A uniform thin rod of mass m and length l is standing on a smooth horizontal surface . A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling .Find the vel of centre of mass of the rod at the instant when it makes an angle theta with horizontal

 THIS SUM CAN BE SOLVED USING THIS METHODS SEE THE EASIER METHOD IS THIS THE POTENTIAL ENERGY LOST BY THE ROD IS MGL/2(1-SIN@) THIS IS EQUAL TO THE GAIN IN ENERGY OF THE INSTANEOUS AXIS OF ROTATION THE MOMENT OF INERTIA OF THE INSTANEOUS AXIS OF ROTATION IS ml 2 / 12 + ml 2 /4 = ml 2 /3 SO W=UNDER ROOT 3G(1-SIN@)/L

The theorem states that when 2 non parallel velocities are given at any 2 points then , draw perpendiculars from them and the point of intersection of these perpendiculars gives us the location of INSTANTANEOUS CENTRE of zero velocity )IC . 

It can be followed in the figure given below

Now we can proceed by conserving energy as told above and using the fact that

V of COM = w * l /2 [cos theta]

a fiitjee favourite.....

dekho the net force in the horizontal direction on the rod=0 so the cm will follow a straight line motion in a down ward dirn....if given a velocity in a horizontally...it wud have assumed the path of a projectile....but as V-->0 it will assume a vertically down wards path....now to get a relation between the vel and ang vel abt the pt of contact with the smooth floor....let the cm have gone down by a distance 'x' so wat we have is (l/2 - x) = (l/2)sin@ ....@ being the inclination with the horizontal at ne instant of time 't'... do diff. we have -v =wLcos@ /2-------1)

so now u can conserve energy summing up the problem thus...

mgL/2 = 1/2m {v(@)}² + 1/2 (mL²/3) {W(@)}² + mgH(@) ------------2)

but frm 1) we have w(@) =(-2vsec@ /L)------3)

and H(@) = (L/2)*sin@ -------------4)

so using 3) and 4) in 2) one can solve for v(@) .....

pardon ne silly mistakes from my side....im in the habbit of making of many.....

hope it helps...

rate if useful....