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Statics and Rotational Dynamics

  1. Apr 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A thin walled spherical shell with a mass of 0.605 kg and a radius of 0.0402 is released, from rest, at the top of an incline. The spherical shell rolls down the incline without slipping. The spherical shell takes 7.49 s to get to the bottom of the incline.

    A solid sphere with a mass of 0.127 kg and a radius of 0.1123 is released, from rest, at the top of the same incline. The solid sphere rolls down the incline without slipping. How much time does it take for the solid sphere to reach the bottom of the incline?

    2. Relevant equations
    thin sphere I = (2mr2)/3
    solid sphere I = (2mr2)/5


    3. The attempt at a solution
    I do not have an attempt yet as I do not even know where to begin
     
  2. jcsd
  3. Apr 26, 2017 #2
    You should start with a free body diagram and sum the forces. You also need to sum the torques on the spheres. You also need to relate angular acceleration with linear acceleration. I know that is fairly vague, but it really is a pretty good starting point.
     
  4. Apr 26, 2017 #3
    When you do your calculations, make sure that you distinguish between mass and radius for the shell versus mass and radius for the solid sphere.
     
  5. Apr 26, 2017 #4

    kuruman

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    This problem gives a lot of unnecessary information. All you need is the kind of rolling object (shell, sphere) and the time it takes for one of them to reach bottom. Radii and masses are irrelevant. For example, two solid spheres of different masses and radii will reach bottom at the same time.

    I would recommend that you use mechanical energy conservation to find expressions for the final speeds and see why the above is true. Then use the kinematic equations to relate the time of the sphere to the given time of the shell. Hint: If v2 = 2 a s for constant acceleration, what is v2sphere/v2shell?
     
  6. Apr 26, 2017 #5

    haruspex

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    Yes, a very good exercise. It is the fastest route to solving the problem, and a useful result for the future.
     
  7. Apr 27, 2017 #6
    I beg to differ. By my analysis, the solid sphere arrives quicker, with a total time = 6.8648 s. Please see my analysis in the attached PDF for the full argument.

    Moderator note: The PDF has been removed as providing full solutions before the OP has arrived at a correct and satisfactory solution by their own work is not allowed by forum rules.
     
    Last edited by a moderator: May 3, 2017
  8. Apr 27, 2017 #7
    I don't think @kuruman was saying that the thin shell sphere and solid sphere would arrive at the same time. He was saying that 2 solid spheres of different mass and different radii would arrive at the same time, and 2 thin shell spheres with different mass and different radii would arrive at the same time.

    P.S. I got the same answer as you for the solid sphere. Edit: However, your work looked very much nicer than mine. :)
     
  9. Apr 27, 2017 #8

    haruspex

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    I would put it more strongly. It is quite clear @kuruman did not say that.
     
  10. Apr 27, 2017 #9

    kuruman

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    Thank you @TomHart for interpreting correctly what I said. My numerical answer agrees with both of you. I will upload a pdf of my solution, when I have it, as part of a conversation with both of you. I am apprehensive about violating forum rules by posting a solution, albeit in pdf, when OP has not gotten to its completion.
     
  11. Apr 27, 2017 #10

    kuruman

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    Thanks @haruspex. I will add you to the recipients of my solution although I have a suspicion you already know what it looks like.
     
  12. Apr 27, 2017 #11
    A part of my interpretation was based on kuruman's statement that
    "This problem gives a lot of unnecessary information."
    I am anxious to see how the problem can be worked without using all of the given information.
     
  13. Apr 27, 2017 #12

    haruspex

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    Dimensional analysis is helpful here.
    The acceleration can only depend on the slope, the mass, the radius, the radius of inertia (k), and gravity.
    k is dimensionless, so we have
    a = ckmαrβgγ for some constant ck.
    Matching up the dimensions we find α=β=0, γ=1.
    Hence all the facts about the masses and radii are irrelevant to finding the time. All that can matter is slope, distance, g and k.
     
  14. Apr 29, 2017 #13
    Kuruman is correct when he says that all of the given masses and radii are are not essential. The problem can be solved without using all of this information. However, there is no error in using it since it is given data. To solve the problem without using all of it requires a particular insight into the form of the MMOI expressions.
     
  15. Apr 29, 2017 #14

    haruspex

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    No, it only requires following the style I strongly recommend to all students: ignore the numbers. Write all the equations in purely symbolic form, creating variables for the given constants as necessary. Only plug in numbers at the end. This has many advantages. In the present case, several vanish.
     
  16. Apr 29, 2017 #15
    @ haruspex: I really do not see what you are disagreeing about. I agree entirely about writing the whole solution in symbolic form until the last steps where numbers are used. This is exactly what I did in the PDF solution that I posted.

    The only point where I did not follow your prescription is in regard to writing the MMOI in the form I = kappa * M * R^2, to use the notation that kuruman used and then taking advantage of the cancellations that result.

    I stand by the statement that there is no error in using the numbers since they are given data for this problem.

    What is there to disagree about?
     
  17. Apr 29, 2017 #16

    haruspex

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    This:
     
  18. Apr 29, 2017 #17
    Whatever ....
     
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