if a particle moves in a straight line is its angular momentum zero with respect to an arbitrary axis is its angular momentum zero with respect to any specified origin
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Answer:
It can be a bit off-putting at first, but yes, you can have angular momentum even when an object is moving in a perfectly straight line.
Angular momentum is related to the position r and linear momentum p by the vector cross product:
L=r×p
Hence, if we have an object starting out at r0=(0,y0,0)T , and moving along the x axis at a steady velocity v , it's position at a time t is:
r(t)=⎛⎝⎜vty00⎞⎠⎟
The momentum is simply the mass times the velocity vector:
p=⎛⎝⎜m v00⎞⎠⎟
The angular momentum is therefore:
L=⎛⎝⎜v ty00⎞⎠⎟×⎛⎝⎜mv00⎞⎠⎟=⎛⎝⎜00−m v y0⎞⎠⎟
Hence, even though the velocity was purely linear, we do indeed have a non-zero angular momentum.
In general, we can see that the angular momentum is non-zero in general for any motion, since under a frame change r→r′=r+δr , the angular momentum changes as:
L→L′=(r+δr)×p
The cross product is distributive, so:
L′=r×p+δr×p
Hence:
L′=L0+δr×p
This shows that if the angular momentum in one frame ( L0 ) is zero, then it is only zero in frames with an origin chosen such that δr×p=0 . Otherwise, it will in general be a non-zero quantity.
The only places where the angular momentum would be measured to be zero are:
At the particle itself ( r′=0 )
At a position such that r′ and p are parallel or antiparallel.
Angular momentum is dependent on the choice of coordinates (but within a given coordinate choice, it is preserved).
This might seem odd, but if you think in terms of a rotation, then a rotation around the point (0,0) gives you a totally different result to the rotation around (10,3) — so rotations are also origin-dependent!
Therefore L=0, when r=0 i.e. when point lies on the straight line. When a given point does not lie on the line,
Therefore,
