Calculating Angular Momentum of a Rotating Rod | Mass M, Angular Velocity ш

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SUMMARY

The angular momentum of a rod rotating about one end, with mass M and angular velocity ω, can be calculated using the formula L = Iω, where I is the moment of inertia. For a rod of length L, the moment of inertia about its center of mass is I = (1/12)mL², and when adjusted for rotation about one end, the angular momentum becomes L = (1/3)mL²ω. This calculation involves integrating the contributions of differential mass elements along the length of the rod.

PREREQUISITES
  • Understanding of angular momentum and its relation to rotational motion
  • Familiarity with moment of inertia concepts
  • Basic knowledge of calculus for integration of mass elements
  • Proficiency in using physics equations related to rotation
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  • Study the derivation of the moment of inertia for various shapes, including rods and disks
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  • Explore the application of angular momentum conservation in rotational dynamics
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kuyt
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What is the angular momentum of a rod rotating about one end (mass M and angular velocity ш),about its center of mass?
 
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kuyt said:
What is the angular momentum of a rod rotating about one end (mass M and angular velocity ш),about its center of mass?
Is the rod rotating about one end or its center of mass?
 
about one end
 
But this is something fundamental,calculating angular momentum of a system about arbitary points in space.:confused:
 
In the second link the 'bubbles' are hot links.
Navigate through "Mechanics", then "Rotation", then "Moment of Inertia".
Click on "Common Forms" for Enlightenment!
 
Find the angular momentum of a differential length (dm*v*r) and integrate from r=0 to L.
 
kuyt said:
But this is something fundamental,calculating angular momentum of a system about arbitary points in space.:confused:
Yes, it is, and you can answer it by calculating the angular momentum of individual point about the end of the rod, then integrating them along the length of the rod. The result of doing that would be the formula at http://scienceworld.wolfram.com/physics/MomentofInertiaRod.html
that you could get to following the links you were given in tadchem's response.
 
HallsofIvy said:
Yes, it is, and you can answer it by calculating the angular momentum of individual point about the end of the rod, then integrating them along the length of the rod. The result of doing that would be the formula at http://scienceworld.wolfram.com/physics/MomentofInertiaRod.html
that you could get to following the links you were given in tadchem's response.

it gives moment of inertia not angular momentum! Can we use the general formula(if its correct):angular moment abt any point=angular moment of a fictitious particle (of mass m at the position of COM)abt that point + angular moment of the body abt com ?
 
  • #10
hello kuyt, i think you should have a look at the angular momentum equation ie., L= r X P and v=rw. So, use r=l/2 where l=length of rod. hope that makes sense..
 
  • #11
No,it not that easy I guess.
P.S can anyone giveme the final answer(in terms of angular velocity,mass and length)and ofcourse the explanation,instead of links
 
  • #12
It's (1/3)wmL^2 for angular speed w.
 
  • #13
^but that about one of the ends ,not com
 
  • #14
Ah, so you're trying to find the angular momentum about the com, in a system where the rod is rotating about one end.
 
  • #15
yes :approve:
 
  • #16
hmm. Tricky one. Well, I'm pretty sure that the whole point of saying the angular momentum "about a point" is equivalent to calculating the angular momentum, given that the origin is the point about which we want to find the angular momentum.

So I think the angular momentum about the COM is simply angular momentum, given that our origin is the COM. And in our original reference frame, the rod was rotating around the end. So in a reference frame where the COM stays at the origin, the angular momentum will simply be
\omega \frac{mL^2}{12}
(in other words, same as what the angular momentum would be for a system where the rod is rotating around it's COM.)
 
  • #17
The links provide helpful information.

Students are expected to demonstrate effort and show their work. We do not spoon feed students with answers.
 

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