Question

Two uniform circular rough disc of moment of inertia $l_1$ and $\frac{l_1}{2}$ are rotating with angular velocity ${\omega }_1$ and $\frac{{\omega }_1}{2}$ respectively in same direction. Now one disc is placed the other disc co-axially. The change in kinetic energy of the system is :

• A. $-\frac{1}{24}{l}_1{\omega }_1^2$
• B. $\frac{1}{24}{l}_1{\omega }_1^2$
• C. $\frac{1}{12}{l}_1{\omega }_1^2$
• D. $-\frac{1}{12}{l}_1{\omega }_1^2$

Answer 1

The correct option is A $-\frac{1}{24}{l}_1{\omega }_1^2$

${\overrightarrow{L}}_i={\overrightarrow{L}}_f$
$l_1{\omega }_1+\frac{l_1}{2}\frac{{\omega }_1}{2}=l_1{\omega }_f+\frac{l_1}{2}{\omega }_t$
$\frac{5l_1{\omega }_1}{4}=\frac{3}{2}{l}_1{\omega }_f\omega f=\frac{5}{6}{\omega }_1$
$\Delta K.E.=(\frac{1}{2}{l}_1{\omega }_f^2+\frac{1}{2}\frac{l_1}{2}{\omega }_f^2)-(\frac{1}{2}{l}_1{\omega }_1^2+\frac{1}{2}\frac{l_1}{2}{\left(\frac{{\omega }_1}{2}\right)}^2)$
$=\frac{1}{2}.\frac{3}{2}{l}_1\frac{25}{36}{\omega }_1^2-\frac{1}{2}.\frac{9}{8}{l}_1{\omega }_1^2$
$=\frac{75l_1{\omega }_1^2}{144}-\frac{9}{8}{l}_1{\omega }_1^2$
$=\frac{75-81}{144}{l}_1{\omega }_1^2$
$\Delta K.E=-\frac{1}{24}{l}_1{\omega }_1^2$