Question

Two discs have moments of intertia $I _{1}$ and $I _{2}$ about their respective axes perpendicular to the plane and passing through the centre. 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. $\frac{ I _{1} I _{2}}{\left( I _{1}+ I _{2}\right)}\left(\omega_{1}-\omega_{2}\right)^{2}$
• B. $\frac{\left( I _{1}- I _{2}\right)^{2} \omega_{1} \omega_{2}}{2\left( I _{1}+ I _{2}\right)}$
• C. $\frac{ I _{1} I _{2}}{2\left( I _{1}+ I _{2}\right)}\left(\omega_{1}-\omega_{2}\right)^{2}$
• D. $\frac{\left(\omega_{1}-\omega_{2}\right)^{2}}{2\left( I _{1}+ I _{2}\right)}$

Answer 1

Answer: (C)

From conservation of angular momentum we get
$ I _{1} \omega_{1}+ I _{2} \omega_{2}=\left( I _{1}+ I _{2}\right) \omega $
$\omega=\frac{ I _{1} \omega_{1}+ I _{2} \omega_{2}}{ I _{1}+ I _{2}} $
$ k _{ i }=\frac{1}{2} I _{1} \omega_{1}^{2}+\frac{1}{2} I _{2} \omega_{2}^{2} $
$ k _{ f }=\frac{1}{2}\left( I _{1}+ I _{2}\right) \omega^{2} $
$k _{ i }- k _{ f }=\frac{1}{2}\left[ I _{1} \omega_{1}^{2}+ I _{2} \omega_{2}^{2}-\frac{\left( I _{1} \omega_{1}+ I _{2} \omega_{2}\right)^{2}}{ I _{1}+ I _{2}}\right]$
Solving above we get
$k _{ i }- k _{ f }=\frac{1}{2}\left(\frac{ I _{1} I _{2}}{ I _{1}+ I _{2}}\right)\left(\omega_{1}-\omega_{2}\right)^{2}$