A thin circular ring of mass M and radius r


A thin circular ring of mass $\mathrm{M}$ and radius $\mathrm{r}$ is rotating about its axis with an angular speed $\omega$. Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:

  1. $\omega \frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}$

  2. $\omega \frac{\mathrm{M}+2 \mathrm{~m}}{\mathrm{M}}$

  3. $\omega \frac{\mathrm{M}}{\mathrm{M}+2 \mathrm{~m}}$

  4. $\omega \frac{\mathrm{M}-2 \mathrm{~m}}{\mathrm{M}+2 \mathrm{~m}}$

Correct Option: , 3



Using conservation of angular momentum

$\left(\mathrm{Mr}^{2}\right) \omega=\left(\mathrm{Mr}^{2}+2 \mathrm{mr}^{2}\right) \omega^{\prime}$

$\omega^{\prime}=\frac{M \omega}{M+2 m}$

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