a 60 kg child sits on on the outer edge a playground merry-go-around with radius of 4 meters that spins around at 1 revolution every 5 secon

Question

a 60 kg child sits on on the outer edge a playground merry-go-around with radius of 4 meters that spins around at 1 revolution every 5 seconds, mercilessly bringing the child ever closer to nausea. What is the child’s angular momentum? ( Remember to include the units.) What is the angular momentum of another child with the same mass who is sitting half way out from the center?

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Amity 5 years 2021-09-04T20:10:11+00:00 1 Answers 14 views 0

Answers ( )

  1. RobertKer
    0
    2021-09-04T20:11:42+00:00
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    Answer:

    a) 1209.6 kg m^2/s

    b) 302.4 kg m^2/s

    Explanation:

    a)

    The angular momentum of an object in circular motion is given by the formula

    L=m\omega r^2

    where

    m is its mass

    \omega is its angular velocity

    r is the distance of the object from the axis of rotation

    For the child in this problem, we have:

    m = 60 kg is his mass

    r = 4 m is the radius of the merry-go-around

    Here the child completes 1 revolution every 5 seconds; so his frequency is

    f=\frac{1}{5}=0.2 Hz

    And so his angular speed is

    \omega=2\pi f=2\pi 0.2=1.26rad/s

    Therefore, the angular momentum is

    L=(60)(1.26)(4)^2=1209.6 kg m^2/s

    b)

    For a child sitting halfway of the merry-go-around, his distance from the axis of rotation is

    r=\frac{4 m}{2}=2 m

    The mass of the child is the same as before,

    m = 60 kg

    And the angular speed is the  same as well: in fact, the merry go around is a rigid body, so all its points cover the same angle in the  same time; so, they all have the same angular speed. Therefore,

    \omega=1.26 rad/s

    Therefore, the angular momentum here is:

    L=m\omega r^2 = (60)(1.26)(2)^2=302.4 kg m^2/s

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