Question

A smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size, each of mass $m$, which can slide freely along the rod. Initially, the two beads are at the centre of the rod and the system is rotating with angular velocity ${\omega }_0$ about an axis perpendicular to the rod and passing through the mid-point $C$ of the rod. There are no external forces. When the beads reach the ends of the rod, then angular velocity of the system is

• A. $\left(\frac{M}{M+3m}\right){\omega }_0$
• B. $\left(\frac{M}{M+6m}\right){\omega }_0$
• C. $\left(\frac{M+6m}{M}\right){\omega }_0$
• D. ${\omega }_0$

Answer 1

The correct option is B $\left(\frac{M}{M+6m}\right){\omega }_0$

When the beads are at the centre of the rod then let the M.O.I of the system about axis of rotation is $I_1$.
So, $I_1=\frac{M{L}^2}{12}+0+0=\frac{M{L}^2}{12}$
[$r_{\perp }=0$ for beads from axis of rotation ]
After sometime, when beads reaches the respective ends of rod, then let the M.O.I of the system about axis of rotation is $I_2$.
So, $I_2=\frac{M{L}^2}{12}+m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2$
$=\frac{(M+6m)L^2}{12}$
For the system (beads + rod), net external torque is zero.
Applying law of conservation of angular momentum,
$L_i=L_f$
$\Rightarrow I_1{\omega }_1=I_2{\omega }_2$
$\Rightarrow \frac{M{L}^2}{12}{\omega }_0=\frac{(M+6m)L^2}{12}\times {\omega }_2$
$\therefore {\omega }_2=\left(\frac{M}{M+6m}\right){\omega }_0$