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Rotational inertia and areal velocity

  1. Jul 16, 2009 #1
    One expression for angular momentum is L=2m*(dA/dt),where m is the mass of the body in concern and (dA/dt) is the areal velocity of this rotating body.

    What is the logical physical explanation (or description) for this?

    Is there any other physical importance of this expression (or result) other than the fact that it helps to prove Kepler's second law?
  2. jcsd
  3. Jul 16, 2009 #2


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    Hi Urmi Roy! :wink:

    For a complete explanation and proof, see http://en.wikipedia.org/wiki/Areal_velocity :smile:
    I've never come across areal velocity except in relation to Kepler's laws, so no I don't think so.
  4. Jul 16, 2009 #3
    Hi tiny-tim !!

    Thanks for the link. I went thorugh it,but you see, it wasn't the proof that I was looking for,I would rather like to know if this formula has any 'deep inner meaning and concept about physics' or is it just a formula that physicists derived to help in practical life.

    Again,thanks for confirming that this formula is used mainly for proving the validity of Kepler's 2nd law.
  5. Jul 17, 2009 #4
    According to the diagram and explanation there, if the angular velocity is pi rad/sec, then
    r x v gives zero and no areal velocity, which is wrong. The velocity vector is not as meant there, it is tangential to the path and includes the white portion between the purple path and the green portion.
  6. Jul 17, 2009 #5


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    Delta t is not equal to 1 second, though.

    And, velocity is the change in position over some increment of time.

    In the case of limits, that increment of time is much, much smaller than the time it takes to travel halfway around the circle no matter how fast you make the angular velocity.

    If there's any real discrepancy in the diagram, it's that r(t + Dt) is too far away from r(t), but you have to draw it wrong or no one could see any difference between the two.

    i.e. - the distance between the two position vectors should be so small that the secant line is tangent to the path.
    Last edited: Jul 17, 2009
  7. Jul 18, 2009 #6
    Thanks for your help in regard to 'areal veocity'.

    By the way, I was just wondering if there is any way to continuously change the linear inertia (mass) of a body,just as it is possible to interchange moment of inertia of a body(in the example of the spinning dancer,she changes her moment of inertia by pulling in or pushing out her arms).

    Please let me know if this is possible at all.
  8. Jul 18, 2009 #7


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    Yes, it is. In fact, that's how a rocket engine works. Mass from the rocket is thrown out the back at incredibly high velocities and, thanks to conservation of linear momentum, the rocket moves the opposite direction. Obviously, the mass of the rocket must be decreasing if it's being thrown out the back.

    There's even a tutorial talking about it in the Math and Sciences Learning Material forum:

    I'm not sure that's exactly what you had in mind, but it is the closest you would get. With linear motion, any reduction in mass has get rid of the mass somewhere.
  9. Jul 18, 2009 #8
    Thanks BobG.

    I had read about this concept,though in much simpler terms when I was in class 11,but it had completely skipped my mind.

    What led me into thinking about this issue was that,since every linear quantity in mechanics has a rotational analogue,(and vice versa),which are similar in their properties,this property of rotational inertia that it can be varied in magnitude,may also exist in linear inertia.

    Ofcourse,this isn't exactly a property as such.Anyway,the example of the rocket provides the best comparison.

    Thankyou very much!!
  10. Jul 19, 2009 #9
    I've just 2 more regions in regard to rotational inertia to clear up.The first one,I'd like to speak of here.....
    During the derivation of various expressions related to rotational mechanics (e.g. the expression for net rotational kinetic energy of a body, derivation for net torque acting on a body),as done in my book, the expression of moment of inertia (I=mr^2) suddenly springs out,after we sum the individual kinetic energies or torques acting upon the infinitesimal particles (depending on the derivation we're doing).

    Then,the book says,that since the formulae for rotational mechanics and linear mechanics are similar,except that 'mass' is replaced by 'moment of inertia',moment of inertia is called the rotational analogue of mass.

    Is this just by coincidence,since as I said, the expression for moment of inertia just springs up incidentally in almost all derivations--or is there any bigger reason for this?

    I've been told that these can be explained by lagrangian dynamics and noether's theorem,but surely there must be a simpler and more 'physical' way of explaining it!
    I just want a logical, intuitive idea of it,that's all.
  11. Jul 20, 2009 #10
    I'll just put forward my last problem,just in case anyone can help me with this one instead....I have brief descriptions of the parallel and perpendicular axes,and then my doubts.

    "The parallel axis theorem allows you to compute the moment of inertia of an object more easily. In many cases you'll consider there is a particularly easy axis about which to compute the moment of inertia, which is parallel to the axis you actually want to find the m.o.i about. Well by the parallel axis theorem you do the easy one and then 'shift' the axis to the one you want."

    ------This 'shift' occurs only by adding Md^2 to the m.o.i about the original axis,(where M is mass of object and d is distance between original axis and new one).

    Except for the mathematical proof,there seems to be no substantiation for this theorem.What is the term Md^2, separately?Is there no way we can arrive at this theorem by physical reasoning?

    "Similarly for the perpendicular axis theorem, for planar (ie 2d) objects sometimes its easier to compute the moi about axes in the plane and then use the theorem to get the moi for an axis perpendicular. In the case of symmetric planar objects you get things like I_x = I_y = (1/2)I_z, which means you only need to find the moi about one axis and you have them about the other 2 perpendicular directions"

    -----Again,is there any way we can arrive at this by reasoning...I mean,what would one say if I asked why the m.o.i about one axis is equal to the sum of the moi's about two of its perpendicular axes?

