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Thread: Rotational energy problem

  1. #1
    s3a is offline
    Super Member
    Nov 2008

    Rotational energy problem


    (Correct) Answers:
    (a) 0.146938776m
    (b) 0.102857143m

    My work:

    First off, my problem likely lies with the fact that it is rotating instead of being translated like in regular translations (like a ball being thrown etc). I remember my teacher saying something of the sort and I checked my notes and it seems to be about moment of inertia which has me lost. I have a feeling that in the following my mistake lies with the initial kinetic energy.

    Ok so what I did for (a) which gives me half the right answer is:
    Ui = 0, Ki = 1/2 * m * v^2, Kf = 0, Uf = mgh
    Ef = Ei
    mgh = 1/2 * m * v^2
    gh = 1/2 * v^2
    h = (1/2 * v^2)/g
    h = 1/2 * 1.2^2 / 9.8
    h = 0.073469388

    as for (b), it seems to be very similar.

    Any help would be GREATLY appreciated!
    Thanks in advance!
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Aug 2008
    If a body slips,it has only translational Kinetic energy

    and its $\displaystyle KE = \frac{ M{V_{cm}}^2}{2}$

    If it doesnot slip than it has Rotational energy and translational energy as well

    the $\displaystyle KE = \frac{ M{V_{cm}}^2}{2}+\frac{I{\omega}^2}{2} $

    I for ring is (mr^2)/ 2

    for hollow sphere its (2mr^2)/3

    And for solid sphere its 2(mr^2)/5


    you will get correct answers now..feel free to ask incae you still find it difficult to digest.....
    Last edited by ADARSH; Apr 27th 2010 at 01:49 AM. Reason: ...
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