hi,
i am getting 1.4456 metres(used the equation i said earlier)
did u get that? (sorry, but your writing is appearing very small to read)
say at height h from ground, ball lose contact.ie.N=0
let theta be angle from horizontal(through centre of circle) to the point where ball lose contact.
so at that point , N(=0)+ mg cos(90- theta)=mv^2/R gives v^2=mgR sin theta
now, change in K.E= work done by gravity- work done by friction
1/2 mgRsin theta= mg(2-h)- 0.22 mg cos 53(2/sin 53)
replaced sin theta=(h-R)/R
AND GOT h=1.4456
now what is ur idea?
*
Last edited by r.samanta; Dec 28th 2009 at 05:41 AM.
* |