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Why doesn't a bullet tumble?

  1. May 26, 2010 #1
    Had a quick search, didn't find anything.


    So, we have been told on our course, that a spinning extended object will "re-align" itself such that it is spinning along a principle axis with the highest moment of inertia, such as to conserve angular momentum, but minimise rotational kinetic energy.

    By this process, why does a bullet fired from a rifled barrel not "want" to tumble, rather than fly straight and stable?
     
  2. jcsd
  3. May 26, 2010 #2

    Andrew Mason

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    The rifle barrel has grooves that cause the bullet to rotate on its long (principal) axis so it leaves the barrel spinning. This spinning gives the bullet angular momentum. The angular momentum points along the principal axis. In order to change the direction of that angular momentum, a torque has to be applied to the bullet:

    [tex]\vec{\tau} = \frac{d\vec{L}}{dt}[/tex]

    The greater the spin, the greater the angular momentum and, hence, the greater the torque that is required to change the angular momentum.

    When a bullet passes through something, that spin may be lost. When the non-spinning bullet emerges it will begin to tumble.

    AM
     
  4. May 26, 2010 #3

    Filip Larsen

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    I'm not expert in firearms, but I do think that most projectiles to some extend precess (that is, the symmetry axis rotates around the velocity vector) and that it is aerodynamic forces (that is, center of pressure being aft of center of mass) that keeps this motion from growing into a tumble. However, I don't think the initial precession is evolved from pure rotation around a minor principal axis - the time frame seems much too short for that - but more is a result of the dynamics during the acceleration phase in the barrel.

    If the bullet were to be traveling in vacuum it would, like you have been told, eventually begin to tumble after some time (hours?) depending on rate that the non-pure rotation is turned into heat for the given material and geometry.
     
  5. May 26, 2010 #4
    Yes, this seems to be the answer. Someone else said the same thing to me.
     
  6. May 26, 2010 #5
    I don't follow this. In a vacuum there would be no friction or drag to alter the orientation or speed of the bullet. What is this non-pure rotation you are takling about, and why would it turn into heat?
     
  7. May 26, 2010 #6

    Filip Larsen

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    Pure (or torque-free) rotation of a rigid body is when the rotation axis coincide with one of the principal axis which means that the rotation axis also coincide with the angular momentum vector and that the rotation itself therefore does not produce any torque.

    In non-pure rotation (when the body is rotating around an axis that is not a principal axis) the rotation will induce a torque on the body. If the rotation axis is fixed (like with a dynamically unbalanced wheel), the "bearings" of the rotation axis will react with an opposite torque, but if the body is rotating freely in space, there is nothing to counter the torque and the direction of the rotation axis in inertial space will rotate around the angular momentum vector and this is called tumbling. This also means, that different parts of the body will experience different acceleration forces over time.

    Now, if you look a rigid body made of a real physical material, the body is not really absolutely rigid, but only semi-rigid, which means that while the parts of the body may overall remain at fixed positions relative to each other they will shift with friction relative to each other at least a little bit if the parts experience different forces (that is, if the internal forces between the parts are non-zero). If the internal forces are oscillating like they are in the non-pure rotation described above, a little bit of power is lost as thermal energy thus converting a bit of rotational energy into heat in the body.

    So, finally, we have a semi-rigid projectile which does not rotate exactly around its symmetry axis, meaning that the rotation is slightly non-pure. Since for an oblong body like a projectile the symmetry axis is also the minor (as opposed to major) principal axis, this non-pure rotation will result in the direction of the rotation axis slowly diverging from the minor principal axis and converging on a major principal axis. This is so since rotation around a major principal axis will represent less rotational energy (reduced by friction) for the same angular momentum (which is conserved in free rotation in space). Or, in other words, rotation around a minor principal axis is unstable for a semi-rigid body (unstable like a ball resting on a hill-top) whereas rotation around a major principal axis is stable (like a ball resting a the bottom of a bowl), and rotation "in between" the principal axis (tumbling) makes the body convert rotational energy into heat due to flexing of the body.

    Tumbling is actually very easy to see. Just pick up a box sized body like a brick or a hardcover book (with a rubber band around) and toss it the air. Such a body has three different principal axis and only rotation around the major and minor of them will appear pure. Toss it around the middle principal axis (that is, rotate it around an axis parallel to the direction the text lines would have if it is a book) and the box will tumble wildly in the air.

