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$R/2$ 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

Step 1: Given data

Mass of the circular disc $=M$

Radius of the disc$=R$

Angular speed of the circular disc $={\omega }_1$

Radius of the stationary disc $=\frac{R}{2}$

Both disc's constant angular speed $={\omega }_2$

Step 2: To find the value of $p$

The moment of inertia of the bigger disc $I=\frac{M{R}^2}{2}$

Moment of inertia of the small disc $I_2=\frac{M{\left({\displaystyle \frac{R}{2}}\right)}^2}{2}=\frac{I}{4}$

By conservation of angular momentum

$I{\omega }_1+\frac{I}{4}\left(0\right)=I{\omega }_2+\frac{I}{4}{\omega }_2$

${\omega }_2=\frac{4{\omega }_1}{5}$

Initial kinetic energy $K_1=\frac{1}{2}I{{\omega }_1}^2$

Final kinetic energy $K_2=\frac{1}{2}\left(I+\frac{I}{4}\right)\left(\frac{4{\omega }_1}{5}\right)=\frac{1}{2}I{{\omega }_1}^2\left(\frac{4}{5}\right)$

$\begin{array}{rcl}p\%& =& \frac{K_1-K_2}{K_1}\times 100\%\\ & =& \frac{1-4/5}{1}\times 100\\ & =& 20\%\end{array}$

The energy lost in the process is$p\%$ of the initial energy.

Therefore, the value of $p$ is $\mathbf{20}$.