MHT CET · Physics · Rotational Motion
Two discs of moment of inertia ' \(\mathrm{I}_1\) ' and ' \(\mathrm{I}_2\) ' and angular speeds ' \(\omega_1{ }^{\prime}\) and ' \(\omega_2\) ' are rotating along the collinear axes passing through their centre of mass and perpendicular to their plane. If the two discs are made to rotate together along the same axis. The rotational kinetic energy of the system will be
- A \(\frac{\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2}{2\left(\mathrm{I}_1+\mathrm{I}_2\right)^2}\)
- B \(\frac{\left(I_1 \omega_1-I_2 \omega_2\right)^2}{2\left(I_1+I_2\right)}\)
- C \(\frac{\left(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2\right)^2}{2\left(\mathrm{I}_1-\mathrm{I}_2\right)}\)
- D \(\frac{\left(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2\right)^2}{2\left(\mathrm{I}_1+\mathrm{I}_2\right)}\)
Answer & Solution
Correct Answer
(D) \(\frac{\left(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2\right)^2}{2\left(\mathrm{I}_1+\mathrm{I}_2\right)}\)
Step-by-step Solution
Detailed explanation
\(L_{initial} = I_1 \omega_1 + I_2 \omega_2\) \(L_{final} = (I_1 + I_2) \omega_{final}\)
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