Physics - Rotation Question with Solution | TestHub

By the principle of conservation of angular momentum, since no external torque acts on the system about the axis of rotation, the initial angular momentum equals the final angular momentum:

$L_i=L_f$

$I_i{\omega }_i=I_f{\omega }_f$

Initially, the two point masses are at the centre $O$, so their distance from the axis of rotation is zero. The initial moment of inertia of the system is just that of the ring:

$I_i=M{R}^2$

Let the distance of the second mass from $O$ be $r_2$. The final moment of inertia of the system when one mass is at $r_1=\frac{3}{5}R$ and the other is at $r_2$ is:

$I_f=M{R}^2+\frac{M}{8}{\left(\frac{3}{5}R\right)}^2+\frac{M}{8}{r}_2^2$

$I_f=M{R}^2+\frac{9M}{200}{R}^2+\frac{M}{8}{r}_2^2$

Substituting the values into the conservation of angular momentum equation:

$M{R}^2\omega =\left(M{R}^2+\frac{9M}{200}{R}^2+\frac{M}{8}{r}_2^2\right)\left(\frac{8\omega }{9}\right)$

Dividing both sides by $M\omega$:

$R^2=\left(R^2+\frac{9}{200}{R}^2+\frac{r_2^2}{8}\right)\frac{8}{9}$

Multiplying by $\frac{9}{8}$:

$\frac{9}{8}{R}^2=R^2+\frac{9}{200}{R}^2+\frac{r_2^2}{8}$

$\frac{9}{8}{R}^2-R^2=\frac{9}{200}{R}^2+\frac{r_2^2}{8}$

$\frac{1}{8}{R}^2=\frac{9}{200}{R}^2+\frac{r_2^2}{8}$

Multiplying the entire equation by $200$ to clear the denominators:

$25R^2=9R^2+25r_2^2$

$16R^2=25r_2^2$

$r_2^2=\frac{16}{25}{R}^2$

$r_2=\frac{4}{5}R$

Answer: $\frac{4R}{5}$