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Homework Help: Simple harmonic motion of ornament on tree

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data
    A holiday ornament in the shape of a hollow sphere with mass 2.0×10−2 kg and radius 5.5×10−2 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.
    Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5MR^2/3.)
    Take the free fall acceleration to be 9.80

    2. Relevant equations
    T = 2pi(square root of (I/mgd))
    I = (5MR^2)/(3)

    3. The attempt at a solution[/b
    First of all, I used the moment of inertia they gave me in the problem in the equation. I don't know if that is ok or am i supposed to use the moment of inertia for a hollow sphere equation. Also, the distance, I do not have, so what would I do with that. I was using L/2 and assuming L was the radius but I don't know if that is ok. Could I use the parallel axis theorem? If someone could tell me or guide me in the right path I would appreciate it. Thanks.
  2. jcsd
  3. Oct 28, 2007 #2
    First of all: we always solve idealized problem (at least at this level) .. so always look out for possible assumption in the question.
    Yes, you can use parallel axis theorem, but what's the need when they have already given you "I".
    I have assumed 'R' as the radius of the (thin) hollow sphere.
    I think, you didn't understand (or, read carefully) the question. "hung from a tree limb by a small loop of wire attached to the surface of the sphere" -- note that loop is small, so we can assume that hinge point is at the surface itself. It is more than confirmed by the value of 'I' given in question. (Hollow sphere :2MR^2/3. Use parallel axis for an axis touching its surface: 2MR^2/3 + MR^2 = 5MR^2/3.)
    Now, you will realize that, d = R (where 'd' is the one to be used in the formula).
    I don't think you will need any further assistance.
  4. Oct 28, 2007 #3
    thanks alot!
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