Rotation Ques 50
- A thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of the same dimensions but of mass $M / 4$ is placed gently on the first disc coaxially. The angular velocity of the system now is $2 \omega / \sqrt{5}$.
$(1986,3 M)$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 50.$F$
Solution:
- $I _1 \omega _1=I _2 \omega _2$
$$ \therefore \omega _2=\frac{I _1}{I _2} \cdot \omega _1=\frac{\frac{M R^{2}}{2}}{\frac{M R^{2}}{2}+\frac{M}{4} \cdot \frac{R^{2}}{2}} \omega=\frac{4}{5} \omega $$
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