Two discs of same moment of inertia rotating | vedclass

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Question

Initial angular momentum = $I{\omega _1} + I{\omega _2} $

Let $\omega $ be angular speed of the combined system.

Final angular momentu

Vedclass · Accepted Answer

$\;\frac{I}{4}{\left( {{\omega _1} - {\omega _2}} \right)^2}$

Vedclass · Answer

Initial angular momentum = $I{\omega _1} + I{\omega _2} $

Let $\omega $ be angular speed of the combined system.

Final angular momentum $=2I$$\omega $

$	herefore $   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 

$E = \frac{1}{2}I\left( {\omega _1^2 + \omega _2^2} ight)$

Final rotational kinetic energy 

$\begin{array}{l}
{E_f} = \frac{1}{2}\left( {2I} ight){\omega ^2} = \frac{1}{2}\left( {2I} ight){\left( {\frac{{{\omega _1} + {\omega _2}}}{2}} ight)^2}\
\,\,\,\,\,\,\,\, = \frac{1}{4}I{\left( {{\omega _1} + {\omega _2}} ight)^2}\
	herefore \,Loss\,of\,energy\,\Delta E = {E_i} - {E_f}\
\,\,\,\,\,\, = \,\frac{1}{2}\left( {\omega _1^2 + \omega _2^2} ight) - \frac{I}{4}\left( {\omega _1^2 + \omega _2^2 + 2{\omega _1}{\omega _2}} ight)\
\,\,\,\,\,\, = \frac{I}{4}\left[ {\omega _1^2 + \omega _2^2 - 2{\omega _1}{\omega _2}} ight] = \frac{I}{4}{\left( {{\omega _1} - {\omega _2}} ight)^2}
\end{array}$

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