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Simple Mechanics Problem Holds Politicians In Disarray ?

  1. Jun 12, 2005 #1
    Hello

    This is a question from the recent national exams in Physics in Greece:

    Two equal sized coins (1 euro) are placed on one edge of a frictionless flat table.
    Coin "A" has a continuous force F applied at it's centre and pointing to the opposite edge of the table. Coin "A", as a result, moves and takes a time Ta to reach the other edge.
    Coin "B" has a continuous force of the same magnitude, F, applied at it's rim, tangentially and also pointing to the opposite edge.
    Will coin "B" get there (a) faster , (b) slower (c) at the same time ?


    You can have a look at the free-body diagram at

    http://www.cosmicway.net/freebody.gif


    Although I suspect the answer is (c), my dynamics are a little rusty.
    Very few students gave the answer and the political left are "up in arms" because this question (and some others) were too difficult.
     
  2. jcsd
  3. Jun 12, 2005 #2

    OlderDan

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    (c) is correct, but could seem paradoxical. It is tempting to think that both forces act over the same distance, and do the same amount of work, so both coins arrive at the other edge with the same energy. Since one arrvies with rotational energy, you might conclude that it must have lower speed, and so took more time to get there. In fact, the work done is not the same. The tangential force must be applied over a greater distance because the coin rotates (imagine a string wound around the coin). More work is done on the rotating coin.
     
  4. Jun 12, 2005 #3

    pervect

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    It seems like a fair question to me, though tricky. I was just about to give the wrong answer, though, when I read the previous correct response, which is 'c', they will arive at the same time, causing me to rethink my aswer in more detail (by writing the equations of motion) and saving me some embarassment.

    One coin will rotate, the other coin will not. But clearly, one can write the acceleration of the center of mass as F/m in both cases, meaning that they follow the same trajectory.

    This follows directly from Newton's second law, F = dP/dt. The angular momentum of the coin will not contribute to it's linear momentum. Thus the total linear momentum of the coin will be the velocity of it's center of mass multiplied by the mass.

    The force on any point particle of the coin will be the sum of the external forces and the internal forces. The internal forces, however, will not contribute to the momentum of the coin. Thus the total applied force is F.

    Thus, letting v_cm equal the velocity of the center of mass, we have dp/dt = d/dt(m * v_cm) = m dv_cm/dt = F, or

    d v_cm/dt = F/m

    for both cases.
     
    Last edited by a moderator: Jun 19, 2005
  5. Jun 16, 2005 #4
    Isn't the correct answer, b) slower, only true if slower equals never?

    A force applied tangential to the circumference of coin b will only spin the coin. Even if the coin were standing on edge, which was not stated, the surface is frictionless and no friction means no rolling. Coin B wil stay in place and spin faster and faster as long as the force is applied.
     
  6. Jun 16, 2005 #5

    Doc Al

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    No. Reread the posts by OlderDan and pervect.

    Not true. Any net force on the coin, regardless of where it is applied, will accelerate the center of mass of the coin.
     
  7. Jun 16, 2005 #6

    Chi Meson

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    One point that is not clear in the question is, does the force remain tangential to the edge of the coin? After pondering the question, it appears that force B remains pointing to the opposite edge of the table, but hey I can see how many students will be misled by an expectation of a circular motion question.

    This one is not "too hard." It is a conceptual, rather than kinematic question. The teachers or texts should have, at some point, made it clear that "net force" and "torque" are not exclusive; that is, a force that produces a torque does not stop being a linearly accelerating force.
     
  8. Jun 16, 2005 #7

    OlderDan

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    That seems pretty explicit to me. Lot's of problems are "too hard" if you start thinking about them the wrong way and don't have a chance to reflect on what you are doing. I just took my first written exam in about 30 years on Saturday and missed a couple of really easy questions because of time constraints and getting started in the wrong direction. Testing is an inexact science and always will be, in my opinion, in spite of what the specialists would like to achieve.
     
  9. Jun 18, 2005 #8
    Force equals mass * acceleration or F = ma. The total acceleration of coin b, both linear and angular, must be equal to the strictly linear acceleration of coin a. For the two coins to travel the same distance in the same time the linear accelerations would have to be the same. This cannot be true if coin b undergoes any angular acceleration.
     
  10. Jun 18, 2005 #9

    Chi Meson

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    Oh but yes it is true. First of all, you cannot add linear acceleration to angular acceleration, so such a "sum" is meaningless. As I stated, the fact that the force causes a torque does not take away fromt he fact that it is a linear force.

    I belive you are confusing accelration with energy: the force on coin B causes some angular motion, so it will also have rotational kinetic energy along with the linear kinetic energy. Therefore, at certain times coin B will have more total KE than coin A. But after the coin takes 1/4 turn, the force will cause a torque in the opposite angular direction and the angular rotation will slow, cease, and then reverse.

