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Two cylinders: free to fall and rotate

  1. Mar 27, 2015 #1
    Problem: Find the acceleration of the blue cylindrical disk. Both, red and blue, cylinders have same mass and radius, and string is massless.

    image.png

    Attempt:
    Identified three equations: one for linear acceleration of blue cylinder, and two of angular accelerations (one each for blue and red cylinder).

    1) mg - T = ma
    2) Tr = I*alpha (blue)
    3) -Tr = -I*alpha (fred)

    I have chosen downward and anti-clockwise to be positive.

    Trouble: I obtain T from equation 2 and substitute it in eqn. 1 and solve for 'a' (=2*g). This answer has been marked red.

    Would appreciate advice to correct course.

    Regards,
    wirefree
     
  2. jcsd
  3. Mar 27, 2015 #2

    TSny

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    I'm a little confused with your signs. You said that you are taking anti-clockwise as positive. The red cylinder rotates anti-clockwise. Yet your equation 3 has negative signs. I think we would need to see more of your work to make sure you are treating the signs correctly.

    You have not indicated how you relate the angular acceleration of the blue cylinder to its linear acceleration. This is an important part of this problem and it's a little bit tricky.
     
    Last edited: Mar 27, 2015
  4. Mar 27, 2015 #3
    EDIT: My equations were mislabeled. Properly, they are:

    1) ) mg - T = ma
    2) Tr = I*alpha (red)
    3) -Tr = -I*alpha (blue)

    NOTE: Apologies; could not edit original Usenet post
     
  5. Mar 27, 2015 #4

    TSny

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    Can you please show your steps that lead to your answer? That way we can pinpoint any mistakes.
     
  6. Mar 27, 2015 #5

    haruspex

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    At a guess, you have the relationship between a and alpha wrong.
     
  7. Mar 30, 2015 #6
    Appreciate all responses.

    My attempt, as requested, was:

    1) obtain T from equation 2
    - for 'alpha' in the T expression, I inserted the no-slipping constraint: alpha*r=a
    2) substitute T in eqn. 1 and solve
    3) a = 2*g

    This answer has been marked red.

    Please advise.

    Regards,
    wirefree
     
  8. Mar 30, 2015 #7

    TSny

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    The constraint is a little more complicated due to the string unwinding off of the top cylinder. The linear acceleration of the bottom cylinder depends on both the angular acceleration of the top cylinder and the angular acceleration of the bottom cylinder.

    The diagram shows a specific point of the string in green, a specific point on the rim of the lower cylinder in blue, and a specific point on the red cylinder in red. During a small time interval, the point on the string moves down from ##a## to ##a'## while the points on the cylinders rotate from ##b## to ##b'## and ##c## to ##c'##. This diagram might help you relate the angular displacements ##\theta_{red}## and ##\theta_{blue}## of the cylinders to the vertical drop of the center of the blue cylinder from ##a## to ##d##.
     

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  9. Jun 14, 2015 #8
    Greatly appreciate the direction, TSny.

    My key take-away from all discussions above is that the no-slipping condition alone is not sufficient in determining acceleration of the blue cylinder because, to quote TSny, "linear acceleration of the bottom cylinder depends on both the angular acceleration of the top cylinder and the angular acceleration of the bottom cylinder."

    To the above extend, it was suggested I start with the relationship between angular displacements and proceed.

    My unsuccessful attempt at that aside, a confirmation of my understanding of the basic premise - a = alpha_b*R + alpha_r*R - would be appreciated.

    Regards,
    wirefree
     
  10. Jun 14, 2015 #9

    TSny

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    Yes, that's the correct relation.
     
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