MHT CET · Physics · Rotational Motion
A thin uniform circular disc of mass ' \(\mathrm{M}\) ' 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 \(\left(\frac{M}{2}\right)\) is placed gently on the first disc co-axially. The new angular velocity will be
- A \(\frac{2}{3} \omega\)
- B \(\frac{4}{5} \omega\)
- C \(\frac{5}{4} \omega\)
- D \(\frac{3}{2} \omega\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3} \omega\)
Step-by-step Solution
Detailed explanation
Angular momentum \(=\mathrm{I} \omega\) By conservation of angular momentum, \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2\)
Here, \(I_1=\frac{M^2}{2}, I_2=\frac{(M+M / 2)}{2} R^2=\frac{3 M^2}{4}\)
\(\therefore \frac{\mathrm{MR}^2}{2} \omega_1=\frac{3 \mathrm{MR}^2}{4} \omega_2\)
\(\therefore \omega_2=\frac{2}{3} \omega_1\)
Here, \(I_1=\frac{M^2}{2}, I_2=\frac{(M+M / 2)}{2} R^2=\frac{3 M^2}{4}\)
\(\therefore \frac{\mathrm{MR}^2}{2} \omega_1=\frac{3 \mathrm{MR}^2}{4} \omega_2\)
\(\therefore \omega_2=\frac{2}{3} \omega_1\)
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