skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular momentum?

Question

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular momentum?

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Huyền Thanh 4 years 2021-08-17T15:46:39+00:00 1 Answers 9 views 0

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  1. minhkhang
    0
    2021-08-17T15:48:22+00:00
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    Answer:

    L=11.3\ kg-m^2/s

    Explanation:

    Given that,

    Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

    The moment of inertia of the skater, I = 0.6 kg-m²

    We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

    L=I\omega

    Substitute all the values,

    L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

    So, its angular momentum is equal to 11.3\ kg-m^2/s.

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