Problem: A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angular

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Problem: A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angular speed wi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR2.

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Kiệt Gia 5 years 2021-07-14T20:39:42+00:00 1 Answers 21 views 0

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  1. quangkhai
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    2021-07-14T20:41:34+00:00
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    Answer:

    The final angular velocity is w_f = \frac{MR^2}{MR^2+ mr^2} w

    Explanation:

    From the question we are told that

         The mass of the first disk is  m

          The radius of the first  disk is  r

          The mass of  second disk is  M

          The radius of second disk is R

           The speed of rotation is w

           The moment of inertia of second disk is  I = \frac{1}{2} MR^2

    Since the first disk is at rest initially

            The initial angular momentum would be due to the second disk  and this is mathematically represented as

           L_i = Iw = \frac{1}{2} MR^2 w

    Now when the first disk is then dropped the angular momentum of the whole system now becomes

           L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f

    This above is because the formula for moment of inertia is the same for every disk

           According to the law  conservation of  angular momentum

                    L_f = L_i

        ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w

    =>              w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}

                      w_f = \frac{MR^2}{MR^2+ mr^2} w

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