A thin, circular hoop with a radius of 0.32 m is hanging on a nail. Adam notices that the hoop is oscillating back and forth through small a

Question

A thin, circular hoop with a radius of 0.32 m is hanging on a nail. Adam notices that the hoop is oscillating back and forth through small angles like a physical pendulum. The moment of inertia of the hoop for the rotational axis passing through the nail is I = 2mr^2. What is the period of the hoop?

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Eirian 5 years 2021-08-22T22:55:21+00:00 1 Answers 39 views 0

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  1. Ben
    0
    2021-08-22T22:56:29+00:00
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    Answer:

    1.605 s

    Explanation:

    radius of hoop = 0.32 m

    moment of inertia I = 2mr^{2}

    period P of a hoop is given as

    P = 2\pi \sqrt{\frac{I}{mgr} }

    since  I = 2mr^{2}

    \frac{I}{mgr} = \frac{2r}{g}

    therefore the period of the hoop reduces to

    P = 2\pi \sqrt{\frac{2r}{g} }

    where  = acceleration due to gravity = 9.81 m/s^2

    P = 2*3.142 \sqrt{\frac{2*0.32}{9.81} } = 1.605 s

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