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Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $\omega_1$ and $\omega_2$ They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
$I{\left( {{\omega _1} - {\omega _2}} \right)^2}$
$\frac{I}{8}{\left( {{\omega _1} - {\omega _2}} \right)^2}$
$\;\frac{I}{2}{\left( {{\omega _1} + {\omega _2}} \right)^2}$
$\;\frac{I}{4}{\left( {{\omega _1} - {\omega _2}} \right)^2}$
Solution
Initial angular momentum = $I{\omega _1} + I{\omega _2} $
Let $\omega $ be angular speed of the combined system.
Final angular momentum $=2I$$\omega $
$\therefore $ According to conservation of angular momentum
$I{\omega _1} + I{\omega _2} = 2I\omega \,\,or\,\,\omega = \frac{{{\omega _1} + {\omega _2}}}{2}$
Initial rotational kinetic energy
$E = \frac{1}{2}I\left( {\omega _1^2 + \omega _2^2} \right)$
Final rotational kinetic energy
$\begin{array}{l}
{E_f} = \frac{1}{2}\left( {2I} \right){\omega ^2} = \frac{1}{2}\left( {2I} \right){\left( {\frac{{{\omega _1} + {\omega _2}}}{2}} \right)^2}\\
\,\,\,\,\,\,\,\, = \frac{1}{4}I{\left( {{\omega _1} + {\omega _2}} \right)^2}\\
\therefore \,Loss\,of\,energy\,\Delta E = {E_i} – {E_f}\\
\,\,\,\,\,\, = \,\frac{1}{2}\left( {\omega _1^2 + \omega _2^2} \right) – \frac{I}{4}\left( {\omega _1^2 + \omega _2^2 + 2{\omega _1}{\omega _2}} \right)\\
\,\,\,\,\,\, = \frac{I}{4}\left[ {\omega _1^2 + \omega _2^2 – 2{\omega _1}{\omega _2}} \right] = \frac{I}{4}{\left( {{\omega _1} – {\omega _2}} \right)^2}
\end{array}$
