Two discs have moments of intertia I _ 1 and I _ 2 about their...

Two discs have moments of intertia

{I}_{1}

and

{I}_{2}

about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds,

{\omega}_{1}

and

{\omega}_{2}

respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

Options

A $\dfrac{{I}_{1}{I}_{2}}{\left({I}_{1} + {I}_{2}\right)}$ { $\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$
B $\dfrac{{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}}$ {2 $\left({I}_{1} + {I}_{2}\right)}$
C $\dfrac{{I}_{1}{I}_{2}}{2\left({I}_{1} + {I}_{2}\right)}$ { $\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$
D $\dfrac{{\left({I}_{1} - {I}_{2}\right)}^{2}{\omega}_{1}{\omega}_{2}}$ {2 $\left({I}_{1} + {I}_{2}\right)}$

Correct answer

C. $\dfrac{{I}_{1}{I}_{2}}{2\left({I}_{1} + {I}_{2}\right)}$ { $\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$

Step-by-step solution

Angular momentum conservation

${I}_{1}{\omega}_{1} + {I}_{2}{\omega}_{2} = \left({I}_{1} + {I}_{2}\right)\omega$

$\omega = \dfrac{{I}_{1}{\omega}_{1} + {I}_{2}{\omega}_{2}}{{I}_{1} + {I}_{2}}$

Loss $= \dfrac{1}{2}{I}_{1}{\omega}_{1}^{2} + \dfrac{1}{2}{I}_{2}{\omega}_{2}^{2} - \dfrac{1}{2}\left({I}_{1} + {I}_{2}\right){\omega}^{2}$

$= \dfrac{1}{2}{I}_{1}{\omega}_{1}^{2} + \dfrac{1}{2}{I}_{2}{\omega}_{2}^{2} - \dfrac{1}{2}\left({I}_{1} + {I}_{2}\right){\left(\dfrac{{I}_{1}{\omega}_{1} + {I}_{2}{\omega}_{2}}{{I}_{1} + {I}_{2}}\right)}^{2}$

$= \dfrac{1}{2}\dfrac{{I}_{1}{I}_{2}}{\left({I}_{1} + {I}_{2}\right)}{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}$

${E}_{i} - {E}_{f} = \dfrac{{I}_{1}{I}_{2}{\left({\omega}_{1} - {\omega}_{2}\right)}^{2}}{2\left({I}_{1} + {I}_{2}\right)}$

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