Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round is 14 feet an

Question

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round is 14 feet and it takes 6 seconds for it
to make a complete revolution? Round the solutions to two decimal places.

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Trung Dũng 4 months 2021-09-05T12:31:19+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-05T12:32:20+00:00

    Step-by-step explanation:

    Given that,

    The diameter of the merry-go-round, d = 14 feet

    Time taken, t = 6 seconds

    Radius, r = 7 feet

    The linear speed of the merry-go-round is given by :

    v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s

    Also,

    v=r\omega

    Where

    \omega is the angular speed

    So,

    \omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s

    Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

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