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Freely moving mass

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Given: mass m, moment of inertia I and a impulse applied at a distance x form the center of mass (c.m.).

    Find the angular speed and the speed of the c.m.


    2. Relevant equations

    [tex]\sum M_O = I \ddot{ \theta} [/tex]

    [tex] F\Delta t = \sum m_i v_i [/tex]

    3. The attempt at a solution

    I can solve for the acceleration but do I have to use a momentum balance to solve for angular speed?
  2. jcsd
  3. Jul 13, 2008 #2


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    Hi dirk_mec1! :smile:

    (Do you mean acceleration? It's an impulse … the acceleration is taken to be irrelevant/infinite :confused:)

    For the angular speed, find the torque of the impulse, then use the angular version of Newton's second law. :smile:
  4. Jul 13, 2008 #3
    Hi tim!

    Yes, this solves for the acceleration as shown in angular version of newton second law:

    \sum M_O = I \ddot{ \theta}

    But I presume you mean:

    \sum M_O = I \ddot{ \theta} = I \omega^2 r

  5. Jul 13, 2008 #4


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    Hi dirk_mec1! :smile:
    No … wrong formula … that's something like the formula for a continuous centripetal force making the centre of mass move in a large circle.

    When the force ends, the circular motion stops!

    This is an impulse, an instantaneous force, that makes the body spin on it own axis, with the centre of mass moving in a circle of radius zero.

    When the force ends, the circular motion continues! :smile:

    Newton's second law for impulses takes the forms:

    impulse = momentum after minus momentum before;

    torque of impulse = angular momentum after minus angular momentum before. :smile:
  6. Jul 14, 2008 #5
    Re: impulse

    So in formula form:

    F \Delta t = \Delta p [/tex]

    [tex] \mathbf{L}=\sum_i \mathbf{r}_i\times m_i \mathbf{V}_i [/tex]

    So if I understand you correctly the torque is :

    [tex] F \Delta t \cdot x [/tex]

    And how do I proceed?
    Last edited: Jul 14, 2008
  7. Jul 14, 2008 #6


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    I'm Pulse

    Hi dirk_mec1! :smile:

    (have a delta: ∆)
    Yes and no …

    F ∆t is correct, if you're given F (a force) and ∆t (a time interval).

    But the question doesn't give you those …

    it gives you an impulse …

    let's call it P (for "I'm Pulse" :biggrin:) …

    Then the formula is P = ∆p. :smile:

    And the torque is Pxx
    Well, that formula for angular momentum L is correct in principle (though you could have written it L = ∑ri x pi) …

    but again you're not given the individual ri and pi

    you're given the moment of inertia, I …

    so L = … ? :smile:

    and then ∆L = … ? :smile:
  8. Jul 14, 2008 #7
    Re: I'm Pulse

    Well there´s another formule here in my book it reads:

    [tex] \mathbf{L}= I \mathbf{\omega} [/tex]

  9. Jul 14, 2008 #8


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    Re: I'm Pulse

    Looks good! :smile:
  10. Jul 14, 2008 #9

    [tex] Px=I \omega \rightarrow \omega = \frac{P \cdot x}{ I} [/tex]

    right, Tim?
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