Angular momentum of a thin spherical shell

  • Thread starter slr77
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  • #1
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Homework Statement


[/B]
A thin spherical shell of radius R = 0.50 m and mass 15 kg rotates about the z-axis through its center and parallel to its axis. When the angular velocity is 5.0 rad/s, its angular momentum (in kg ⋅ m2/s) is approximately:

a . 15
b. 9.0
c. 12
d. 19
e. 25

Homework Equations


[/B]
L = r x mv
L = Iω
v = rω
Moment of inertia of thin spherical shell = (2/3)mr^2

The Attempt at a Solution


[/B]
I seem to be getting different answer using the equations above and I can't figure out why.

v = 0.5 * 5 = 2.5
L = (0.5)(15)(2.5) = 18.75

and

L = (5)(2/3)(15)(0.5)^2 = 12.5

The answer is the first one (18.75) but where did I go wrong with the using L = Iω?
 

Answers and Replies

  • #2
gneill
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Hi slr77,

You didn't go wrong using the correct formula for the moment of inertia of a thin spherical shell. The correct answer to the problem is 12.5 kg m2 s-1.

L = r x mv applies to a point mass. The shell is not a point mass, and can't be treated as such as most of it is distributed at distances other than r from the axis of rotation.
 
  • #3
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Hey gneill,

The solution listed d (19) as the correct answer so I guess it's mistaken.

This has uncovered some confusion about this topic though. I thought a thin-walled spherical shell had all it's mass distributed a distance r from the center. A thin-walled how cylinder has I = MR^2 (which means that all it's mass distributed at a distance r from the axis of rotation, right? I could use r x mv in that case because then Iω = rmv). How does a spherical shell differ from this? Or am I thinking of thin-walled how cylinders incorrectly too?

edit: Ok, I think I understand. The axis of rotation of the sphere passes through the middle so there are points on the wall of the sphere that are less than r away from the axis of rotation. In the cases of the hollow cylinder, every point on the wall is r away from the axis of rotation
 
Last edited:
  • #4
gneill
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2,867
Hey gneill,

The solution listed d (19) as the correct answer so I guess it's mistaken.

This has uncovered some confusion about this topic though. I thought a thin-walled spherical shell had all it's mass distributed a distance r from the center.
Correct. But the center is not an axis of rotation. The axis may pass through the center.
A thin-walled how cylinder has I = MR^2 (which means that all it's mass distributed at a distance r from the axis of rotation, right?
Right. All the mass is located at distance R from the axis of rotation.
I could use r x mv in that case because then Iω = rmv). How does a spherical shell differ from this? Or am I thinking of thin-walled how cylinders incorrectly too?
No, cylinder good, shell bad :smile:

Sketch the profile of a cylindrical shell and its axis of rotation. Draw perpendiculars from the axis to the shell. Are they all the same length? How about the case of a spherical shell?
 
  • #5
rude man
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L = (5)(2/3)(15)(0.5)^2 = 12.5

The answer is the first one (18.75) but where did I go wrong with the using L = Iω?
Your answer is correct, theirs isn't.
A spherical shell and a right circular cylinder are fundamentally different geometries. Don't try to extrapolate from one to the other.
 

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