A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spi

Question

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.77 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? kg-m2 2)What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? kg-m2 3)What is the final angular velocity of the disk? rad/s 4)What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) J 5)What is the centripetal acceleration of the person when she is at R/3? m/s2 6)If the person now walks back to the rim of the disk, what is the final angular speed of the disk? rad/s

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Thái Dương 4 years 2021-07-17T11:05:18+00:00 1 Answers 82 views 0

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  1. nguyetanh
    0
    2021-07-17T11:07:01+00:00
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    Answer:

    1) 883 kgm2

    2) 532 kgm2

    3) 2.99 rad/s

    4) 944 J

    5) 6.87 m/s2

    6) 1.8 rad/s

    Explanation:

    1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

    I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

    If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

    I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

    2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

    I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

    3) Since there’s no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

    I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

    \omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

    \omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

    4)Kinetic energy before:

    E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

    Kinetic energy after:

    E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

    So the change in kinetic energy is: 2374 – 1430 = 944 J

    5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

    6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

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