Question

A disc of the moment of inertia $I_1$ is rotating in a horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed ${\omega }_1$ Another disc of the moment of inertia $I_2$ having zero angular speed is placed coaxially on a rotating disc. Now both the disc are rotating with the constant angular speed ${\omega }_2$ . The energy lost by the initial rotating disc is

• A. $\frac{1}{2}\left[\frac{I_1+I_2}{I_1{I}_2}\right]{\omega }_{\text{1}}^2$
• B. $\frac{1}{2}\left[\frac{I_1{I}_2}{I_1-i_2}\right]{\omega }_{\text{1}}^2$
• C. $\frac{1}{2}\left[\frac{I_1-I_2}{I_1{I}_2}\right]{\omega }_{\text{1}}^2$
• D. $\frac{1}{2}\left[\frac{I_1{I}_2}{I_1+I_2}\right]{\omega }_{\text{1}}^2$

Answer 1

Answer: (D)

$I_1{\left(\omega \right)}_1=\left(I_1+I_2\right){\left(\omega \right)}_2$
$\frac{{\omega }_2}{{\omega }_1}=\frac{I_1}{I_1+I_2}$
$E_1-E_2$
$=\frac{1}{2}{I}_1{\omega }_1^2-\frac{1}{2}(I_1{\left(+I\right)}_2){\omega }_2^2$
$=\frac{1}{2}{\omega }_1^2\left[I_1-\left(I_1+I_2\right)\frac{{\omega }_2^2}{{\omega }_1^2}\right]$
$=\frac{1}{2}{\omega }_1^2\left[I_1-\left(I_1+I_2\right)\frac{I_1^2}{{\left(I_1+I_2\right)}^2}\right]$
$=\frac{1}{2}{\omega }_1^2\left[\frac{I_1^2+I_1{I}_2-I_1^2}{I_1+I_2}\right]=\frac{1}{2}\left[\frac{I_1{I}_2}{I_1+I_2}\right]{\omega }_1^2$