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feynmann

Hey guys here is my soln to the Bonus problem

Suppose the rod is A P B with AP = l/4 , AB = 2l , P is the last point of contact with the table .

Now mass of AP = mg/8 , mass of PB = 7/8mg

C.M of AP is at a distance of l/8 from P , and C.M. of PB is at a distsnce of 7/8 l from P

Now moment of forces around P = (7/8)^2 mgl - (1/8)^2 mgl = 3/4 mgl

Now m.i. of the rod about P =1/12 m ( 2l ) ^2 +m ( 3/4l )^2

= 43/48 ml^2

So angular accn = = 36/43 g/l

Linear accn of c.m of the whole rod = 36/43 g/l * 3/4 l = 27/43 g

Now applying Newton's second law of motion we get

mg - N = 27/43 mg

N = 16/43 mg