Angular Momentum of a Moving Particle

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A point particle moving in a straight line at constant speed has a nonzero angular momentum with respect to the origin when it is at a distance l. The angular momentum is calculated using the equation L = r x p, where r is the position vector and p is the momentum vector. Although there are no external torques acting on the system, the angular momentum vector changes as the position vector changes over time. Thus, while the magnitude of angular momentum may remain constant, its direction does vary as the particle moves. This confirms that the problem is resolved, as the dynamics of angular momentum are understood.
Mark Zhu
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Homework Statement


A point particle travels in a straight line at constant
speed, and the closest distance it comes to the origin of coordinates
is a distance l. With respect to this origin, does the particle
have nonzero angular momentum? As the particle moves along its
straight-line path, does its angular momentum with respect to the
origin change?

Homework Equations


L = r x p

The Attempt at a Solution


Since there are no external torques applied to this particle system, I assume the angular momentum doesn't change. However, since the position vector of the particle changes over time and the linear momentum vector remains constant, wouldn't the angular momentum vector also change looking at the L = r x p
equation?
 
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Mark Zhu said:

Homework Statement


A point particle travels in a straight line at constant
speed, and the closest distance it comes to the origin of coordinates
is a distance l. With respect to this origin, does the particle
have nonzero angular momentum? As the particle moves along its
straight-line path, does its angular momentum with respect to the
origin change?

Homework Equations


L = r x p

The Attempt at a Solution


Since there are no external torques applied to this particle system, I assume the angular momentum doesn't change. However, since the position vector of the particle changes over time and the linear momentum vector remains constant, wouldn't the angular momentum vector also change looking at the L = r x p
equation?
I see this is marked solved. Please confirm with a post to the thread.
 
Yes, it is solved. Thanks.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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