## 1 Answers

Sourish Ghosh

**·**2014-02-24 07:49:56Let the reference point be intersection of the wall and the floor, call it 'O'.

Note: the locus of the COM of the rod is a circle of radius l/2.

Conserving energy about O,

mgl2(1 - cosÎ¸) = 12mv^{2} + 12I_{com}w^{2}

w = 2vl

Solving,

v = âˆš[3gl(1 - cosÎ¸)4]

Now v_{x} = vcosÎ¸

As Î¸ increases, v_{x} acquires a local maximum, and then it decreases.

In short,

dv_{x}dÎ¸ > 0, Î¸ < cos^{-1}(2/3)

dv_{x}dÎ¸ < 0, Î¸ > cos^{-1}(2/3)

Normal reaction was from the wall was acting to the right till Î¸ = cos^{-1}(2/3) increasing the velocity. When Î¸ > cos^{-1}(2/3), v_{x} starts decreasing => normal reaction is acting to the left which is impossible. Hence it loses contact at that angle.

- Akash Anand Excellent work

·0·Reply·2014-02-24 23:17:15