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Rotational & Kinetic Energy

  1. Mar 25, 2016 #1
    1. The problem statement, all variables and given/known data
    I am looking at part C.

    https://scontent-ord1-1.xx.fbcdn.net/hphotos-xal1/v/t34.0-12/12782422_10205657311624690_1648245885_n.jpg?oh=79b686a8b74392888fab0b070b86627d&oe=56F7E1E4

    3. The attempt at a solution

    Here I am looking at part C. My initial thought is that they have the same kinetic energy. For one puck, this is pure translational, and for the other puck, this is rotational + translational, but regardless the energy which is put into the system with no nonconservative forces acting will be transferred directly into kinetic energy (whether translational or rotational).

    It is worth noting that I am actually the teaching assistant for this course, and I'm quite ashamed that I am getting the answer wrong; apparently puck 1 has more kinetic energy than two because it has rotational energy whereas puck 2 does not. I do not understand this argument... I do not understand why, when the same amount of energy is being applied to both systems, the pucks have two different kinetic energies. Maybe it has just been too long since mechanics :P Also, all systems are assumed to be ideal here (no energy is being "lost" due to rotation).

    I do deserve being made fun of for this... let it out.
     
  2. jcsd
  3. Mar 25, 2016 #2

    Nathanael

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    From a force perspective, it is clear they both have the same acceleration and thus the linear kinetic energy term is the same for both, and so the rotating one has more total energy.

    From a work-energy perspective, it's a bit more subtle. The incorrect line of thought that you are following is that the force F through the distance d does the same work. This is not true though, because although the center moves through a distance d either way, the contact point at which the force F is being applied is actually moving faster than the center, and so the force F effectively acts through a distance larger than d.

    The power generated by F in the central-case is Fv, but in the tangential-case the power is F(v+ωR).
     
  4. Mar 25, 2016 #3
    Ah! Thanks for clearing it up!!
     
  5. Mar 26, 2016 #4
    Could you please tell me if the answer (A), (A) and (A)? I've taken the puck to be a disk btw, so I = mr2/2.
     
  6. Mar 26, 2016 #5

    Nathanael

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    The answers are (same) (same) (puck 1)
     
  7. Mar 26, 2016 #6
    For parts A and B, since C is correct.
    Case 1:
    τ = Iα
    RF = Iα
    F = mR2α/2
    α = a/R
    a = 2F/m
    s=ut+at2/2
    t1 = ##\sqrt{\frac{ms}{F}}##
    v1 = ##2\sqrt{\frac{sF}{m}}##

    Case 2:
    t2 = ##\sqrt{\frac{2ms}{F}}##
    and v2 = ##\sqrt{\frac{2sF}{m}}##

    Could you please point out the error?
     
  8. Mar 26, 2016 #7

    Nathanael

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    This is what led to your false answer. Why should this be true? (It's true when rolling without slipping, but it need not be true in general.)

    The solution to part A and B boils down to this: we have two objects with the same mass and the same force applied on them, therefore they will both have the same center-of-mass-acceleration a=F/m. The motion of their centers will be identical.
     
  9. Mar 26, 2016 #8
    Oh, of course. Thank you so much!
     
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