A disk-shaped merry-go-round of radius 3.03 mand mass 145 kg rotates freely with an angular speed of 0.681 rev/s . A 65.4 kg person running

Question

A disk-shaped merry-go-round of radius 3.03 mand mass 145 kg rotates freely with an angular speed of 0.681 rev/s . A 65.4 kg person running tangential to the rim of the merry-go-round at 3.41 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round’s rim. What is the final angular speed of the merry-go-round?

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Kim Cúc 4 years 2021-08-14T19:01:59+00:00 1 Answers 151 views 0

Answers ( )

  1. huyenthanh
    0
    2021-08-14T19:03:26+00:00
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    Answer:

    \omega_2=0.891\ rev/s

    Explanation:

    Given that

    Radius , r= 3.03 m

    Mass of disk , M= 145 kg

    Initial angular velocity

    ω=0.681 rev/s

    Mass of person , m= 65.4 kg

    Velocity of person , V= 3.41 m/s

    Initial mass moment of inertia

    I_1= \dfrac{M\times R^2}{2}

    I_1= \dfrac{145\times 3.03^2}{2}=665.61\ kg.m^2

    Final mass moment of inertia

    I_2= \dfrac{M\times R^2}{2}+m\times R^2

    I_2= \dfrac{145\times 3.03^2}{2}+65.4\times 3.03^2=1266.04\ kg.m^2

    Final\ angular\ velocity =\omega_2

    By using angular momentum equation

    I_1\times \omega+m\times V\times R=I_2\times \omega_2

    665.61\times 0.681+65.4\times 3.41\times 3.03=1266.04\times \omega_2

    1129.01= 1266.04\times \omega_2

    \omega_2=\dfrac{1129.01}{1266.04}

    \omega_2=0.891\ rev/s

    Thus the angular velocity will be 0.891 rev/s

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