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Moment of inertia confusion

  1. Oct 4, 2009 #1
    My textbook claims that we can describe the kinetic energy of a rigid body as follows:

    K = 1/2 Mv^{2}_{cm} + 1/2 I_{cm}w^{2}

    The first term makes complete sense. Earlier the book showed that we can model an entire body's translational motion (and momentum and gravitational energy) as though its entire mass were concentrated at its center of mass. Naturally this idea carries out to K = 1/2 Mv_{cm}^{2}.

    I don't understand the I_{cm} aspect of this formula. If the object is rotating about an axis that doesn't pass through its center of mass, then we can only describe I_{cm} theoretically through the parallel axis theorem. However, the book derives this formula with the assumption that the object IS in fact rotating about its center of mass.

    I can't help but think I'm conceptualizing rotation all wrong. If I throw an object up in the air, it only has 1 axis of rotation, not infinitely many, right? Like if I say an object is rotating about a point P, I can't also describe it as rotating about a point P' without changing my frame of reference.
  2. jcsd
  3. Oct 4, 2009 #2

    Ben Niehoff

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    A rotation about some axis A that does NOT pass through the center of mass can always be described as circular motion about A plus center of mass rotation. That is,

    [tex]K = \frac12 I_A \omega^2 = \frac12 M r_A^2 \omega^2 + \frac12 I_{cm} \omega^2[/tex]

    This is the parallel axis theorem, in essence. (Notice that the omegas are all the same.)
  4. Oct 4, 2009 #3


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    This is a relevant question: if an object is thrown up in the air, does it only have a single axis of rotation during the flight?

    Let me discuss a different case first, the case without gravitation. Some object is in interstellar space, gravitation is negligable. Then the spinning object is in uniform motion. More precisely, the center of mass of the spinning object is in uniform motion. We think of the object as spinning around a single axis. That is, if we consider the spin we think of that spin as relative to the center of mass of the spinning object. The principle of relativity of inertial motion justifies that. Any velocity of the center of mass is relative, so in that sense there is only a single rotation axis, not an infinite number of rotation axes.

    In the case of an object thrown up in the air the equivalence of inertial and gravitational mass comes into view. If you throw a machanical accelererometer into the air then during the flight the accelerometer will register zero acceleration (actually, air resistance will play up, but you can pump out all the air from a tall cilinder, and then do the falling experiment.)

    Notice that the mechanical accelerometer will register zero acceleration during free fall if and only if inertial mass is equivalent to gravitational mass.

    Given that inertial mass is equivalent to gravitational mass we are justified in treating the spinning object that is in free fall as an object that is in true inertial motion. Therefore we are justified to assign just a single rotation axis to the spinning object during its free fall.

    Note that this applies only if the gravitational field is perfectly uniform. Strictly speaking the Earth's gravitational field isn't uniform, further away from the Earth the field is weaker. But when the vertical travel is no more than couple of meters up and down the tidal effect will be negligable.

  5. Oct 4, 2009 #4
    I'm fairly certain that K = \frac{1}{2} I_{A} \omega^{2} is only valid if the point A is stationary. What if we have something of the form:

    [tex] K = \frac{1}{2}I_{A} \omega^{2} + \frac{1}{2}Mv^{2}_{A}[/tex]

    Last edited: Oct 4, 2009
  6. Oct 5, 2009 #5
    Since I'm not getting an answer, I'm going to assume that you'd get the same result, but with the velocity of the center of mass being different.

    Right now I'm content with the idea that a combination of translational and rotational motion about the center of mass of a rigid object can result in a motion such that the object isn't rotating about its center of mass.
  7. Oct 8, 2009 #6
    The first term is the kinetic energy due to the motion of the center of mass. The second term is the kinetic energy due to the rotation of the object (treated as a rigid body).

    If the body were rotating about the center of mass only (that is, about an axis passing through the center of mass), then the first term would be zero, because v_cm would be zero. For example, a rotating flywheel would have a kinetic energy equal to 1/2 I_{cm}w^{2}, since the center of mass coincides with the center of rotation (by assumption) - therefore it has zero velocity (v_cm = 0) - so the first term in the kinetic energy equation would equal zero.

    If your book assumes that the object is rotating about the center of mass, that can only mean that its considering the rotation (w) of the body separately, perhaps to make it easier to grasp. And then to complete the analysis, the motion of the center of mass is accounted for, with the inclusion of the first term: (1/2 Mv^{2}_{cm}).

