this question may come (my intuition) as comprehension in jee 2010 
1the center of the cylinder lies exaactly in the midway of teh supports........hence teh horizontal component of the velo=v/2
the centre moves in a cirle about one edge of the support....if u is teh speed of teh centre then ucos45=v/2=> u=v/√2
Now Mgcos45-N=Mu2/R
N= Mg/√2 - Mv2/2R
11This can also be done by applying centrifugal force arguments....by considering a particle stationed at the point of contact - which can be thought to be at rest momentarily....
33
work done by N should be zero(internal force)
so Nv/√2=Nvc
Vc=V/√2
now at contact with fixed block mg/√2-m(v/√2)2/R
33same problem
http://targetiit.com/iit-jee-forum/posts/find-normal-force-4410.html
