Discussion Forum

 A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two-point masses each of mass  $\frac{M}{8}$  at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure.

At some instant , the angular speed of the system is  $\frac{8}{9}\omega$  and one of the maases is at a distance of $\frac{3}{5}R$   from O.  At this instant , the distance  of the other mass from O is

Answer:

Explanation:

 Let the other mass at this instant is at a distance of x from the centre O.

  Applying law of conservation of angular momentum, we have

                     $I_{1}\omega_{1}=I_{2}\omega_{2}$

$\therefore$       $(MR^{2})(\omega)$

      $= \left[MR^{2} +\frac{M}{8}\left(\frac{3}{5}R\right)^{2}+\frac{M}{8}x^{2}\right]\left(\frac{8}{9}\omega\right)$

 Solving this equation , we get ,   $x=\frac{4}{5}R$

Note  If we take  identical situations with both point masses, then answer will be (c) .But in that case angular momentum is not conserved.