Rotation Ques 46
- A thin smooth rod of length $L$ and mass $M$ is rotating freely with angular speed $\omega _0$ about an axis perpendicular to the rod and passing through its centre. Two beads of mass $m$ and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be
(2019 Main, 09 April II)
(a) $\frac{M \omega _0}{M+3 m}$
(b) $\frac{M \omega _0}{M+m}$
(c) $\frac{M \omega _0}{M+2 m}$
(d) $\frac{M \omega _0}{M+6 m}$
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Answer:
Correct Answer: 46.(d)
Solution:
- As there is no external torque on system.
$\therefore$ Angular momentum of system is conserved.
$\Rightarrow \quad I _i \omega _i=I _f \omega _f$
Initially,
Finally,
$$ \Rightarrow \quad \frac{M L^{2}}{12} \cdot \omega _0+0=\frac{M L^{2}}{12}+2(m) \frac{L^{2}}{2} \omega $$
So, final angular speed of system is
$$ \Rightarrow \quad \omega=\frac{\frac{M L^{2}}{12} \cdot \omega _0}{\frac{M L^{2}+6 m L^{2}}{12}}=\frac{M \omega _0}{M+6 m} $$
BETA