    I also seem to recall one of my text books on the subject mentioning that a spin-stabilized satellite once were lost when its rotation around the long symmetry axis was "converted" into a rotation around an axis perpendicular to that (relative to the body) due to a delayed separation from its kick stage. With the stage still attached the two combined bodies acted like a prolong body which has unstable rotation around the symmetry axis. After the kick stage was released the satellite was now oblate and the rotation again converted back to coincide with the symmetry axis as the satellite was designed for, but, alas, it ended up rotating the opposite way of the intended direction and could no longer deploy its angular momentum dump device (a yo-yo).
     
    Last edited: May 26, 2010
  8. May 26, 2010 #7

    Andrew Mason

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    I don't see how the body can "induce a torque" on itself. If there is no external torque there can be no change in the angular momentum vector. What is the source of the (external) torque on the bullet?

    AM
     
  9. May 26, 2010 #8
    Also, due to the gyroscope effect, any torque trying to tumble the bullet will actually deflect it.
     
  10. May 27, 2010 #9

    Filip Larsen

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    I probably shouldn't have called it torque. For a torque free rotation the angular momentum vector is of course constant no matter what the rotation axis is.

    In inertial space the relationship between angular momentum, the moment of inertia and the rotation vector of a body is

    (1) [tex] L = I(t) \omega[/tex]

    For a rotating body the coordinates of the moment of inertia matrix will in general be time-varying as a function of the body attitude, for instance expressed as

    (2) [tex] I(t) = \Phi(t) I_b \Phi(t)^T [/tex]

    where [itex]\Phi[/itex] is the rotation matrix of the body system with respect to inertial system and [itex]I_b[/itex] is the moment of inertia in the body system, which is constant for a rigid body.

    Since [itex]L[/itex] is constant for a torque-free rotation but [itex]I(t)[/itex] is time-varying, equation (1) implies that the rotation vector, [itex]\omega(t)[/itex], also may be time-varying in general. In the special case that [itex]\omega(t)[/itex] is an eigenvector of [itex]I(t)[/itex] we can write (1) as

    (3) [tex] L = I(t) \omega(t) = \lambda \omega(t)[/tex]

    where [itex]\lambda[/itex] is a scalar, and in this case [itex]\omega(t)[/itex] is parallel to [itex]L[/itex] and therefore constant. As you may know, such an eigenvector of [itex]I(t)[/itex], when rotated back to the body system, is called a principal axis.

    If [itex]\omega(t)[/itex] is not an eigenvector of [itex]I(t)[/itex] (i.e. the body does not rotate around a principal axis), then equation (1) imply that [itex]\omega(t)[/itex] do indeed vary with time. I'm not sure there is a concise way to explain this, but the resulting motion of the rotation vector relative to the angular momentum vector is that of a slow precession (rotation around L) possibly overlaid with an fast oscillatory motion called nutation. The key point here is that the motion will generate time-varying internal forces in the body.

    When you combine this time-varying internal forces with a semi-rigid body that has friction to those internal forces, you end up with a body that eventually will end up rotating around its major principal axis.
     
  11. May 27, 2010 #10

    Andrew Mason

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    Unless an external torque is acting on a body, the angular momentum vector will not change. Internal friction cannot cause a change in angular momentum.

    If a body is rotating in space with no external torques acting on it, there will be no change in the angular momentum, so there will be no change in the axes about which the object is rotating. It can rotate about different axes at the same time. But that does not mean that the moment of inertia changes with time. So I don't see why I is a function of time if there are no external torques acting.

    AM
     
  12. May 27, 2010 #11

    Filip Larsen

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    I do not claim this and I'm a bit puzzled you think I did. I may not be very good at explaining this, but I would still like to ask you to read my previous post again.

    I suspect that you somehow think that the constancy of I in body coordinates means that it also must be constant in inertial coordinates. If you do, then think about if the moment of inertia for, say a rod, is identical in inertial coordinates no matter what the orientation of the rod is. The moment of inertia is a tensor and coordinate matrix instances of it in one coordinate system can be rotated to another coordinate system with a transform like the one in equation (2).

