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A uniform narrow rod of mass M and length 2 L is kept vertically, along the y-axis (as shown in Fig. ) on a smooth horizontal plane. The lower end of the rod coincides with the origin (0,0). Due to a slight disturbance at time t = 0 , the rod slides along the positive x-axis and begins to fall. Determine (i) the shift in the centre of gravity during the fall, (ii) the equation of the locus of a point ata distance r from the lower end of the rod and also mention the shape of the locus. |
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Answer» Solution :Initially (0,L) was the coordinates of the centre of gravity of the rod. Let B be the point at a distance r, from point O (0,0) , on the rod. So, (0,r) denotes the point B. At t=0, the rod is disturbed and it falls. During this fall there is no other force acting on it except the downward gravitational force. (i) We know that the centre of gravity is influenced only by an external force which is gravity in this case. So the centre of gravity will be SHIFTED vertically downwards from (0,L) ot (0,0). (ii) Let (x,y) be the POSITION of B. at any moment during the fall of the rod, where B. is the point at a distance r, from the lower end O. of the rod. From the Fig. `(OG.)/(O.G.)=(CB.)/(O.B.)or, (a)/(L)=(y)/(r)` Again, `(O.O)/(OC)=(O.G)/(G.B.)or, (OC)/(G.B.)=(O.O)/(O.G.)or, (x)/(r-L) =(b)/(L)` `:. (x^(2))/((r-L)^(2))=(b^(2))/(L^(2))=(L^(2)-a^(2))/(L^(2))=1-(a^(2))/(L^(2))=1-(y^(2))/(r^(2))` or, `(x^(2))/(r-L)^(2)+(y^(2))/(r^(2))=1` This is the EQUATION of the locus of a point (x,y) and it is elliptical with its centre at the origin. 
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