MHT CET · Physics · Rotational Motion
A thin uniform circular dise of mass ' M ' and radius ' \(R\) ' is rotating with angular velocity ' \(\omega\) ' in a horizontal plane about an axis passing through its centre and perpendicular to its plane.
Another disc of same radius but of mass \(\left(\frac{M}{3}\right)\) is placed gently on the first disc co-axially. The new angular velocity will be
- A \(\frac{2}{3} \omega\)
- B \(\frac{3}{4} \omega\)
- C \(\frac{4}{3} \omega\)
- D \(\frac{5}{4} \omega\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{4} \omega\)
Step-by-step Solution
Detailed explanation
Angular momentum \(=\mathrm{I} \omega\)
By conservation of angular momentum,
\(\begin{aligned}
& \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\
& \text { Here, } I_1=\frac{M R^2}{2}, I_2=\frac{(M+M / 3)}{2} R^2=\frac{2 M R^2}{3} \\
& \therefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{2 \mathrm{MR}^2}{3} \omega_2 \\
& \text {...[From(i)] } \\
& \therefore \quad \omega_2=\frac{3}{4} \omega_1=\frac{3}{4} \omega \quad \ldots\left(\because \omega_1=\omega\right)
\end{aligned}\)
By conservation of angular momentum,
\(\begin{aligned}
& \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\
& \text { Here, } I_1=\frac{M R^2}{2}, I_2=\frac{(M+M / 3)}{2} R^2=\frac{2 M R^2}{3} \\
& \therefore \quad \frac{\mathrm{MR}^2}{2} \omega_1=\frac{2 \mathrm{MR}^2}{3} \omega_2 \\
& \text {...[From(i)] } \\
& \therefore \quad \omega_2=\frac{3}{4} \omega_1=\frac{3}{4} \omega \quad \ldots\left(\because \omega_1=\omega\right)
\end{aligned}\)
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