Question

A circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega .$ If another stationary disc having radius $\frac{R}{2}$ and same mass $M$ is dropped co-axially on to the rotating disc. Gradually, both discs attain constant angular speed ${\omega }_f.$ The energy lost in the process is $p\%$ of the initial energy. The value of $p$ is (integer only)

Answer 1

As we know moment of inertia disc, $I_{disc}=\frac{1}{2}M{R}^2$


Using angular momentum conservation,

$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\times {\omega }_f$
$\frac{M{R}^2}{2}\times \omega +0=(\frac{M{R}^2}{2}+\frac{M{R}^2}{8}){\omega }_f$
$\Rightarrow {\omega }_f=\frac{4}{5}\omega$
$\text{Initial K.E.,}{K}_i=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{M{R}^2}{2}\right){\omega }^2=\frac{M{R}^2{\omega }^2}{4}$
$\text{Final K.E.,}{K}_f=\frac{1}{2}(\frac{M{R}^2}{2}+\frac{M{R}^2}{8})\frac{16}{25}{\omega }^2=\frac{M{R}^2{\omega }^2}{5}$

Percentage $\%$ loss in kinetic energy is,

$\%=\frac{{\text{KE}}_i-{\text{KE}}_f}{{\text{KE}}_i}=\frac{\frac{M{R}^2{\omega }^2}{4}-\frac{M{R}^2{\omega }^2}{5}}{\frac{M{R}^2{\omega }^2}{4}}\times 100=20\%=p\%$
Hence, value of $p=20$