In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of gyration k

Question

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of gyration k is defined by the expression I=Mk2.) A child of mass 44.0 kg runs at a speed of 3.00 m/s along a path that is tangent to the rim of the initially stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate

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Đan Thu 3 years 2021-08-14T03:36:17+00:00 1 Answers 11 views 0

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  1. Dau
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    2021-08-14T03:37:52+00:00
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    Answer:

    a) 145.6kgm^2

    b) 158.4kg-m^2/s

    c) 0.76rads/s

    Explanation:

    Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

    (b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

    (c) the angular speed of the merry-go-round and child after the child has jumped on.

    a) From I = MK^2

    I = (160Kg)(0.91m)^2

    I = 145.6kgm^2

    b) The magnitude of the angular momentum is given by:

    L= r × p The raduis and momentum are perpendicular.

    L = r × mc

    L = (1.20m)(44.0kg)(3.0m/s)

    L = 158.4kg-m^2/s

    c) The total moment of inertia comprises of the merry- go – round and the child. the angular speed is given by:

    L = Iw

    158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

    w = 158.6/208.96

    w = 0.76rad/s

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