Discussion Forum



1)

A disc of the moment of interia 'l1' 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 moment of interia 'l2'  having zero angular speed is placed co-axially on a rotating disc . Now, both the discs are rotating with constant angular speed '$\omega_{2}$ . The energy lost by the initial rotating disc is 


A) $\frac{1}{2}\left[\frac{l_{1}+l_{2}}{l_{1}l_{2}}\right]\omega_{1}^{2}$

B) $\frac{1}{2}\left[\frac{l_{1}l_{2}}{l_{1}-l_{2}}\right]\omega_{1}^{2}$

C) $\frac{1}{2}\left[\frac{l_{1}-l_{2}}{l_{1}l_{2}}\right]\omega_{1}^{2}$

D) $\frac{1}{2}\left[\frac{l_{1}l_{2}}{l_{1}+l_{2}}\right]\omega_{1}^{2}$

Answer:

Option D

Explanation:

Net  external torque  on the system is zero. Therefore, angular momentum of  the system will remain  same.

$\Rightarrow $    $l_{1}\omega_{1}=(l_{1}+l_{2})\omega_{2}$

$\frac{\omega_{2}}{\omega_{1}}=\frac{l_{1}}{l_{1}+l_{2}}$ .....(i)

 The energy lost . $E_{1}-E_{2}$

$=\frac{1}{2}l_{1}\omega_{1}^{2}-\frac{1}{2}(l_{1}+l_{2})\omega_{2}^{2}$

$=\frac{1}{2}\omega_{1}^{2}\left[l_{1}-(l_{1}+l_{2})\frac{\omega_{2}^{2}}{\omega_{1}^{2}}\right]$

 $=\frac{1}{2}\omega_{1}^{2}\left[l_{1}-(l_{1}+l_{2})\frac{l_{1}^{2}}{(l_{1}+l_{2})^{2}}\right]$

                              [ $\because  $ Eq.(i)]

 $=\frac{1}{2}\omega_{1}^{2}\left[\frac{l_{1}^{2}+l_{1}l_{2}-l_{1}^{2}}{(l_{1}+l_{2})^{}}\right]$

  $=\frac{1}{2}\left[\frac{l_{1}l_{2}}{l_{1}+l_{2}^{}}\right]\omega_{1}^{2}$