Angular momentum about inconvenient pivot point

In summary, a rod of length L and mass M rotating with angular velocity w has an angular momentum of 1/12ML^2*w with respect to its center of mass. The parallel axis theorem can be applied to calculate the angular momentum for a point 1/4L from the end of the rod as well as for a stationary point.
  • #1
kittyset
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0
A rod of length L and mass M is rotating with angular velocity w about its center of mass. What is its angular momentum with respect to a rotating point 1/4L from the rod's end? What about a stationary point?

Homework Equations


With the respect to the center of mass, angular momentum L = Iw = 1/12ML^2*w. (and L = r*Mv)

The Attempt at a Solution


Since the rod isn't actually rotating around the point (at least that's what I'm visualizing), does the parallel axis theorem still apply? How would calculation of moment of inertia work?
 
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  • #2
It is actually rotating for given point I think and the parallel axis theorem can be applied for your calculation.
 

1. What is angular momentum about an inconvenient pivot point?

Angular momentum about an inconvenient pivot point refers to the rotational motion of an object around a pivot point that is not the center of mass. This means that the object's mass is not evenly distributed and its center of mass does not align with the pivot point.

2. How is angular momentum about an inconvenient pivot point calculated?

The formula for calculating angular momentum about an inconvenient pivot point is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. The moment of inertia takes into account the distribution of mass around the pivot point.

3. Why is angular momentum about an inconvenient pivot point important?

Understanding angular momentum about an inconvenient pivot point is important in many areas of physics and engineering, such as analyzing the motion of rotating bodies, designing structures and machines, and predicting the behavior of celestial bodies.

4. How does angular momentum about an inconvenient pivot point differ from angular momentum about the center of mass?

Angular momentum about an inconvenient pivot point differs from angular momentum about the center of mass because the center of mass is the point around which an object naturally rotates. When the pivot point is not the center of mass, the object experiences a torque that causes it to rotate differently than if the pivot point were at the center of mass.

5. Can angular momentum about an inconvenient pivot point be conserved?

Yes, angular momentum about an inconvenient pivot point can be conserved as long as there are no external torques acting on the system. This means that the total angular momentum before and after an interaction will remain the same, even if the pivot point is not the center of mass.

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