NEET-XII-Physics
10: Rotational Mechanics
- #18A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become
(a)
ωMM+m
(b)
ωMM+2 m
(c)
ωM-2 mM+2 m
(d)
ωM+2 mM.digAnsr: bAns : (b) `` \frac{\omega M}{M+2m}``
No external torque is applied on the ring; therefore, the angular momentum will be conserved.
`` I\omega =I\text{'}\omega \text{'}``
`` \Rightarrow \omega \text{'}=\frac{I\omega }{I\text{'}}...\left(i\right)``
`` I=M{r}^{2}``
`` I\text{'}=M{r}^{2}+2m{r}^{2}``
On putting these values in equation (i), we get:
`` \omega \text{'}=\frac{\omega M}{M+2m}``
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