JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 12)

Two coaxial discs, having moments of inertia I1 and I1/2, are rotating with respective angular velocities $$\omega $$1 and $$\omega $$1/2 , about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef - Ei) is:
$${{{I_1}\omega _1^2} \over {24}}$$
$${{{I_1}\omega _1^2} \over {12}}$$
$${3 \over 8}{I_1}\omega _1^2$$
$${{{I_1}\omega _1^2} \over {6}}$$

Explanation

$${E_i} = {1 \over 2}{I_I} \times \omega _1^2 + {1 \over 2}{I \over 2} \times {{\omega _1^2} \over 4}$$

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

$${I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _1} = {{3{I_1}} \over 2}\omega $$

$$\omega = {5 \over 6}{\omega _1};{E_f} = {1 \over 2} \times {{3{I_1}} \over 2} \times {{25} \over {36}}\omega _1^2$$

$$ = {{25} \over {48}}{I_1}\omega _1^2$$

$$ \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2{{25} \over {49}} - {{ - 2} \over {48}}{I_2}\omega _1^2$$

$$ = {{25} \over {48}}{I_1}\omega _1^2$$

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

$$ = {{ - {I_1}\omega _1^2} \over {24}}$$

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