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Multi-angular momenta

  1. Jun 13, 2008 #1
    I have a Q re angular momentum (L) that's causing some heated discussion. A bullet is fired off center toward a rotatable target with an axis from, say, 100m (like a ballistic pendulum). The bullet has a fixed linear momentum and an L that remains constant as it travels toward the target -- Pxr stays constant as R and the angle change in sync.

    Now say the shooter's aim is off and the bullet will miss the target a bit. In this case the bullet has the same constant numerical value of linear momentum (with a small change in the vector due to the aim) and a similarly constant L, but this L is numerically different from the first. This means that the bullet can have an angular momentum with a different target separated by distance and angle from the first target (but all in the same fixed unmoving coordinate system), and that this L is different numerically from the first.

    The same bullet now has TWO different numerical angular momenta simultaneously. It then follows that the bullet has a simultaneous momenta with MANY (theoretically infinite) "targets". The angular momentum, in this scenario, is not a specific, single, explicit, inherent quality like linear momentum, but rather it is multiple, with each relative to some other object (axis) within the frame. Does anyone agree or disagree? (A rotating body's L is different and seems specific and inherent, but that's not the debate.)
  2. jcsd
  3. Jun 13, 2008 #2
    what you want to say is that the 'moment arm' stays constant i.e. the value [itex]r\sin(\theta)[/itex] stays constant.

    yes. you are right. But the way you state it is wrong. When we talk about angular momentum, or any moment, be it torque or moment of inertia, we need to specify about which axis we are calculating that moment. This is the reason the moment of Inertia for say, a cylinder is different along different axis. About a definite axis, the angular momentum is always unique and an inherent quality.

    However, you would also like to note that since a sphere is symmetric about it's center, for the sphere, the moment of inertia is the same about any axis passing through the center.

    Take the case of linear momentum. If an object having mass 'm' and velocity 'v' has momentum [itex]p = mv[/itex]. Now, let us say you are moving at a velocity 'v', so, for you as an observer the velocity of the object now is 0 and hence the linear momentum [itex]p = 0[/itex]. similarly, depending on the frame of your reference even the linear momentum has different values.

    There is no such thing as 'absolute momentum'. It is always relative to the frame of reference you are viewing it from and for all moments, it is relative to the axis you choose.
  4. Jun 14, 2008 #3
    Thanks, rohanprabhu

    So, in the same frame linear momentum is fixed (sans any force) and constant no mater how one looks at it. Angular momentum (again in a same frame) of any particular mass is relative to whatever axis one chooses to relate it to. It shouldn't have been that hard! Slap! Slap! As you point out (the obvious :redface: ) even singular rotating masses have different Ls depending on which axis one chooses.
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