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Ratio of angular speed with conservation of energy

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    1. A ball rolls down an incline plane without slipping. What is the ratio of its angular velocity at h/3 to its angular velocity at 2h/3?
      1) 1:2
    2. 2) 1:sqrt(2)
    3. 3) 1:1
    4. 4) sqrt(2):1
    5. 5) 2:1

    2. Relevant equations
    Conservation of energy with provisions for rotational and translational motion.

    3. The attempt at a solution
    Hi everyone!

    I got this funky problem today. Here's what I tried to do:

    at h/3 I have potential energy resulting from lowering the ball a height of 2*h/3 that is converted to kinetic energy.

    Thus, I did the following
    m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
    m * g * (2h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

    Rearrange to get an expression with v, and then substitute v = r * omega to get
    (20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height

    OK. then I said that at height 2h/3 I have potential energy from descending h/3 converted to kinetic energy
    m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * I * omega^2]
    m * g * (h/3) = [(1 / 2) * m * v^2] + [(1 / 2 ) * (2 / 5) * m * r^2 * (v^2 / r^2)]

    Rearrange to get an expression with v and then substitute v = r * omega to get
    (10 * g * h) / (21 * r^2 ) = omega squared.

    Thus, my ratio would have been for the two omega squared terms to be:
    (20 * h * g)/ (21 * r^2) = omega^2 for the h/3 height : (10 * g * h) / (21 * r^2 ) = omega squared. for the 2h/3 height

    Simplifying the ratio for the omega squared gives 20 : 10 -----> 2:1.
    Taking the square root to compare just for the omegas gives sqrt(2) : 1.

    Thus, I got 4. The answer according to the writers is 2.

    Why is this the case?

    Thanks a lot for the help in advance!
     
  2. jcsd
  3. May 7, 2016 #2

    haruspex

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    It's not entirely clear what "at h/3" means. Is that h/3 from the bottom or from the top? Is there any further information that clarifies this?
     
  4. May 7, 2016 #3
    They did not provide any information. I believe that 'h' corresponds to the height at which the ball is initially launched on the ramp.
     
  5. May 7, 2016 #4

    haruspex

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    Sure, but that still doesn't nail down the meaning of "at h/3". It could still be measured from top or from bottom. Measured from bottom, your answer is correct.
     
  6. May 7, 2016 #5
    I have no idea... these authors write really crummy questions, which does not help as I have a final on monday. :-(
     
  7. May 7, 2016 #6

    haruspex

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    So you quoted the question exactly as given to you, word for word, and there is no diagram?
     
  8. May 7, 2016 #7
    Yes, sir!
     
  9. May 7, 2016 #8
    Yes, sir! I just assumed they had a ramp at an arbitrary angle, with it vertically ascending "h".
     
  10. May 7, 2016 #9

    haruspex

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    Well, either they were thinking of h/3 etc. as descent, i.e. measured from the top, or they messed up and gave the wrong answer.
     
  11. May 7, 2016 #10
    OK, thanks so much for helping and confirming what I did. Really appreciated!
     
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