    You may also want to search your references for the so-called Newton-Euler equations that is a combination of the differential equations for translational and rotational motion of a rigid body. In that you will notice a fictional force term for the torque that is non-zero when the angular velocity is not parallel to the angular momentum.
     
  13. May 27, 2010 #12

    Andrew Mason

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    Are you talking about a rigid body? If so, please explain how the moment of inertia can change with time.

    AM
     
  14. May 28, 2010 #13

    Filip Larsen

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    Yes, I am talking about a rigid body that have a constant moment of inertia matrix in body coordinates, but which in general will be time varying when transformed to inertial coordinates.

    The equation [itex] L = I \omega[/itex] is a tensor equation that you can choose to coordinate in whatever coordinate system you like. If you, like I did in my previous posts, want to get the coordinates relative to inertial space, then you have to make sure the involved tensors are transformed to inertial coordinates, specifically that a (constant) moment of inertia matrix that is coordinated in body coordinates will have to be transformed to inertial coordinates as the body orientation changes relative to inertial space.

    You could also, just as valid, coordinate the tensors in body coordinates. In this case the L vector would not have constant element, but the moment of inertia matrix would.

    I'm running a little late for work, so I have to end for now. If you have access to Goldstein, I can refer you to the section "Methods of Solving Rigid Body Problems and the Euler Equations of Motion" where Goldstein transforms the equations of rotational motion to the body system and show how you get a fictitious torque, that is, that even with zero external torque the time derivative of the angular momentum is not zero for a body whose rotation vector is not parallel to the angular momentum vector. It is this fictitious torque I originally referred to when I was saying "induced torque".
     
  15. May 28, 2010 #14

    D H

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    Filip is correct, Andrew.

    You have probably seen the rotational analog of F=ma written as τ=I dω/dt. As you know, F=ma is correct only if the mass is constant. Just as F=dp/dt is a more generic way of writing Newton's second law, it is better to write the rotational analog as τ=dL/dt, where L=. The general expression is thus τ=dI/dt ω + I dω/dt, and this reduces to the simple freshman physics τ=I dω/dt only if I is constant from the perspective of an inertial observer. While the inertia tensor of a rigid body is constant in a frame rotating with the rigid body, it is not necessarily constant from the perspective of an inertial observer.

    Imagine a body that is not subject to any external torques and is rotating such that the angular momentum vector is not aligned with one of the body's principal axes. The body will undergo a torque-free precession. From the perspective of an inertial observer, the object's angular momentum vector will be constant, but the angular velocity and inertia tensor will vary with time. From the perspective of a body-fixed observer, the object's inertia tensor will be constant, but the angular velocity and angular momentum will vary with time.

    Going back to the rotational analog of Newton's second law, I'll write it as

    [tex]\boldsymbol{\tau} = \left(\frac {d\mathbf L}{dt}\right)_I[/tex]

    Here I have used the subscript I on the derivative to explicitly denote that this is the time derivative of the angular momentum as observed by an inertial observer. Before going on, I need to take a side journey into the relation between the time derivatives of some vector quantities b]q[/b] as observed by inertial and rotating observers. I'll assume a common origin; there is no reason to generalize to affine+rotating transformations. The time derivatives of some vector quantity q as observed by these inertial and rotating observers are related via1

    [tex]\left(\frac {d\mathbf q}{dt}\right)_I =
    \left(\frac {d\mathbf q}{dt}\right)_R + \boldsymbol{\omega} \times \mathbf q[/tex]

    Applying this to the angular momentum vector yields

    [tex]\left(\frac {d\mathbf L}{dt}\right)_I =
    \left(\frac {d\mathbf L}{dt}\right)_R + \boldsymbol{\omega} \times \mathbf L[/tex]

    The left hand side is the external torque. Assuming a rigid body, the time derivative of the inertia tensor in the rotating frame vanishes. With this assumption,

    [tex]\boldsymbol I \frac{d{\boldsymbol{\omega}}_R}{dt} =
    \boldsymbol{\tau} -
    \boldsymbol{\omega}_R \times (\boldsymbol I \boldsymbol{\omega}_R)[/tex]

    The second term on the right, -ω×() is the inertial torque. It is a fictitious torque, somwhat analogous to the Coriolis force (but without the factor of 2).


