Question

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 $\frac{R}{2}$ and same mass $M$ is dropped 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 ______.

Answer 1

Answer: 20

Let moment of inertia of bigger disc is $I=\frac{M{R}^2}{2}$
$\Rightarrow$ MOI of small disc $I_2=\frac{M{\left(\frac{R}{2}\right)}^2}{2}=\frac{I}{4}$
by angular momentum conservation
$I{\omega }_1+\frac{I}{4}(0)=I{\omega }_2+\frac{I}{4}{\omega }_2$
$\Rightarrow {\omega }_2=\frac{4{\omega }_1}{5}$
initial kinetic energy $K_1=\frac{1}{2}I{\omega }_1^2$
final kinetic energy $K_2$
$=\frac{1}{2}\left(I+\frac{I}{4}\right){\left(\frac{4{\omega }_1}{5}\right)}^2=\frac{1}{2}I{\omega }_1^2\left(\frac{4}{5}\right)$
$P\%=\frac{K_1-K_2}{K_1}\times 100\%$
$=\frac{1-4/5}{1}\times 100=20\%$