Question

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 moveradially outwards along two massless rods fixed on the ring as shown in the figure. At some instant theangular speed of the system is $\frac{8}{9} ω$ and one of the masses is at a distance of $\frac{3}{5} R$ from O. At this instant the distance of the other mass from O is

• A. $\frac{2}{3}R$
• B. $\frac{1}{3}R$
• C. $\frac{3}{5}R$
• D. $\frac{4}{5}R$

Answer 1

Answer: (C), (D)

This question is based on angular momentum conservation
As given in the question initial angular velocity of ring is ω and final angular velocity of system is $\frac{8}{9}\omega$
so,
$L_i = L_f$
$Iω = L_f$
$I_{ring} \,ω = L_f$
$MR^{2} \,\omega = MR^{2} \times \frac{8}{9} \omega + \frac{M}{8} \times \left(\frac{3}{5} R\right)^{2} \times \frac{8}{9}\omega +\frac{M}{8}x^{2} \times \frac{8}{9}\omega$
$R^{2} \times 1 = \frac{8R^{2}}{9} + \frac{1}{8} \times \frac{9}{25} \times \frac{8R^{2}}{9} + \frac{x^{2}}{9}$
$R^{2}\left[1-\frac{8}{9}-\frac{1}{25}\right] = \frac{x^{2}}{9}$
$R^{2} = \left[\frac{25\times9-8\times25-9}{9\times25}\right] = \frac{x^{2}}{9}$
$R^{2} \frac{16}{25} = x^{2}$
$x = \frac{4}{5} R$
So, option (D) is correct
OR
Now we assume ring start to rotate at t = 0 by certain external agent.
Because both rods are frictionless and mass of the particles are equal. At t = 0, both particle starts to move from centre so they will experience same force at all the time. Hence position of particle will be same. If first particle is at a distance $\frac{3}{5}R$ from centre then other will be also at a distance $\frac{3}{5}R$.
So, option (C) is also correct.