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 its axis perpendicular to the rod and passing through its mid point (see figure).There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is:

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

Answer 1

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

Since there is no external torque acting on the system consisting of the rod and two beads, angular momentum will be conserved.

Hence,

$I_i{\omega }_0=I_f{\omega }_f$ ....(1) where $I_i$ is the MI of the system initially and $I_f$ is the MI finally.

$I_i=\frac{M{L}^2}{12}$

$I_f=\frac{M{L}^2}{12}+2m(\frac{L}{2}{)}^2$

Substituting in eqn(1) we get

$\frac{M{L}^2}{12}{\omega }_0=(\frac{M{L}^2}{12}+2m(\frac{L}{2}{)}^2){\omega }_f$

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