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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • 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

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Answer & Solution

Correct Answer

(D) \( - \frac{{{I_1}\omega _1^2}}{{24}}\)

Step-by-step Solution

Detailed explanation

\({E_i} = \frac{1}{2}{I_1} \times \omega _1^2 + \frac{1}{2}\frac{I}{2} \times \frac{{\omega _1^2}}{4}\)

\( = \frac{{{I_1}\omega _1^2}}{2}\left( {\frac{9}{8}} \right) = \frac{9}{{16}}{I_1}\omega _1^2\)

\({I_1}{\omega _1} + \frac{{{I_1}\omega { _1}}}{4} = \frac{{3{I_1}}}{2}\omega \,\,\,;\,\,\,\frac{5}{4}{I_1}{\omega _1} = \frac{{3{I_1}}}{2}\omega \)

\(\omega  = \frac{5}{6}{\omega _1}\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,{E_f} = \frac{1}{2} \times \frac{{3{I_1}}}{2} \times \frac{{25}}{{36}}\omega _1^2\)

\( = \frac{{25}}{{48}}{I_1}\omega _1^2\)

\( \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\frac{{25}}{{49}} - \frac{{ - 2}}{{48}}{I_2}\omega _1^2\)

\( = \frac{{25}}{{48}}{I_1}\omega _1^2\)

\( \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {\frac{{25}}{{48}} - \frac{9}{{16}}} \right) = \frac{{ - 2}}{{48}}{I_1}\omega _1^2\)

\( = \frac{{ - {I_1}\omega _1^2}}{{24}}\)

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