    -----------------------------------------

    1 You can find hand-waving proofs of this in most undergraduate classical physics texts. A non-handwaving proof requires showing that the time derivative of the transformation matrix from the rotating to the inertial frame is

    [tex]\frac{d}{dt}\boldsymbol T_{R\to I} =
    \boldsymbol T_{R\to I}\boldsymbol S(\boldsymbol{\omega}_R)[/tex]

    where S(ωR) is the skew symmetric matrix generated from the angular velocity vector ωR as expressed in rotating frame coordinates.
     
    Last edited: May 28, 2010
  16. May 29, 2010 #15

    Andrew Mason

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    Interesting. That is what is great about PF - one gets exposed to things one either didn't take or slept through in physics classes.

    I am having difficulty understanding exactly what you are saying. Are you saying that the angular velocities about each principal axis changes with time? (we are talking about a rigid body with no external forces or torques acting on it, rotating with three rotational degrees of freedom).

    The angular velocity is a vector that is perpendicular to the plane of rotation about an axis. In three dimensions there are three degrees of rotational freedom, so there are three angular velocities. (The total angular velocity would just be the vector sum of these three angular velocity vectors, which is not a particularly useful quantity). I don't see how any of those three angular velocity vectors would change with time. If they do not change with time, and the moments of inertia about each principal axis does not change with time (rigid body), then the angular momentum about each axis cannot change with time.

    Can you give an example where a free rigid rotating body has a time dependent moment of inertia, and time dependent angular velocities, about each of its principal axes?

    AM
     
  17. May 29, 2010 #16

    D H

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    You must have slept through lecture on why "the polhode rolls without slipping on the herpolhode lying in the invariable plane" (Goldstein, Classical Mechanics). You apparently didn't buy the nerdy T-shirt either.

    Correct.

    I derived the equation for this in post #14,

    [tex]\boldsymbol I \frac{d{\boldsymbol{\omega}}_R}{dt} =
    \boldsymbol{\tau} -
    \boldsymbol{\omega}_R \times (\boldsymbol I \boldsymbol{\omega}_R)[/tex]

    Here the subscript R on the angular velocity vector ω denotes that the angular velocity is expressed in rotating (i.e., body-fixed) coordinates.

    In the absence of any external torques, the above becomes

    [tex] \frac{d{\boldsymbol{\omega}}_R}{dt} = \boldsymbol I^{-1}\left(
    -\boldsymbol{\omega}_R \times (\boldsymbol I \boldsymbol{\omega}_R)\right)[/tex]

    There is an easy to demonstrate this behavior. Put a rubber band around a rectangular (not square) hardcover book. The book will have three unequal moments of inertia. The directions of the principal axes are
    1. Parallel to the binding of the book (but passing through the middle of the book rather than the edge) (i.e., up and down the middle page of the book),
    2. Parallel to the face of the book but perpendicular to the binding of the book (i.e., along the writing of the middle page of the book), and
    3. Normal to the face of the book, passing through the center of the front cover.

    Toss the book in the air three times, once each with a flip about the three principal axes. Nothing untoward will happen when you try to make the book rotate about axis #1 or #3. You will find that the rotation is rather bizarre when you try to flip the book about axis #2.

    :surprised
    That vector is an extremely useful quantity. Even more useful is the angular velocity vector expressed in an arbitrary body-fixed frame. It is the principal axes that in practice are not particularly useful. Think of an airplane. A natural set of airplane-fixed coordinates is +x pointing to the front of the plane along the longitudinal axis of the plane, +y pointing out the right wing, and +z down. Those axes are much more useful than the principal axes. The principal axes change from flight to flight, and even within flight. For example, there might be a rather heavy passenger in seat 9A and a baby in seat 9F. Or maybe the cargo wasn't loaded perfectly. Or the left engines consume slightly more fuel than do the engines on the right.

    Have you ever tried to read a physics text or article written before 1900 or so? It is painful to do so because the authors didn't have the expressibility that modern vector notation provides. Reading how Euler derived Euler's equations is torturous, for example.

    The rotating book described above. While the inertia tensor is constant in a book-fixed frame, it is not constant when expressed in inertial coordinates.

    Google the phrase "the polhode rolls without slipping on the herpolhode lying in the invariable plane". You will find not only T-shirts but also mathematical descriptions and movies.
     
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