    But during the entire procedure of rotaional oscillation, the cm of coin B will continue to accelerate linearly regardless of the angular acceleration.

    This is very easily shown by a dependable simulation program.
     
  11. Jun 18, 2005 #10

    Doc Al

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    Just to be clear: The force applied to both coins is constant in magnitude and direction. (At least that's how I interpret the problem.) That force accelerates the center of mass of each coin equally (a = F/m). For coin B that force also exerts a torque about the center of mass, causing the coin to rotate as well as translate. Coin B will at all times have more total KE than coin A. (As OlderDan points out, this is because the tangential force ends up doing more work than the force applied at the center.)
     
  12. Jun 18, 2005 #11

    pervect

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    Consider the following diagram, where the two masses of m/2 are joined by a rod, and are initally stationary. A force is applied to the top mass.

    Code (Text):

    ---->F (m/2}
             |
             |
           (m/2}
     
    At t=0, the top mass accelerates with an acceleration of 2*F/m, the bottom mass does not accelerate at all. (Hopefully this is obvious? The tension in the rod is w^2/d, where w is the angular frequency of the rod and d is the length of the rod, and while dw/dt is greater than zero, at t=0 w=0, therefore the tension on the rod is zero). Thus there is no force on the bottom mass at t=0, no tension on the rod at t=0, and the top mass has a total force of F, while the bottom mass has a total force of 0.

    Now, the questions:


    1) What is the acceleration of the center of mass?
    2) What is the angular acceleration of the rod (dw/dt)?
    3) Is your quoted statement correct?
     
  13. Jun 19, 2005 #12
    The only way the rotation of the coin would oscillate is if the force were always applied to the same point on the coin as it rotates with the coin. The result of that would be the same as if the force were oscillating back and forth across the coin always parallel to its original direction. In that case the linear acceleration of the coin would not be constant but its average acceleration would be.

    As I understood the question the force is always tangential to the coin. Which, as I understand it, is pretty much the definition of torque. So coin b sees only a torque. This torque then will always be in the same angular direction giving only an angular acceleration.
     
  14. Jun 19, 2005 #13

    Doc Al

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    What the coin "sees" is a force. A net force always produces an acceleration of the center of mass. If it didn't, it would violate Newton's 2nd law. In the case of coin B, that same force also exerts a torque about the center of mass. That torque produces an angular acceleration about the center of mass. Coin B executes both translation and rotation.

    Note: It is not possible for a single force acting on the coin to produce only rotation without translation. But if two equal and opposite forces acted, one on each side of and tangential to the coin, then you could have pure rotation. In that case the net force would be zero, but the torque would be non-zero.
     
  15. Jun 20, 2005 #14

    Chi Meson

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    This is how I saw the problem, hence this is why I thought it would oscillate.


    No, again, linear accelration is NOT diminished byu the fact that the force also applies torque. Linear acceleration remains constant as long as the net force remains constant.


    THis explains the problem. Your understanding is incorrect.
     
  16. Jun 20, 2005 #15
    Would there be any rotational motion at all?

    In order for an object to rotate (say a wheel for example) it needs to have frictional force on the other end to counter the force on one end. On a frictionless surface, the wheel would not rotate at all.

    Since the table in this case is frictionless, wouldn't a tangentional force move the coin in the same manner as a force applied to the center?
     
  17. Jun 20, 2005 #16

    Doc Al

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    Yes.
    Not sure what you are saying here. For something to begin rotating, all that's needed is a net torque.

    Please reread the previous posts describing the motion of each coin.
     
  18. Jun 20, 2005 #17

    pervect

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    I think the simplified example I gave illustrates why the correct solution is correct. Since nobody has answered the questions as I hoped, I will answer them myself in the hope that the answers will be instructive.

    It should be obvious that

    velocity_top = velocity_cm + (d/2)*w
    velocity_bot = velocity_cm - (d/2)*w

    The three velocities are all velocities, obviously, d is the distance between the two masses in the diagram, and w is the angular velocity of the rod.

    Then

    a_top = a_cm + (d/2)*dw/dt
    a_bot = a_cm - (d/2)*dw/dt

    Here a_top is the acceleration of the top mass, a_bot is the acceleration of the bottom mass, a_cm is the acceleration of the center of mass, d is the distance between the masses, and dw/dt is the angular frequency of the rate of rotation of the rod.

    Given a_top = 2*F/m and a_bot =0, we find

    a_cm = F/m
    dw/dt = 2*F/m*d
     
  19. Jun 20, 2005 #18
    The answer will be c. You can replace the force acting tangentially with a force through the center coupled with a clockwise moment of F*r, where r is the radius of the coin. The moment causes the coin to rotate but does not affect its acceleration due to the force acting through the center. The two coins must accelerate at the same rate then.

    Its interesting though that even here there is uncertainty so its not surprising a lot of students were confused, and I would love to see those same politicians try to answer the question.
     
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