    The equation you gave is valid whether or not the object is rotating about its center of mass. There's basically two cases: If the object is rotating about its center of mass, meaning v_cm = 0, then

    K = 1/2 I_{cm}w^{2}

    Second case: If the object isn't rotating at all and is only translating, then w=0, and

    K = 1/2 Mv^{2}_{cm}

    For the general case, the object is experiencing a general state of motion where it is translating and rotating, and you have the equation you started with:

    K = 1/2 Mv^{2}_{cm} + 1/2 I_{cm}w^{2}

    This equation basically says that the kinetic energy of a rigid body consists of translational kinetic energy of the mass center (first term) PLUS the rotational kinetic energy (second term).

    I'm guessing your confusion might have something to do with you thinking in terms of the kinetic energy of a particle, as opposed to the kinetic energy of a rigid body. If a body could be treated as a particle that would mean it has negligible volume, and I_cm would be zero, so you would only have the first term (since the second term would be zero). But for the general case a rigid body has finite size and therefore cannot be treated as a particle, so you need the two terms.
  8. Oct 8, 2009 #7

    D H

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    It has infinitely many axes of rotation. You can arbitrarily pick any point on, or even away from, some translating/rotating object as the center of rotation. The reason for choosing the center of mass as the center of rotation is not because physics is invalid with any other choice. The reason for choosing the center of mass as the center of rotation is because in many cases the mathematics takes on its simplest form with this choice. The translational and rotational equations of motion often are uncoupled from one another in the center of mass formulation.

    There are cases where a choice other than the center of mass makes sense. A couple of examples:
    • An object attached to the Earth at one point but free to rotate in one or more dimensions about this fixed point. In this case, the behavior of the object is purely rotational when one describes the object's behavior from the perspective of the fixed point.
    • A spacecraft that is burnin' and turnin' at a fierce rate. The center of mass will move inside the spacecraft due to fuel consumption. The translational and rotational equations of motion are unavoidably coupled in this case. Ignoring this coupling is a viable option if the center of mass motion is sufficiently small. It is not a viable option for modeling the behavior of a launch escape system. These systems must accelerate and turn at a phenomenal rate to ensure the spacecraft gets away from the exploding rocket underneath the spacecraft. In this case, picking a fixed point on the spacecraft as the center of rotation can make formulating the equations of motion a bit easier.
  9. Oct 9, 2009 #8
    Even the first term doesn't make complete sense at first glance:
    like consider a rod rotating with its end fixed.
    Now, consider a particle at a distance r, its velocity is (rw) so its K = 0.5 * m* (rw)^2
    k=1/2 mr[tex]^{2}[/tex]w[tex]^{2}[/tex]
    k = 1/2 I w[tex]^{2}[/tex]
    Now, I=mr[tex]^{2}[/tex]/3
    Icm = mr[tex]^{2}[/tex]/12
    vcm = r/2 * w
    put that in the equation, mysteriously, it gives the same result.
    Now can somehow please explain this phenomenon
  10. Oct 9, 2009 #9

    Doc Al

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    The rod rotating about a fixed end can be considered to be in pure rotation about that end, thus: KEtotal = KErot(about the end)

    That KEtotal also must equal KErot(about the cm) + KEtrans(of the cm), which is always true.
  11. Oct 9, 2009 #10
    Why is that so? is the question I am asking
  12. Oct 9, 2009 #11
    Wait I think I got it mathematically!
    but I am not good at all this Latex stuff..
    I have to find another way......
  13. Oct 9, 2009 #12

    D H

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    A somewhat snarky answer: We use this expression because it yields the right answer. If it didn't yield the right answer we wouldn't use it.

    A less snarky answer: The parallel axis theorem derives the transformation needed to make the answer come out right.
  14. Oct 9, 2009 #13
    I can't figure it out, perhaps if you could give me some pointers..
  15. Oct 9, 2009 #14
    I think I might've found the mathematical answer to my earlier question:

    Going by what Ben Niehoff said,


    K = \frac{1}{2} Mv^{2}_{A} + \frac{1}{2} I_{A} w^{2} = \frac{1}{2} Mv^{2}_{A} + \left( \frac{1}{2} M(r_{A}w)^{2} + \frac{1}{2} I_{cm} w^{2} \right) [/tex]

    We can use [tex] v_{cm} = r_{A}w[/tex] to get

    [tex]K = \frac{1}{2} Mv^{2}_{A} + \frac{1}{2} Mv^{2}_{cm} + \frac{1}{2} I_{cm} w^{2} = \frac{1}{2} M ( v^{2}_{A} + v^{2}_{cm} ) + \frac{1}{2} I_{cm} w^{2}[/tex]

    Now, the stated v_{cm} is only true if the object weren't translating about its axis of rotation. Thus the true velocity of the center of mass v'_{cm} has to have the magnitude of v^{2}_{A} + v^{2}_{cm}. I can't see how this is generally true, though in one instance the two vectors are perpendicular, so pythagoras would give the correct magnitude. Thankfully, energy is conserved so we can use this special case to our advantage :)
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