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Homework Help: Angular momentum of a thin spherical shell

  1. Feb 14, 2015 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

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

    3. The attempt at a solution

    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


    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?
  2. jcsd
  3. Feb 14, 2015 #2


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    Staff: Mentor

    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.
  4. Feb 14, 2015 #3
    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: Feb 14, 2015
  5. Feb 14, 2015 #4


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    Staff: Mentor

    Correct. But the center is not an axis of rotation. The axis may pass through the center.
    Right. All the mass is located at distance R from the axis of rotation.
    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?
  6. Feb 14, 2015 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    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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