Question

Two extremely small blocks of mass $m$ are lying on a smooth uniform rod of mass $M$ and length $L$. Initially the blocks are lying at the centre. The whole system is rotating with an angular velocity ${\omega }_0$ about an axis passing through the centre and perpendicular to the rod. When the blocks reach the ends of the rod, the angular velocity of the rod will be :

• A. $\frac{M{\omega }_0}{M+2m}$
• B. $\frac{M{\omega }_0}{M+4m}$
• C. $\frac{M{\omega }_0}{M+6m}$
• D. $\frac{M{\omega }_0}{M+8m}$

Answer 1

Answer: (C)

$\frac{M{\omega }_0}{M+6m}$

From the principle of conservation of momentum we have $I_0{\omega }_0=I\omega$
Also we have $I_0=\frac{M{L}^2}{12}$
and
$I=\frac{M{L}^2}{12}+\frac{m{L}^2}{4}+\frac{m{L}^2}{4}=\frac{L^2}{12}(M+6m)$
Thus
$\frac{M{L}^2}{12}{\omega }_0=\frac{(M+6m)\omega L^2}{12}$
or
$\omega =\frac{M{\omega }_0}{(M+6m)}$