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Zero Angular Momentum

 
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Dec6-12, 03:52 PM   #1
 

Zero Angular Momentum


Hello,

I am currently reading about angular momentum. A sentence in my textbook sort of confused me: "when the translational velocity of the particle is along a line that passes through the axis, the particle has zero angular momentum with respect to the axis."

I was wondering if someone could explain what the author is trying to convey.
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Dec6-12, 04:03 PM   #2
 
Are they basically saying that the linear momentum of the particle is parallel to the perpendicular distance between the particle and axis of rotation? When would a situation like that ever occur? That seems kind of strange.
Dec6-12, 04:06 PM   #3
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They are saying that a particle moving in radial direction has no angular momentum. Which is entirely correct.
Dec6-12, 04:36 PM   #4
 

Zero Angular Momentum


It sounds like he is talking about angular momentum (of a particle), which is a little different. A Google search for "momentum of a particle" will find plenty of info.

The angular momentum of a particle with respect to some point, is the cross product of the position vector of that particle with the vector of its (linear) momentum. In the case you mention, zero translational velocity in a particular direction of course means zero translation momentum in that direction, and therefore the cross product which gives the angular momentum will be zero.
Dec6-12, 04:46 PM   #5
 
Quote by Bashyboy View Post
Hello,

I am currently reading about angular momentum. A sentence in my textbook sort of confused me: "when the translational velocity of the particle is along a line that passes through the axis, the particle has zero angular momentum with respect to the axis."

I was wondering if someone could explain what the author is trying to convey.
The particle has zero angular momentum with respect to any point on the axis.

The angular momentum of a particle with respect to a point is the cross product of the displacement vector from the point to the particle, and the linear momentum of the particle with respect to that point. If the displacement vector is parallel to the linear momentum, the cross product of the two vectors is a zero vector.
Dec6-12, 05:12 PM   #6
 
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Quote by MikeGomez View Post
It sounds like he is talking about angular momentum (of a particle), which is a little different.
Nope, I don't think so: the book refers to translational motion and relates that to angular momentum - implying momentum (angular and linear) wrt the reference axis and due to its translational motion and not its 'rotation'. I think you are looking for more in this than there is.
Dec6-12, 06:13 PM   #7
 
Angular momentum for a particle is its times its angular velocity, right? Angular velocity is the rate of change of angular position with respect to some axis. A particle heading straight towards that axis never changes angle, so has zero angular velocity and zero angular momentum.

For example, you are facing due north at a mate, who kicks a football towards you. Being a physicist, he's lousy at sport and the ball goes wide. If you watch the ball, it'll pass from being due north of you, through NE, then E, then SE. If you extend a hand to stop it, the impact will spin you round - the ball had angular momentum with respect to an axis through your center of mass, and transferred it to you.

Your mate kicks the ball again and, defying stereotypes, sends the ball straight at you. This time, the ball is always due north of you. When it hits you it knocks you backwards, but doesn't spin you round. It was coming directly towards the axis through your center of mass, so had zero angular momentum, so you end up with zero angular momentum.
Dec6-12, 06:34 PM   #8
 
If you shoot a bullet into a block of wood, hitting the side, then it will start spinning. If you shoot a bullet at the center of mass of the wood, then it won't spin.
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