- JEE Mains 2019
- Hard
- MCQ
Two coaxial discs, having moments of inertia \(I_1\) and \(\frac{I_1}{2}\) are a rotating with respectively angular velocities \(\omega_1\) and \(\frac{\omega_1}{2}\), about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If \(E_f\) and \(E_i\) are the final and initial total energies, then \((E_f -E_i)\) is
Answer & Solution
(D) \( - \frac{{{I_1}\omega _1^2}}{{24}}\)
Step-by-step Solution
Detailed explanation
\( = \frac{{{I_1}\omega _1^2}}{2}\left( {\frac{9}{8}} \right) = \frac{9}{{16}}{I_1}\omega _1^2\)
\({I_1}{\omega _1} + \frac{{{I_1}\omega { _1}}}{4} = \frac{{3{I_1}}}{2}\omega \,\,\,;\,\,\,\frac{5}{4}{I_1}{\omega _1} = \frac{{3{I_1}}}{2}\omega \)
\(\omega = \frac{5}{6}{\omega _1}\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,{E_f} = \frac{1}{2} \times \frac{{3{I_1}}}{2} \times \frac{{25}}{{36}}\omega _1^2\)
\( = \frac{{25}}{{48}}{I_1}\omega _1^2\)
\( \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\frac{{25}}{{49}} - \frac{{ - 2}}{{48}}{I_2}\omega _1^2\)
\( = \frac{{25}}{{48}}{I_1}\omega _1^2\)
\( \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {\frac{{25}}{{48}} - \frac{9}{{16}}} \right) = \frac{{ - 2}}{{48}}{I_1}\omega _1^2\)
\( = \frac{{ - {I_1}\omega _1^2}}{{24}}\)
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