Question

Two discs have moments of inertia $I_{1}and{I}_2$ about their respective axes perpendicular to the plane and passing through the center. They are rotating with angular speeds, ${\omega }_{1}and{\omega }_2$ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

• A. $\frac{I_1{I}_2}{(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2$
• B. $\frac{I_1{I}_2}{2(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2$
• C. $\frac{({\omega }_1-{\omega }_2{)}^2}{2(I_1+I_2)}$
• D. $\frac{(I_1+I_2{)}^2{\omega }_1{\omega }_2}{2(I_1+I_2)}$

Answer 1

The correct option is B $\frac{I_1{I}_2}{2(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2$

Let their final common angular velocity is $\omega$.
From conservation of angular momentum we have :
$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$
$\Rightarrow \omega =\frac{(I_1{\omega }_1+I_2{\omega }_2)}{(I_1+I_2)}...\left(1\right)$
Initial kinetic energy of the system :
$K{E}_i=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2$
Final kinetic energy of the system :
$K{E}_f=\frac{1}{2}(I_1+I_2){\omega }^2$
Loss in kinetic energy ;
$\Delta KE=K{E}_i-K{E}_f=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2-\frac{1}{2}(I_1+I_2){\omega }^2$
From $\left(1\right)$ :
$\Delta KE=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2-\frac{1}{2}\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2)}$
$\Rightarrow \Delta KE=\frac{1}{2}\left(\frac{I_1{I}_2}{I_1+I_2}\right)({\omega }_1-{\omega }_2{)}^2$