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Hardest Freshman Physics Question Ever!

  1. Jan 22, 2014 #1
    Yes, very catchy title. But it also happens to be true! Here's the windup: This question comes from the Halliday and Resnick edition that was used in the 80's. I was a grad student at the time at a very prestigious department and somehow this question came to my attention. I couldn't figure it out. Soon, there was a group of four grad students losing sleep over it. I even took it to a very famous theoretical physicist. He smiled and laughed out loud, because he didn't see an obvious answer. Decades later, I ran into the author of another widely used freshman textbook and he just said well if the famous guy couldn't figure it out, neither could he! So this has bothered me for going on three decades. Hopefully somebody here can put my mind at ease. Remember, this is supposed to be freshman level.

    It is a "qualitative" question.

    Drum roll....

    If you take a flat object (like a book) and slide it along a surface where there is friction, why does it stop translating and rotating at the same time?

    That's all there is to it. You can quickly verify that it is true. But why?

    I'm sorry if you lose some sleep!
     
  2. jcsd
  3. Jan 22, 2014 #2
    Won't that be because the force of friction works in the opposite direction of the total movement, not seperately for the translational and rotational?

    What I mean is, if you take one point on the corner of the spinning book, it has a movement due to the rotation around the center of mass, and one due to the translation of the COM. The total movement for that point is the sum of those, and this is the movement of which friction would be directly opposite.

    Say the center of mass of the book is moving with velocity 1 in the x direction, while one corner at that instant is rotating with velocity 4 in the y-direction on top of that. The total velocity vector would be (1,4). Since the negative acceleration due to friction would be the opposite, it would be some vector -a(1,4) / ||(1,4)||, therefore doing proportionaly more work on the element due to rotation (or less if it was lower). Anyway, apply the thought experiment for each point in the book, and each pair of coordinates should reach (0,0) at the same time.
     
  4. Jan 22, 2014 #3

    berkeman

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    It's not true for all initial values of spins and linear velocity. As you said, you can quickly verify that it is true. :smile:

    (Try a slow linear push with a strong spin -- the book spins still after the linear motion stops)
     
  5. Jan 22, 2014 #4

    berkeman

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    @tehrv -- So a *much* more interesting and challenging question is for you to calculate the initial conditions to have the following things happen:

    -1- translation stops before rotation

    -2- rotation stops before translation

    -3- both stop at the same time

    Please show your work... :smile:
     
  6. Jan 23, 2014 #5
    It is supposed to be a 'qualitative' problem, so I don't think they wanted you to write much. Something like gralla's argument may be on the right track. Berkeman, the high spin case exception you mention is probably because the surfaces are not perfectly flat. Anyway, it is not my question, so it isn't really up to me to tweak it.

    We could never figure out what was going on. Really, some brilliant grad students were stumped. I remember one guy was trying to consider a point at infinity. lol.
     
  7. Jan 23, 2014 #6

    sophiecentaur

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    When I first read that I believed you (stands to reason dun nit?). But does it really? Have you (/we) actually measured this or just imagined it? Would we spot a slow forward motion of a fast spinning book? I could believe that we wouldn't. (Are we going to argue with eminent Professors????? lol)
    I guess it must be to do with the difference between dynamic and static friction and as long as you have motion in one mode or the other, the friction is dynamic in both cases and the retarding force / torque will be tied to the relative speeds between all points on the object and the surface. So static friction kicks in at the same time for both modes. The mode with the higher relative speeds involved will lose energy faster than the mode with the lower relative speeds (Power = force times speed). Arm waving allowed here by the original problem, I'm pleased to say - but the Maths would not be beyond a bright student, I'm sure. It works, I'm sure, if the retardation is proportional to speed and not constant.
     
  8. Jan 23, 2014 #7
    I don't think static friction kicks in before the book stops moving completely.
    There's no way that a part of the book can stop moving, except for a single point, and the fricton on a single point is 0.
    So I don't think static friction is relevant at all.
     
  9. Jan 23, 2014 #8
    If the dynamical friction force on a surface element of the book is linearly dependent on the velocity of that surface element, and if we only consider a purely translational motion for simplicity, then the momentum of the book would decay exponentially it seems: the smaller the momentum, the smaller the force that acts to diminish it. This would imply an infinite time interval would pass before the book would come to a complete rest given any nonzero initial momentum.

    That would indicate that below some velocity threshold the nature of friction would have to be different, or that we need to consider a different relation between friction and velocity altogether.
     
  10. Jan 23, 2014 #9

    sophiecentaur

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    You are right. Static friction is not relevant - except in as far as it lurks in our subconscious when we think about friction problems.
     
  11. Jan 23, 2014 #10

    Office_Shredder

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    It's not though, the dynamic friction is constant (just a different constant value from when the book is not moving).


    I don't understand the statement of the problem in the context of pushing a book forwards with only linear momentum; certainly that seems possible to do theoretically, or is this example (and where you rotate it with no linear momentum) supposed to be excluded?
     
  12. Jan 23, 2014 #11
    Yes, the two limiting cases are excluded. You just sorta fling the book across the suface so that it both translates and rotates. Maybe somebody has the book (big and green with yellow waves as I recall) and can give us the exact wording.
     
  13. Jan 23, 2014 #12
    Ah, you're right of course - my apologies. I somehow completely misinterpreted sophiecentaur's post. Dissipation power is proportional to velocity, but dynamic friction force is constant.
     
  14. Jan 23, 2014 #13

    jbriggs444

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    The answer is simple and has already been given in this thread. You just have to take the arguments to their logical conclusion.

