Question

Two beads (each of mass $m$) can move freely in a frictionless wire, whose rotational inertia with respect to the vertical axis is $I$. The system is rotated with an angular velocity ${\omega }_0$, when the beads are at a distance $r/2$ from the axis. What is the angular velocity of the system when the beads are at a distance $r$ from the axis?

• A. $\omega =\left(\frac{2I+m{r}^2}{I+2m{r}^2}\right)\frac{{\omega }_0}{2}$
• B. $\omega =\left(\frac{2I+m{r}^2}{I+2m{r}^2}\right){\omega }_0$
• C. $\omega =\left(\frac{I+2m{r}^2}{2I+m{r}^2}\right)2{\omega }_0$
• D. $\omega =\left(\frac{I+2m{r}^2}{2I+m{r}^2}\right){\omega }_0$

Answer 1

The correct option is A $\omega =\left(\frac{2I+m{r}^2}{I+2m{r}^2}\right)\frac{{\omega }_0}{2}$

Consider (wire + two beads) as a system. There is no external torque acting on it. Hence, angular momentum will be conserved.

$L_i=L_f$

$I_i{\omega }_0=I_f\omega$

$[I+m{\left(\frac{r}{2}\right)}^2+m{\left(\frac{r}{2}\right)}^2]{\omega }_0=[I+m{r}^2+m{r}^2]\omega$

$\left(\frac{2I+m{r}^2}{2}\right){\omega }_0=(I+2m{r}^2)\omega$

$\therefore \omega =\left(\frac{2I+m{r}^2}{I+2m{r}^2}\right)\frac{{\omega }_0}{2}$

Hence, $\left(A\right)$ is the correct answer.