Question

A disk of mass $m$ and radius $r$ rotates about an axis passing through its center and perpendicular to its plane with angular velocity $\omega$. Find the percentage change in the kinetic energy when an identical disc is placed over the first disc and both the discs rotate about an axis passing through their center and perpendicular to the plane with same angular velocity $\omega$. Assume there is no friction between the surfaces.

• A. $25\%$
• B. $50\%$
• C. $75\%$
• D. $100\%$

Answer 1

The correct option is D $100\%$

When single disc is rotating, kinetic energy is given by
$K.E_1=\frac{1}{2}{I}_1{\omega }^2$
$MOI$ of disc, $I_1=\frac{m{r}^2}{2}$
$\therefore K{E}_1=\frac{1}{2}\left(\frac{m{r}^2}{2}\right){\omega }^2=\frac{m{r}^2{\omega }^2}{4}$


When two discs are rotating, then kinetic energy is given by
$K.E_2=\frac{1}{2}{I}_2{\omega }^2$
$I_2=2\times \frac{m{r}^2}{2}=m{r}^2$
i.e $K{E}_2=\frac{1}{2}\left(m{r}^2\right){\omega }^2=\frac{m{r}^2{\omega }^2}{2}$


Hence, percentage change in kinetic energy,
$\Delta KE\%=\frac{K{E}_2-K{E}_1}{K{E}_1}\times 100$
$=\frac{\left(\frac{m{r}^2{\omega }^2}{2}\right)-\left(\frac{m{r}^2{\omega }^2}{4}\right)}{\left(\frac{m{r}^2{\omega }^2}{4}\right)}\times 100$
$=\frac{0.5-0.25}{0.25}\times 100$
$=100$%

So, percentage change in the kinetic energy is $100\%$.