I've already tried your advice and obviously made an error somewhere. Can you point out the error please?
     
  11. Jan 29, 2014 #10

    lightgrav

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    Forces: F1 = - F2
    torques: τ1 = R1F1 = - R1F2 = - R1 (R2/R2) F2 = (R1/R2) R2F2 = (R1/R2) τ2

    now you multiply τ∙Δt = I Δω
     
  12. Jan 29, 2014 #11

    haruspex

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    You dropped the minus sign, which might be confusing. Just to clarify, the two radii in question are, as vectors, from the disc centres to the point of contact. Hence one is a negative multiple of the other.
    ##\vec \tau_1 = \vec R_1 \times \vec F_1 = -\vec R_1 \times \vec F_2 = -\frac{-r_1}{r_2}\vec R_2 \times \vec F_2 = \frac{r_1}{r_2}\vec \tau_2 ##
     
  13. Jan 31, 2014 #12
    Hello Agrasin

    Sorry for the late response .I was out of town for a couple of days.

    The torques on the two discs are not equal,rather the magnitude of frictional force is equal .

    fr1 = I1ω1/t
    fr1 = m1r12ω1/(2t)
    fr1 = ρπr141/(2t)
    So,f = ρπr131/(2t) (1)

    Similarly,for the second disc f = ρπr232/(2t) (2)

    Equating the two , we get ω1r13 = ω2r23
     
  14. Jan 31, 2014 #13
    Right and just to clarify above, the density times volume is substituted for mass since the material is the same.
     
  15. Nov 12, 2014 #14
    So, just to clarify, in problems like this, where two spinning cylinders rub together and bring each other to rest:

    1) Energies aren't equal, so one cylinder's energy can't be equated with the other's

    2) Momenta aren't equal (?), so the cylinders' momenta can't be equated.

    3) Neither energy nor momentum is conserved because of friction, a nonconservative force.

    4) The frictional forces on each cylinder ARE equal by Newton's third law, so that's the best way to solve the problem.

    Feel free to correct me or elaborate on a point. I think I understand this now.
     
  16. Nov 13, 2014 #15

    haruspex

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    There is no basis for supposing they are equal. Energy does not cancel other energy, and positive work done does not cancel other positive work done. Sometimes you read about an object skidding to a stop, described as work done by the object equalling work done by friction. This is not correct. They are equal in magnitude but opposite in sign.
    They could have happened to be equal.
    There is no basis for supposing they are equal here. They could have happened to be equal.
    Strictly speaking energy, in a closed system, is always conserved. Work is not necessarily conserved. Heat energy is not work. That is the sense in which friction is non-conservative.

    Linear momentum is not of interest here. Angular momentum could have been, but unlike linear momentum and work, angular momentum is only meaningful in the context of a specified axis (though in certain simple cases, it is the same for all axes in a specified orientation).
    It turns out that the angular momentum of the two discs as a system is not conserved no matter which of the obvious axes you pick. The reason is that there are forces exerted by the two axles, so no matter which axle you pick there is still a force, with moment about that axis, exerted by the other axle. If you pick some other axis then both axles exert relevant forces. (There may well be some axis about which angular momentum is conserved, but there is no easy way to know where that is.)

    The friction between the two discs does not alter their combined momentum or angular momentum. This is because the force on one disc is equal and opposite to the force on the other.
    Equal and opposite.
     
  17. Nov 13, 2014 #16
    Thank you, @haruspex.

    4) Yup, equal and opposite-- oops.

    3) I have some questions about angular momentum. I understand it's only meaningful if you choose an axis. Say you choose the axis of Cylinder 1. Then L = Iω. But how does the spin of Cylinder 2 add into the total angular momentum about axis 1?

    Also, I thought conservation of angular momentum always held given there are no EXTERNAL forces/torques. If you calculate the angular momentum of the system, there may be internal forces/torques between the axles, but isn't it true that there are no external forces/torques on the system?

    2) Also, I was confused about this one. Cylinders rubbing together seems to be the rotational analog of a collision with linear momentum. So if they cancel each other out, shouldn't they be equal?


    Sorry, it seems I may be lacking some important intuition. Feel free to direct me to some site or source that can explain everything in-depth. I would appreciate it.
     
  18. Nov 13, 2014 #17

    haruspex

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    I mentioned that there are cases where the angular momentum of an object only depends on the orientation of the chosen axis. It will take the same value about any axis parallel to that. An object rotating about its CoM and having no linear motion in the frame of reference is that case. Thus each rotating disk here has the same angular momentum about its own axis as about the other disk's axis.
    Quite so, but something is holding the axles steady. That is an external force to the system consisting of the two disks and their axles.
    That would be true if there were no external force. For the present case, the linear frictional analogy might be one block sliding along the top of another, but the lower one is either fixed to the ground or has friction with the ground.
     
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