In the figure `(i)` a disc of mass `M(kg)` and radius `R(m)` is rotating smoothly about a fixed vertical axis `AB` with angular speed `26 rad//s` A rod rod `CD` of length `(R)/(2)(m)` and mass `M(kg)` is hinged at one end at point `'D'` on the disc. The rod remains in vertical position and rotates along with the disc about axis `AB`. At some moment the rod `CD` gets a very smal impulse at point `'C'` due to air due to which the rod falls on the disc along one radius and sticks to the disc as shown in figure `(ii)`. Now find the angular velocity of the disc in `rad//s`.

Answer

36

Explanation

`MI` of the system when rod is vertical
`I=(1)/(2)mR^(3)+mR^(2)`
`MI` of system when rod is horizontal
`I'=(1)/(2)mR^(2)+[(m((R)/(2))^(2))/(12)+m((3R)/(4))^(2)]`
`=(13)/(12)mR^(2)`
from conservation of angular momentum of system about axis `AB` is
`I omega =I' omega'`
or `omega'=((3)/(2)m R^(2)omega)/((13)/(12)mR^(2))=(18)/(13)omega`