Question

A hollow straight tube of length $2 l$ and mass m can turn freely about its centre on a smooth horizontal table. Another smooth uniform rod of same length and mass is fitted into the tube so that their centres coincide. The system is set in motion with an initial angular velocity $\omega_{0}$. Find the angular velocity of the tube at the instant when the rod slips out of the tube :

• A. $\frac{\omega_{0}}{4}$
• B. $\frac{\omega_{0}}{5}$
• C. $\frac{\omega_{0}}{7}$
• D. $\frac{\omega_{0}}{2}$

Answer 1

Answer: (C)

Using angular momentum conservation
$2 \times \frac{m(2 l)^{2}}{12} \omega_{0}=I^{\prime} \omega^{\prime}$

$\frac{m(2 l)^{2}}{12}+\frac{m(2 l)^{2}}{12}+m(2 l)^{2}$
$=\frac{m l^{2}}{3}+\frac{13 m l^{2}}{3}=\frac{14 m l^{2}}{3}$
$\frac{2}{3} m l^{2} \omega_{0}=\frac{14}{3} m l^{2} \omega^{\prime}$
$\Rightarrow \omega^{\prime}=\frac{\omega_{0}}{7}$