spherical star spinning at angular velocity omega suddenly collapses to half of its original radius without any loss of mass. Assume the sta

Question

spherical star spinning at angular velocity omega suddenly collapses to half of its original radius without any loss of mass. Assume the star has uniform density before and after the collapse. What will the angular velocity of the star be after the collapse

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Khoii Minh 3 years 2021-07-20T21:15:24+00:00 1 Answers 197 views 0

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  1. Nem
    1
    2021-07-20T21:16:50+00:00
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    Answer:

    The angular velocity of the star after the collapse will be four times greater than its initial angular velocity.

    Explanation:

    Given:

    The initial angular velocity of a spherical star is \omega.

    The final radius of the star is half of its initial radius.

    The conservation of the angular momentum is given by

    I \omega = constant

    where I is the moment of inertial of the star.

    Consider that the initial moment of inertia of the star is I_{1}, the final moment of inertia is I_{2} and the final angular velocity is \omega_{f}.

    The moment of inertia of a sphere is given by

    I = \dfrac{2}{5}MR^{2}

    where M is the mass of the sphere and R is the radius of the sphere.

    The expression for the conservation of angular momentum for the star is given by

    ~~~~~&& I_{1} \omega = I_{2} \omega_{f}\\&or,& (\dfrac{2}{5}MR^{2}) \omega = (\dfrac{2}{5}M(\dfrac{R}{2})^{2}) \omega_{f}\\&or,& \omega_{f} = 4 \omega

    Thus, the angular velocity of the star after the collapse will be four times greater than its initial angular velocity.

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