Q. A uniform rod pivoted at its upper end hangs vertically. it is displaced through an angle of 60 degree and then released. find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37 degree with the vertical....

 

Ans. i did like dis:-

conserving energy

mgl(cos 37 - cos 60) = 1/2 mv^2

v^2 = 3gl/5

now at 37 degree posn.

dm . v^2/l = F(in radial dir.) = 3 . dm.g/5

F(in tangential dir.) = dm.g.sin 37 = 3dm.g/5

 

Req. F = (F(in rad.) ^2 + F(in tang.)^2)^1/2

       = 0.6dm.g.2              

 

but answer given is 0.9 dm.g 2

 

plzzz tell me where i m wrong.