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Rate of change of angular velocity

  1. Feb 25, 2013 #1
    Let a particle of mass m be rotating in a circle, radius be r, axis of rotation through the center and perpendicular to the plane of motion.
    Now if the radius is made decrease somehow at some rate until it get's zero, how do I find the rate of change in angular velocity?
    Also, the angular momentum being conserved, dL/dt=0, ie, no torque acts on it. Yet since the angular velocity is changing, there has to be an angular acceleration. So how can there not be a torque? Is it because w*dI/dt exactly cancels I*dw/dt?
  2. jcsd
  3. Feb 25, 2013 #2
    I haven't studied this yet but I'll give it a shot

    Using [itex]L=rmv[/itex] you can work out [itex]\frac{dv}{dr}[/itex], since [itex]L[/itex] and [itex]m[/itex] are constants.

    Using that equation and a given rate of change of the radius ([itex]\frac{dr}{dt}[/itex]) you can work out [itex]\frac{dv}{dt}[/itex].


    Edit: As for your other questions ... idk what torque or interia are :3
  4. Feb 26, 2013 #3
    dv/dr works out to -v/r...
    dv/dt works out to -v/r(dr/dt)...
    is that it?
    i really don't think that's all!
    let's wait for some other opinions!
  5. Feb 26, 2013 #4


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    That really is pretty much about it.

    If you want to visualize what's going on better, and see exactly how the angular velocity can change without any torque being applied, consider a single moment frozen in time. The mass is moving with linear velocity v in a direction perpendicular to the radius. Apply more centripetal force to reduce r, and v doesn't change.
  6. Feb 26, 2013 #5
    ok got it... thanx to both of you.
  7. Feb 27, 2013 #6


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    In the diagram the dotted line represents a section of a typical trajectory when a circling mass is pulled closer to the central axis.

    The dark grey arrow represents the centripetal force.

    To understand the angular acceleration it helps to think of that centripetal force as decomposed in two perpendicular components:
    - one at right angles to the instantaneous velocity
    - one parallel to the instantaneous velocity.

    The parallel component increases the angular velocity.

    The general expression is 'contraction of a rotating system'. The example you ask about, a mass circling a central point, is a specific case of that.
    When a rotating system is contracting the centripetal force is doing work, and the overall kinetic energy of the system increases.

    The only case where the centripetal force is not doing work is when the motion is perfectly circular.
  8. Feb 27, 2013 #7


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    To find the angular acceleration as a function of the velocity in radial direction. First some remarks:

    I will use the following names:

    [itex]v_r[/itex] velocity component in radial direction
    [itex]v_c[/itex] velocity component at right angles to the radial vector (circling component).
    [itex]\omega[/itex] Angular velocity

    As pointed out earlier in this thread:
    Angular momentum L is given by:
    [tex]L = mv_{c}r[/tex]
    So you can compare two states:
    1 - before contraction
    2 - after contraction

    You start with circular motion (state 1), you reduce the radial distance by half, from there onwards circular motion again (state 2). Then the following applies:
    [tex]v_{c,1}r_{1} = v_{c,2}r_{2} [/tex]
    From state 1 to state 2:
    - when the radial distance is halved the tangential velocity is doubled.
    - when the radial distance is halved the angular velocity is quadrupled.

    In reply #4 Nugatory offered the suggestion that 'the angular velocity changes because the tangential velocity stays the same'. Obviously that suggestion is wrong, as it violates [itex]v_{c,1}r_{1} = v_{c,2}r_{2} [/itex].

    To find the angular acceleration as a function of the velocity in radial direction.

    We know that [itex]\omega r^2[/itex] is a constant, so the derivative of that expression with respect to time is zero:
    [tex] \omega r^2 = constant \qquad \Rightarrow \qquad \frac{d(\omega r^2)}{dt} = 0 [/tex]
    [tex]r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0[/tex]
    I want to work towards an expression with the radial velocity component (dr/dt) so I use the chain rule.
    [tex]r^2 \frac{d\omega}{dt} + 2 r \omega \frac{dr}{dt} = 0[/tex]
    Dividing both sides by r2, and rearranging:
    [tex]\frac{d\omega}{dt} = - \frac{2 \omega}{r} \frac{dr}{dt}[/tex]
  9. Feb 28, 2013 #8
    that's understandable... but tell me this... if the angular acceleration, α[itex]\propto[/itex]-r-1, then what will happen th it at r=0?
    in that case, α approaches ∞ and the angular velcity, ω, should also become ∞.
    but seeing this problem using our 'common sense', isn't it so that ω will have a finite value? the radius from the axis becoming 0, the motion then transforms to pure rotation... with K.E. [itex]\frac{1}{2}[/itex]Iω2 being equal to [itex]\frac{1}{2}[/itex]mv2 plus the work done to reduce r!!!
    the two solutions appear contradicting...
  10. Mar 27, 2013 #9
    Hello?? Anyone there?? It's been a long time with no replies.
  11. Mar 27, 2013 #10


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    The math applies to a point mass. A mass sphere will have a finite ω when pulled to the center.
  12. Mar 27, 2013 #11
    A hypothetical point particle has no angular moment of inertia which is why ω goes to infinity.
  13. Mar 27, 2013 #12
    I don't get that. I thought all particles have moment of inertia.
  14. Mar 27, 2013 #13


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    A classical point particle has no intrinsic spin in the way,for example, a sphere can have a spin angular momentum about its center of mass. If a classical point particle is in say a circular orbit of radius ##r## then, after orienting the coordinates as needed, its orbital angular momentum can be written in coordinates as ##L = mrv\hat{z} = mr^{2}\omega \hat{z}##. It doesn't of course have a spin angular momentum in the above sense, on account of it being a classical point particle.
  15. Mar 28, 2013 #14
    I was thinking along the lines that his imaginary particle had no dimensions, so it wouldn't have any angular inertia. What happens at a quantum level is another story.
  16. Mar 28, 2013 #15
    hmm... pretty complicated stuff! will get back to these things when i'll get to know all these things.
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