A Uniform rod of mass m=2kg and l=7.8 m is lying at rest on a smooth horizontal surface in vertical position. Due to slight disturbance, it starts falling with the lower end remain in contact to the surface. Find the speed of the lower end of the rod when it makes an angle of 30 degree with the horizontal.

by energy conservation :

mgl/2=1/2I(w²)+mgl/4

w=(3g/2l)^(1/2)

v=(3gl/8)^(1/2)

so..by putting the values..

value of v =5.40.

(is this answer correct or not.plzz tell me.)

By the method of energy conservation,

= mgl(1- sin x)/2 (Here x is the angle made by the stick to the horizontal)=I(omega)^2/2

= 2mgl(1- sin x)/2 = ml^2/12 (Omega^2)

= 12mgl(1- sin x)/ml^2 = (Omega^2)

= 12g(1-sin x)/l = (Omega^2)

= root(12g(1-sin x)/l)= Omega

= v(Linear Velocity) = root(12g(1-sin x)/l) *l/2

= root(3gl(1-sin x))

However if the rod is in touch with the floor ,the only the horizontal component of the rods velocity will be taken into account  = root(3gl(1-sin x)) sin x =1/2 root(3gl/2)

Put the value of x and g and l for getting the answer