System of Particles and Rotational Motion - Result Question 42
45. A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be
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(a) $\frac{(M-4 m) \omega}{M+4 m}$
(b) $\frac{M \omega}{4 m}$
(c) $\frac{M \omega}{M+4 m}$
(d) $\frac{(M+4 m) \omega}{M}$
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Answer:
Correct Answer: 45. (c)
Solution:
- (c) Applying conservation law of angular momentum, $I_1 \omega_1=I_2 \omega_2$
$I_2=(M r^{2})+4(m)(r^{2})=(M+4 m) r^{2}$
$ \begin{aligned} & .\quad \text{ (Taking } \omega_1=\omega \text{ and } \omega_2=\omega_1) \\ \Rightarrow & M r^{2} \omega=(M+4 m) r^{2} \omega_1 \\ \Rightarrow & \omega_1=\frac{M \omega}{M+4 m} \end{aligned} $
