A circular disc of mass M and radius R is rotating about its axis with angular speed ${\omega _1}$ . If another stationary disc having radius ${R \over 2}$ and same mass M is droped co-axially on to the rotating disc. Gradually both discs attain constant angular speed ${\omega _2}$ the energy lost in the process is p% of the initial energy. Value of p is __________.

${I_f}{\omega _f} = {I_i}{\omega _i}$<br><br>${I_i} = {{M{R^2}} \over 2}$<br><br>${I_f} = {{M{R^2}} \over 2} + {{M{{(R/2)}^2}} \over 2}$<br><br>$= {5 \over 4}.{{M{R^2}} \over 2}$<br><br>$$\left[ {{{M{R^2}} \over 2} + {M \over 2}{{\left( {{R \over 2}} \right)}^2}} \right]\omega ' = \left( {{{M{R^2}} \over 2}} \right).\omega $$<br><br>$\Rightarrow$ $$\left[ {{{M{R^2}} \over 2}.\left( {{5 \over 4}} \right)} \right]\omega ' = {{M{R^2}} \over 2}\omega $$<br><br>$\omega = {4 \over 5}\omega$<br><br>loss of K.E. = ${{Loss} \over {{K_i}}} \times 100$ <br><br>= $${{{1 \over 2}I{\omega ^2} - {1 \over 2}\left( {{5 \over 4}I} \right){{\left( {{4 \over 5}\omega } \right)}^2}} \over {{1 \over 2}I{\omega ^2}}}$$ $\times$ 100 <br><br>= $${{{\omega ^2} - {{16} \over {25}}{\omega ^2}\left( {{5 \over 4}} \right)} \over {{\omega ^2}}}$$ $\times$ 100 = $\left( {1 - {{80} \over {100}}} \right) \times 100$<br><br>= 20%