# Q7.25     Two discs of moments of inertia $I_{1}$ and $I_{2}$  about their respective axes (normal to the               disc and passing through the centre), and rotating with angular speeds $\omega _{1}$ and $\omega _{2}$               are brought into contact face to face with their axes of rotation coincident.              (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take $\omega _{1}\neq \omega _{2}$ .

The initial kinetic energy is written as :

$K_i\ =\ \frac{1}{2}I_1\omega_1^2\ +\ \frac{1}{2}I_2\omega_2^2$

or                                             $K_i\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )$

Now the final kinetic energy is :

$K_f\ =\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$

Put the value of final angular velocity from part (a).

We need to find :

$K_i\ -\ K_f\ =\ \frac{1}{2}\left ( I_1\omega_1^2\ +\ I_2\omega_2^2 \right )\ -\ \frac{1}{2}\left ( I_1\ +\ I_2 \right )\omega^2$

Solve the above equation, we get :

$=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1^2\ +\ \omega_2^2\ -\ 2\omega_1\omega_2)$

or                                           $=\ \frac{I_1 I_2}{2(I_1\ +\ I_2)}(\omega_1\ -\ \omega_2)^2$

$>\ 0$                                                                             (Since none of the quantity can be negative.)

Thus initial energy is greater than the final energy.  (due to frictional force).

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