    Answers:

    1. The only condition in which translation stops before rotation is when there was no translation to start with.

    2. The only condition in which rotation stops before translation is when there was no rotation to start with.

    3. In all remaining cases, they both stop at the same time.

    Work:

    The key observation (from Gralla55) is that for a force that opposes velocity and does not scale with velocity, the component of the force in one direction scales down with the component of velocity in the direction at right angles.

    Quantitatively: fx = ftot√(1-vy/vtot)

    Qualitatively, what matters is that when the vx << vy, fx is proportional to vx. As Brinx has noted, this means that the decay of vx is approximately exponential.

    For sake of clarity, let us assume that this object is sliding from North to South...

    Suppose that the rate of translation reaches zero while the rate of rotation is non-zero. It follows that for some time prior to this occurring, almost all of the points on the object would have been rotating east/west much faster than they were translating north/south. [More formally, for any desired value of "almost all" and any desired ratio of "much faster", there is a time after which the rate of translation is small enough that both conditions are upheld]

    During this time interval the north/south force is approximately proportional to north/south velocity and rate of decay of translational velocity is exponential. It takes infinitely long for it to reach zero. During this same time interval the opposite condition is not upheld. The resistance to rotation is not reduced significantly by the low translation rate. The rotation rate must be therefore be decreasing approximately linearly and will reach zero in finite time. If the "almost all" and "much faster" conditions continue to be upheld, the rotation rate will reach zero before the translation rate does.

    Contradiction. So the rate of translation cannot reach zero first.

    The same argument applies if the object is translating much faster than it is rotating. So the rate of rotation cannot reach zero first.

    The only consistent resolution is that the two rates must reach zero together (or that one or the other were zero to start). QED.
     
    Last edited: Jan 23, 2014
  15. Jan 23, 2014 #14

    AlephZero

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    Do you have a proper reference to this? I can't find anything with Google, except a Gralla who is working in quantum cosmology...

    From what you posted, I can't decide between two options:
    1. It doesn't make sense
    2. I don't understand it
    (of course there may be other alternatives, and those two options may not be mutually exclusive :smile:)
     
    Last edited: Jan 23, 2014
  16. Jan 23, 2014 #15

    jbriggs444

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    The observation from Gralla55 is in post #2 of this thread.
     
  17. Jan 23, 2014 #16

    AlephZero

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    OK - I took it the complete argument was from a paper by Gralla.

    There is a simpler explanation of why the translational and rotational decelerations will converge to a have consistent relationship between them, independent of the starting conditions and also independent of the details of the friction model.

    Suppose the book is translating north, and suppose it has a "low" rotation speed compared with the translation speed. Relative to the table, both the east and west sides of the book are sliding north. So the moments of those frictional forces about the CM of the book will cancel, but the translational friction forces north will add. For fast translation and slow rotation, the friction mostly decelerates the translation, not the rotation.

    Now, suppose the rotation speed is "high" relative to the translation. The opposite occurs: one side of the book is sliding north relative to the table, but the other side is sliding south. The translational components of the friction cancel out, but the moments about the CM add. For slow translation and fast rotation, the friction mostly decelerates the rotation, not the translation.

    So, for any "reasonable" friction model and any "reasonable shaped" book, the relative translation and rotation speeds will tend to a fixed ratio of each other, and both reach zero at the same time.

    (Note to pedants: feel free to argue at length about the definition of "reasonable" in the previous paragraph :smile:)
     
  18. Jan 23, 2014 #17
    How about this?

    The frictional force on each point is opposite to the direction of that point's velocity.

    So (by Newton's Third Law) the deceleration on each point is opposite to the direction of that point's velocity.

    But that point's velocity can be written as a sum of a translational and a rotational part.

    So, we can rewrite the deceleration of each point as a sum of a translational and a rotational part.

    Moreover, and this is the key part, we see that the ratio of the magnitudes of the deceleration of the translational part and the deceleration of the rotational part is PROPORTIONAL to the ratio of the magnitudes of their respective velocities. Call this the PROPORTIONALITY CONSTRAINT:

    Deceleration of the translational component/Deceleration of the rotational component
    =Speed of the translational component/Speed of the rotational component.

    For all the points, except a subset with zero total mass, the translational and rotational velocities at the beginning are non-zero.

    At each moment, a bit of work gets done on each point. The translational velocity decreases a bit. And so does the rotational velocity. The greater of the two decreases the most. But because of the PROPORTIONALITY CONSTRAINT, the velocities and decelerations all go to zero at the same time.

    It's not as though there is some funky relation between the velocities allowing for strange behaviors. Rather there is this simple proportionality constraint.

    Needs some polishing, but something like that.
     
  19. Jan 23, 2014 #18

    AlephZero

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    True.
    That is not true for a rigid body. The deceleration depends on the resultant force on "the point", including the internal forces (stresses) in the body. The simplest way to deal with the internal forces correctly, is to find the resultant forces and moments at the center of mass of the body.

    (And I think you meant Newton's second law, not the third).
     
  20. Jan 23, 2014 #19
    Right. I should have known it was too easy. It's been a few decades since I've done any real physics!

    And right again. I kept thinking about the force being opposite to the velocity, so I had "opposite" on my tongue.
     
    Last edited: Jan 23, 2014
  21. Feb 4, 2014 #20
    So can somebody give a complete answer without making reference to other posts in the thread? I'm sorry, but I still don't see a full solution here. :(
     
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