Two discs of moments of inertia I_1 and I_2 a | vedclass

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Question

Let the moment of inertia of disc $I$ be $I$, and

the angular speed of disc $I$ be $\omega_{1}$.

Let the moment of inertia of disc $\Pi$ be $I_{

Vedclass · Accepted Answer

$\frac{{{I_1}{I_2}{{({\omega _1} - {\omega _2})}^2}}}{{2({I_1} + {I_2})}}$

Vedclass · Answer

Let the moment of inertia of disc $I$ be $I$, and

the angular speed of disc $I$ be $\omega_{1}$.

Let the moment of inertia of disc $\Pi$ be $I_{2}$ and the angular speed of disc $\Pi$ be $\omega_{2}$

Angular momentum of disc $i$ be $\mathrm{L}_{1}=\mathrm{I}_{1} \omega_{1}$ and angular momentum of disc $II$ be $L_{2}=I_{2} \omega_{2}$

The initial angular momentum of two discs is given by $\mathrm{Li}=\mathrm{I}_{1} \mathrm{\omega}_{1}+\mathrm{I}_{2} \mathrm{\omega}_{2}$

When two discs are brought into contact face to face (one on top of other) and their axis of rotation coincide, the moment of inertia $I$ of

the system is equal to the sum of their individual moments of inertia.

i.e. $\mathrm{I}=\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{2}$

Let $\omega$ be the final angular speed of the system.

The final angular momentum of the systemis given by

$\mathrm{L}_{t}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega$

Acconding to law of conservation of angular momentum, we get

$\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{t}}$

$\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega$

According to law of conservation of angular momentum, we get

$\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{t}}$

$\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega$

$\omega=\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}$        $...(i)$

Initial kinetic energy of two discs is

$\mathrm{K}_{1}=\frac{1}{2} \mathrm{I}_{1} \omega_{1}^{2}+\frac{1}{2} \mathrm{I}_{2} \omega_{2}^{2}$

Final kinetic energy of the system is

$\mathrm{K}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega^{2}$

$=\frac{1}{2}\left(I_{1}+I_{2}\right)\left(\frac{I_{1} \omega_{1}+I_{2} \omega_{2}}{I_{1}+I_{2}}\right)^{2}$         $(Using \,(i))$

$=\frac{1}{2} \frac{\left(\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}\right)^{2}}{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}$

Loss in kinetic energy of the system

 $\Delta \mathrm{K}=\mathrm{K}_{i}-\mathrm{K}_{\mathrm{t}}$

$=\frac{1}{2}\left(\mathrm{I}_{1} \omega_{1}^{2}+\mathrm{I}_{2} \omega_{2}^{2}\right)-\frac{\left(\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}\right)^{2}}{2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}$

$=\frac{1}{2} \mathrm{I}_{1} \omega_{1}^{2}+\frac{1}{2} \mathrm{I}_{2} \omega_{2}^{2}-\frac{1}{2} \frac{\mathrm{I}_{1}^{2} \omega_{1}^{2}}{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}$

$-\frac{1}{2} \frac{\mathrm{I}_{2}^{2} \omega_{2}^{2}}{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}-\frac{1}{2} \frac{2 \mathrm{I}_{1} \mathrm{I}_{2} \omega_{1} \omega_{2}}{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}$

$=\frac{1}{\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)}\left[\frac{1}{2} \mathrm{I}_{1}^{2} \omega_{1}^{2}+\frac{1}{2} \mathrm{I}_{1} \mathrm{I}_{2} \omega_{1}^{2}\right.$

$+\frac{1}{2} \mathrm{I}_{1} \mathrm{I}_{2} \omega_{2}^{2}+\frac{1}{2} \mathrm{I}_{2}^{2} \omega_{2}^{2}-\frac{1}{2} \mathrm{I}_{1}^{2} \omega_{1}^{2}$

$\left.-\frac{1}{2} \mathrm{I}_{2}^{2} \omega_{2}^{2}-\mathrm{I}_{1} \mathrm{I}_{2} \omega_{1} \omega_{2}\right]$

$=\frac{I_{1} I_{2}}{2\left(I_{1}+I_{2}\right)}\left(\omega_{1}^{2}+\omega_{2}^{2}-2 \omega_{1} \omega_{2}\right)$

$=\frac{I_{1} I_{2}\left(\omega_{1}-\omega_{2}\right)^{2}}{2\left(I_{1}+I_{2}\right)}$

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