Ring has mass M and radius R. Initial angular speed of ring = $$\omega$$. Two point masses, each of mass are at rest at O. Initial angular momentum of ring and point masses system,

$${L_i} = {I_R}\omega + {I_m}\omega + {I_m}\omega $$

$$ = M{R^2}\omega + 0 + 0 = M{R^2}\omega $$

After some time, situation is changed as shown in the figure.

Angular speed of the system, $$\omega ' = {8 \over 9}\omega $$; $$OA = {{3R} \over 5}$$; OB = r = ?

Moment of inertia about O of point mass at A,

$${I_A} = {M \over 8} \times {{9{R^2}} \over {25}}$$

Moment of inertia about O of point mass at B, $${I_B} = {M \over 8}{r^2}$$

Fina angular momentum of the system

$${L_f} = M{R^2}\omega ' + {I_A}\omega ' + {I_B}\omega '$$

$$ = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$$

As there is no external torque acting on the system so its angular momentum will be conserved, L_i = L_f

$$M{R^2}\omega = M{R^2} \times {{8\omega } \over 9} + {M \over 8} \times {{9{R^2}} \over {25}} \times {{8\omega } \over 9} + {M \over 8}{r^2} \times {{8\omega } \over 9}$$

$${R^2} = {{8{R^2}} \over 9} + {{{R^2}} \over {25}} + {{{r^2}} \over 9} \Rightarrow {{{r^2}} \over 9} = {{16} \over {225}}{R^2}$$ $$\therefore$$ $$r = {4 \over 5}R$$