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 $\frac{8}{9}\omega$ 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,
$L_i=L_f$
$I\omega =L_f$
$I_{ring}\omega =L_f$
$M{R}^2\omega =M{R}^2\times \frac{8}{9}\omega +\frac{M}{8}\times {\left(\frac{3}{5}R\right)}^2\times \frac{8}{9}\omega +\frac{M}{8}{x}^2\times \frac{8}{9}\omega$
$R^2\times 1=\frac{8R^2}{9}+\frac{1}{8}\times \frac{9}{25}\times \frac{8R^2}{9}+\frac{x^2}{9}$
$R^2\left[1-\frac{8}{9}-\frac{1}{25}\right]=\frac{x^2}{9}$
$R^2=\left[\frac{25\times 9-8\times 25-9}{9\times 25}\right]=\frac{x^2}{9}$
$R^2\frac{16}{25}=x^2$
$x=\frac{4}{5}R$
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.