A holiday ornament in the shape of a hollow sphere with mass 0.015 kg and radius 0.055 m is hung from a tree limb by a small loop of wire at

Question

A holiday ornament in the shape of a hollow sphere with mass 0.015 kg and radius 0.055 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 5MR²/3.)
Take the free-fall acceleration to be 9.80 m/s². Express your answer using two significant figures.

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Farah 6 months 2021-07-15T19:53:57+00:00 1 Answers 16 views 0

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    2021-07-15T19:55:28+00:00

    Answer: 0.61 s

    Explanation:

    Given

    Mass of object, m = 0.015 kg

    Radius of object, r = 0.055 m

    Acceleration of object, g = 9.8 m/s²

    In a pendulum,

    T = 2π * √[I /(mgd)]

    The moment of Inertia, I of a hollow sphere is given by

    I(sphere) = 2/3MR² + MR²

    I(sphere) = 5/3MR²

    Also, d = R

    Substituting these into the first equation, we have

    T = 2π * √[(5/3MR²) / (mgr)]

    T = 2π * √[(5/3r) / (g)]

    T = 2 * 3.142 * √(5/3 * 0.055) / (9.8)]

    T = 6.284 * √(0.092/9.8)

    T = 6.284 * √0.00939

    T = 6.284 * 0.097

    T = 0.6095 s

    To 2 significant figures,

    The period is 0.61 s

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