A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other end of the rod is hinged at point A. If the string is massless, then the reaction of hinge at the instant when string is cut, is (Take, g=10 ms^(-2))

A rod of mass 5 kg is connected to the string at point B. The span of

1.	A rod of mass 5 kg is connected to the string at point B. The span of rod is along horizontal. The other end of the rod is hinged at point A. If the string is massless, then the reaction of hinge at the instant when string is cut, is (Take, g=10 ms^(-2))
Answer» 10.1 N 12.5 N 5 N 15 N Solution :When string is cut, the weight of the rod CONSTITUTES a torque about the HINGE POINT A. So, `tau_(A)="mg"(l)/(2)`….(i) Also from Newton's law, `tau_(A)=I alpha` where, `alpha=` angular acceleration of the rod and `I=` moment of inertia of rod, From EQS. (i) and (ii), we get `I alpha="mg"(l)/(2)rArr alpha="mg" (l)/(2)//I` As, `I=m (l^(2))/(3)` So, `alpha=(mg(l//2))/((ml^(2))/(3))=(3g)/(2l)` Now, acceleration of centre of mass of rod, `a_(CM) = alpha.r` Here, r = distance between hinge point A and centre of the rod `rArr a_(CM)=(3)/(2)(g)/(l)(l)/(2)=(3g)/(4)rArr mg-R_(A)=ma_(CM)` where, `R_(A)=` REACTION at A `rArr mg-R_(A)=mxx(3g)/(4)rArr R_(A)=(mg)/(4)=(5xx10)/(4)=12.5 N`.