Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end?

Two ends of a uniform rod weighing W, are placed on supports so that t

1.	Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end?
Answer» Solution :Let the length of the rod = `l`cm, its weight = W = Mg , where M is the mass of the rod. When the support at one end is removed suddenly, centre of gravity of the rod FALLS downwards with an ACCELERATION a. Let R = reaction force at the end with the suppot. Hence if the C.G. now falls with an acceleration a, the rod will turn about the point P. The torque on the rod = `Mg.(l)/(2)` ALSO, Mg-R = Ma or, `a = ("Mg-R")/(M)` Here MOMENT of inertia `I= (I)/(3) Ml^(2)` = moment of inertia of the rod about the perpendicular axis passing through the end of the rod and the angular acceleration , `ALPHA=(alpha)/(l//2)=(2a)/(l)` `:. (1)/(3) Ml^(2)alpha=Mg""(l)/(2)[because tau=Ialpha]` or,`(1)/(3)Ml^(2).(2a)/(l)="Mg"(l)/(2)or,(2)/(3)a=(g)/(2)or, (2)/(3)(("Mg-R)/M)=(g)/(2)` `:. R= ("Mg")/(4)=(W)/(4)` Therefore, when one support is removed the support at the other end will exert a reaction force of `(W)/(4)`.