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Homework Help: Change in angular momentum

  1. Dec 28, 2009 #1
    1. The problem statement, all variables and given/known data
    a steel disk 400mm in diameter and a mass of 48.98 kg accelerates from 6.28rad/s to 43.98rad/s calculate the change in angular momentum.


    2. Relevant equations
    final angular momentum = intial momentum

    (I) final x (w) final = (i) initial x (w) intial

    I= mr _2


    3. The attempt at a solution

    (I) initial = 48.98 x 200x10^-3_2 = 1.95 kgm_2/s


    (I) final = (I) initial x (w) initial/ (w) final = 1.95 x 6.28/ 43.98 = 0.278 kg/m_2


    change in angular momentum = final - initial

    (I) initial x (w) initial - (I) final - (w) final

    = 1.95 x 6.28 - 0.278 x 43.98 = 19.56 x 10 -3 kgm_2/s

    can someone check if this correct thanx again....
     
    Last edited: Dec 28, 2009
  2. jcsd
  3. Dec 28, 2009 #2
    You calculate a change in angular momentum in part 3 after assuming the angular momentum is constant in part 2.

    The angular momentum is I * ω, and since the moment of inertia I is constant, the change in angular momentum is I * Δω, where

    ( http://en.wikipedia.org/wiki/List_of_moments_of_inertia )

    I = m * r^2 / 2 = 48.98 kg * ( 0.4 m / 2 )^2 / 2

    and

    Δω = ( 43.98 - 6.28 ) rad/s.
     
  4. Dec 28, 2009 #3
    so I = 3.918kgm_2/s and Δω = 37.7 rad/s

    and the change in angular momentum is = I * Δω = 3.918 x 37.7 = 147.70 kgm_2/s

    thanx for putting me right....
     
    Last edited: Dec 28, 2009
  5. Jan 15, 2012 #4
    This is wrong, from the handout I have the Angular Momentum = Iw(omega)

    There is no division of 2. Wikipedia though helpful, isn't the best source of evidence to use because anyone can edit it.

    Where did you get 3.918 kgm^2/s from calculating this:

    48.98 kg * ( 0.4 m / 2 )^2 / 2


    I'd like to ask for help since I have the exact same question to do.

    Using the same values I got this below:

    Angular Momentum = Iw

    I = mr^2 = 48.984 x (0.2)^2 = 1.95936 = 1.96 kgm^2/s

    w2 - w1 = 43.98 - 6.28 = 37.7 rad/s^2

    therefore: 1.96 x 37.7 = 73.892 kgm^2/s


    The next question asks for the change in angular kinetic energy.

    This is what I got:

    Equation - Angular ke = 1/2Iw^2

    1/2x(1.96 x (37.7)^2) = 1.39x10^3

    Angular ke = 1.39 kJ

    Is this correct??
     
    Last edited: Jan 15, 2012
  6. Jan 16, 2012 #5
    The 1/2 is for the moment of inertia of a solid disk about its center perpendicular to the plane of the disk, not angular momentum. I used Wikipedia because it's a known result, and I didn't want to do this:

    \begin{equation} I = \int dm \ r^2 = \int \left(\frac{M}{\pi R^2} r d\theta dr\right) r^2 = \frac{M}{\pi R^2} \int_0^{2 \pi} d\theta \int_0^R dr r^3 = \frac{1}{2} M R^2 \end{equation}
     
  7. Jan 17, 2012 #6
    Yes my calculations were wrong however I did calculate them again and got the write answers through a little research :

    K^2 = d^2/8 ( This is for a soild Disk i.e Steel Disk)

    Where I=mk^2

    Where Iw2 - Iw1 ( change in angular momentum)

    As for change in angular kinetic energy you are using the correct formula just wrong values.
     
  8. Jan 18, 2012 #7
    @EricAngle - appolagies haha.

    @series111 - I've actually figured it all out now.

    My method is wrong because above I do this: (w2-w1)^2

    But the correct method is (w2^2-w1^2)

    Therefore: 1/2 I (w2^2-w1^2).

    But glad you replied. Cheers.
     
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