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Acceleration of a rotational and translational system

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the acceleration of each object. Note: the question mark in the diagram should be θ (angle of inclined plane)
    Untitled1.jpg

    2. Relevant equations

    τ = F . r
    Στ = I . α
    α = a / R
    ΣF = m . a


    3. The attempt at a solution
    For object A:
    ΣF = m . a
    TA - WA sin θ = mA . aA
    TA - mA . g. sin θ = mA . αA . r1

    Στ = IA . αA
    TA . r1 = IA . αA


    For object B:
    ΣF = m . a
    WB - TB = mB . aB
    mB . g - TB = mB . αpulley . r3


    For pulley:
    Στ = Ipulley . αpulley
    TB . r3 - TA . r4 = Ipulley . αpulley


    My questions:
    1. Are my equations above correct?

    2. Is it correct that αA ≠ αpulley?

    3. Can I state that aA = αA . r1 = αpulley . r4?

    Thanks
     
  2. jcsd
  3. Nov 28, 2016 #2

    TSny

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    If object A rolls without slipping, there is a force on A that helps prevent slipping which you left out.

    Yes! That's correct.

    No.

    For the first equality, you have aA = αA r1. Is r1 the radius that you should use here? Consider the condition for rolling without slipping of a wheel.

    For the second equality, you have aA = αpulley r4. The relation between aA and αpulley is a little more complicated than this. (I assume that aA represents the acceleration of the center of object A.) It might help to consider the two colored spots painted on the sloping string shown below
    upload_2016-11-28_21-13-1.png


    Are the linear accelerations of the red and blue spots the same?
     
    Last edited: Nov 28, 2016
  4. Nov 28, 2016 #3
    You mean friction between A and the plane? Can the torque for rolling being produced by tension A only?

    So, αA will be the same as αpulley?

    I am not sure I get the hint. Is it correct if I use r2 instead of r1 because r2 is the radius of rolling?

    I don't think the linear accelerations of the red and blue spots are the same. The linear acceleration of blue spot should be higher than red spot because blue spot will cover more distance in one rotation. The ratio of their linear accelerations will be the same as the ratio of their radius? (ared / ablue = r1 / r4?


    Can I also state that aA / aB = r1 / r3?

    Thanks
     
  5. Nov 28, 2016 #4

    TSny

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    Yes.
    For general values of the masses, etc., it seems unlikely that the friction force would be zero.

    No, they won't be the same. Sorry, when I read this part of your post I saw the symbol =, whereas you actually had ≠. I edited my earlier post.

    Yes, that's right.

    Note that there is a fixed length of string between the two dots. We assume the string cannot stretch.

    No.
     
  6. Nov 28, 2016 #5
    If friction is considered, the equation will be:
    ΣF = m . a
    TA - mA . g. sin θ - friction = mA . aA
    TA - mA . g. sin θ - friction = mA . αA . r2

    Στ = IA . αA
    TA . r1 + f . r2 = IA . αA

    Are their linear accelerations the same? They both have different angular accelerations and radius so it is possible for them to have same linear acceleration?

    Thanks
     
  7. Nov 28, 2016 #6

    TSny

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    Yes.

    Yes. Yes.
     
  8. Nov 28, 2016 #7
    Object with radius r3 and r4 will have same α, so:
    ablue / aB = r4 / r3 ; ablue = ared = αA . r2

    aB = αA . r2 . r3 / r4
     
  9. Nov 28, 2016 #8

    TSny

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    OK

    ared ≠ αA . r2
    αA . r2 gives the acceleration of the red dot relative to the center of object A. But ared denotes the acceleration of the red dot relative to the earth.
     
  10. Nov 29, 2016 #9
    So aA = αA . r2 that is used in the equation of Newton's law (ΣF = m.a) is the acceleration with respect to the center of object A, not with respect to earth?

    aB = αpulley . r3
    ablue . r3 / r4 = αpulley . r3
    ablue = αpulley . r4
    ared = αpulley . r4

    ared / aA = r1 / r2 ?

    Thanks
     
  11. Nov 29, 2016 #10

    TSny

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    aA is the acceleration of the center of object A relative to the earth. For rolling without slipping, aA = αA . r2 .
    upload_2016-11-29_11-40-12.png

    All of this looks good.

    This is not correct.
    One way to get the relation between ared and aA is to use the idea of relative motion. You can think of the (tangential) acceleration of the red point with respect to the earth as the (tangential) acceleration of the red point relative to the center of object A plus the acceleration of the center of object A with respect to the earth:
    ared = ared/A + aA.
     
  12. Nov 29, 2016 #11
    ared/A = αA . r1

    ared = ared/A + aA
    ared = αA . r1 + aA
    ared = (aA / r2) . r1 + aA
    aA = ared . r2 / (r1 + r2) ; ared = ablue = aB . r4 / r3
    aA = aB . r2 . r4 / ((r1 + r2) . r3)

    Is this correct?

    Thanks
     
  13. Nov 29, 2016 #12

    TSny

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    Yes. Good.
     
  14. Nov 29, 2016 #13
    and the relation between αA and αpulley:

    αpulley = αA . (r1 + r2) / r4 ?

    I have another question similar to this one. I will do it first and then post it again. Thanks
     
  15. Nov 29, 2016 #14
    Untitled.png

    Let the red spot and blue spot the same as question before:
    ared = ablue = aC

    ared = aA . (r1 + r2) / r2
    aC = aA . (r1 + r2) / r2

    aB / r3 = aC / r4
    aB = aA . (r1 + r2) . r3 / (r2 . r4)

    αpulley = αA . (r1 + r2) / r4

    Is this good?

    Thanks
     
  16. Nov 29, 2016 #15

    TSny

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    Looks right.
     
  17. Nov 29, 2016 #16

    TSny

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    I think that's all correct.
     
  18. Nov 29, 2016 #17
    Untitled2.png

    Let red spot is the center of object and blue spot is the point where the tension touches object B

    ared = ablue = aA

    aA = αA . rA

    ablue = 2 aB , so aA = 2 aB

    Then:
    aA = 2 aB
    αA . rA = 2 . αB . rB

    and: αpulley = αA . rA / rP


    Is this correct? Sorry for the trouble. My teacher didn't teach about this type of question but this is given as practice for the test today. Thanks
     
  19. Nov 29, 2016 #18

    TSny

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    It all looks correct to me. Good luck on your test!
     
  20. Dec 3, 2016 #19
    The question that came out was the third picture I posted and unfortunately the teacher aA = aB without any further explanation why...


    Thank you very much for your help TSny
     
  21. Dec 3, 2016 #20
    Because of string constraint , the acceleration of center of A is equal to the acceleration of the topmost point of B .
     
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