A 200kg beam 2.0 m in length slides broadside down the ice with a speed of `16 m//s` (figure). A 50 kg small favicol can (guranteed sticking ) at rest sticks at one end as it goes past, both go spinning down the ice. Assume frictionless motion. What is the angular velocity of combined mass in rad/s.

Answer

6

Explanation

After collision let velocity of centre rod is v' and angular velocity `omega`.
Applying conservation of angular momentum about point 'P' on ground (see figure) we get
`(MvL)/2=(Mv'L)/2+((ML^(2))/12)omega.....(1)`

from conservation of momentum we get
`Mv=ubrace(Mv')_("momentum of rod")+ubrace(m(v'+L//2omega))_("momentum of particle")....(2)`
On solving (1) and (2) we get
`omega=6 rad//sec`