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Q 69 let l be the lenght of rod m be mass of rod by energy conservation 1/2I  2-0=mg/2(cos 37- cos 60) 1/2m 2/3=mg1/2(4/5-1/2) 2=9g/10l m  =mg(1.2 sin 37) =mg(1.2)*3/5 =0.9(g/l) is angular accleleration therefore to find the force acting on tip of particle Fc=centrifugal force =(dm)l 2=0.9(dm)g Ft= tangential force =(dm) l=0.9(dm)g so,total force will be F=  (Fc )2+(Ft) 2 =0.9 ]2(dm)g answer
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Moderation Team
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