A uniform rod of mass `m` and length `l` is released from rest from its vertical position by giving a gently push. In consequence, the end of the rod collides at `P` after rotating about the smooth horizontal axis `O`. If the coefficient of restitution `e = (1/2)`. Find the: a. angular speed of the rod just after the impact.

b. energy loss during collision. c. maximum angle rotated by rod after collision.

Answer

`5/8`

Explanation

Since `/_\k=-/_\U`
`1/2((ml^(2))/3)omega^(2)=(mgl)/2(1-1/2)`
`omega =sqrt((3g)/(2l))`
a. The angular speed just after collision is `omega'l=eomegal`

b. `/_\K'=K'-K'=mgl/4 (1-1/4)=(3mgl)/16` (`:'omega'=omega/2`)
c. Let the angle rotated be `phi`
`/_\U+/_\K=0`
or `mgh-K'=0`
or `mgh-K'=0`

where `h=1/2cosbeta-1/2cos60^@`
and `K'=(mgl)/6`
Then `beta=cos^(-1)5/8`
Hence `phi=60^@-beta=60^@-cos^(-1)5/8`