Question

Two point masses are lying on a smooth uniform mass $M$ and length $L$. Initially the masses are in the middle of the rod. The system is rotating about an axis passing through the center and perpendicular to the rod with angular velocities $\omega$. No external force acts on the system. When the masses reach the ends of the rod, then the angular velocity of the system 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
$\omega =\frac{M{\omega }_0}{(M+6m)}$