A uniform slender rod AB is released from rest on a horizontal surface with the rod inclined at angle 60 degrees and end A in contact with the floor. Find the minimum static coefficient of friction betweeen the rod and thge surface so that the rod does not slip. What is the instantaneous acceleration of (a) the centre of mass (b) end B under no slip condition?

answer in terms of g (acc. due top gravity)

look ill give u the soln. to a general case....

the free body diagram is as posted below.......let the mass be 'm' the length be 'l'........let 'N' be the normal force which the rod experiences from the bottom..............." " be the coeff of friction..............

analysing rotational motion abt the pt rotation....ie the pt of contact with ground............

mgcos *l/2 =ml²/3 * ...........................1)     Torque abt pt P

so we have =3gcos /2l.......................2)

analysing accleration of centre of mass in horizontal direction.....

ma_x=N..............3)

analysing acceleration in vertical dirn....

mg-N= ma_y........................4)

but we know that ay is nothing but component of accn of c.m in vertical dirn........and as no slipping.....

accn of c.m= *l/2........

thus a_x=*l/2cos................................5) 

and a_y=*l/2sin ..................................6)

so using 1, 2, 3, 4, 5 and 6 and solving for

we get....=

here u can substitute theta as pi/3 to get the reqd. result....

from the above eqn.s solve for  and the 2 accns wud be l/2 and l respectively...(interms of g)

ill try to post the free body diagram...

hope it helps...

rate if useful....