    I found the proofs in my book...but as you probably realise,that's not exactly what I'm looking for.

    (m.o.i= moment of inertia)
  12. Jul 21, 2009 #11


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    parallel axis theorem

    No, there is only one parallel axis for which adding md2 works … the axis through the centre of mass. :wink:
  13. Jul 21, 2009 #12
    Right,I understand that md2 is added to only the moi about the central axis,but again,referring back to post 10,what is this 'md^2'---its not the moment of inertia about the new axis,surely.
  14. Jul 21, 2009 #13


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    Yes, it is …

    provided you pretend that the whole mass is concentrated at the c.o.m. :wink:

    So the MI about any axis equals the MI of the same mass concentrated at the c.o.m, plus a "spread-out-ness" MI, of the actual distribution of mass about the parallel axis through the c.o.m.

    (This is like the energy of a solid body: it equals the energy of the same mass with the velocity of the c.o.m, plus the rotational energy)​
  15. Jul 21, 2009 #14

    If we have the original axis through the c.o.m, and consider that the entire mass is concentrated there,the m.o.i should be zero,as r=0 for this case.....am I missing anything here?
  16. Jul 21, 2009 #15


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    No, I mean the MI about the new axis. :smile:
  17. Jul 21, 2009 #16
    So what we're doing, is that (the steps)
    1.we find out the MI of the body about the axis through its centre of mass,like we usually do.(by summing up the MIs of the individual particles constituting the body)

    2. We assume that the entire mass is now concentrated at the c.o.m and,calculate the MI of this point mass about the new axis.

    But,here, we're adding two unlike quantities--in the first step,we are not assuming the body's mass to be concentrated at one point,like a point mass,unlike the 2nd step.

    Also,a few days ago,I was doing a sum,where basically I had to calculate the MI of a body about an axis,and I,by mistake, proceeded by finding the body's c.o.m and treated it like a point mass,of which I calculted the MI about the axis.

    Obviously,my answer was wrong (and I was scolded badly for it!)--so in the asumption that we made in the 2nd step--that the whole mass is concentrated at the c.o.m,we should face a similar problem.We can't find the MI of a rigid body by simply considering it a point mass at the c.o.m

    Is there a fundamental difference between my sum and what you're stating?
  18. Jul 21, 2009 #17


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    No, we're adding two quantities of exactly the same kind:

    they're both moments of inertia.
    That's right, not by simply considering it a point mass …

    we consider it as a point mass and then we add an extra term for the spread-out-ness.

    For example, consider two identical bodies on the end of identical rods …

    bash one of the bodies about so that it changes shape but not mass …

    obviously, we expect the MIs to be very similar, but not identical.

    The extra term includes the information about the shape.

    (and if the rod is long enough, we can ignore the shape, and treat the body as it it was a point mass … for example, the MI of a planet round the sun usually treats the radius of the planet as zero)
  19. Jul 22, 2009 #18
    I think I'm getting it,but I have a teeny-weeny problem left.......in the process that we find the 'spread-out-ness' MI of the mass, we're calculating the 'spread-out-ness' MI about the axis through the c.o.m---but in the end, we're trying to calculate the total MI about the new axis-----then how should the 'spread-out-ness' MI about the c.o.m axis matter at all?

    When we spin the body about the new axis, the spatial arrangement of the individual particles constituting the body about the new axis matters--that's what affects the MI about the new axis,(it's not the spatial arrangement about the c.o.m axis that matters here).

    In your analogy with the energy of a solid mass, suppose we have a cylinder rolling down a ramp,we calculate the velocity of the c.o.m (for linear kinetic energy) and the rotational velocity about a particular axis,suppose the axis through the edge of the ramp.

    Here,all the calculations are made w.r.t the frame of the ramp---however,in the parallel axis theorem,we're adding the moments of inertia about the c.o.m axis and the new axis separately,and not one particular axis.

    How is it that the total MI about the new axis is equal to the spread-out-ness MI about the c.o.m axis and the MI of a point mass (of equal mass of the body) about the new axis??
  20. Jul 22, 2009 #19


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    "spread-out-ness" is "spread-out-ness" … it doesn't depend on where the axis is …

    the relative spatial distribution is all that matters.
    i] the rotational velocity is the same for all parallel axes.

    ii] yes, all the calculations are made w.r.t the frame of the ramp …

    but the linear energy and the rotational energy are treated entirely separately, just as the "point mass" MI and the "spread-out-ness" MI are treated entirely separately. :wink:
  21. Jul 22, 2009 #20
    Thanks tiny-tim for your clarification. I really appreciate your continuous and expert help!!

    I also found a website which supports our discussion,but it contains two lines that I didn't understand(the bold lines).Perhaps if you explained it to me,it would help me finally get things clear--after all,the more I read,the clearer things become.

    It goes like this......"The parallel axis theorem you can think of like this: if you're rotating about an axis that doesn't go through the center of mass,you have to rotate the object about its center of mass,plus you have to rotate the center of mass about the axis. If you had a point mass M distance d from the axis, you'd get I=Md^2. That's just to move (translate) the object itself about the axis, but doesn't actually rotate the object."

    Sorry to keep bothering you like this, it won't take much longer!
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