A) 1.5 m
B) 2.0 m
C) 2.5 m
D) 3.0 m
Correct Answer: A
Solution :
Key Idea: Position of centre of gravity of a continuous body is given by \[{{r}_{CG}}=\frac{\int{rdm}}{M}.\] A rod lying along any of coordinate axes serves for us as continuous body. Suppose a rod of mass M and length L is lying along die x-axis with its one end at x = 0 and the other at x = L. Mass per unit length of the rod \[=\frac{M}{L}\] Hence, the mass of the element PQ of length dx \[=\frac{M}{L}\,dx\] The co-ordinates of the element PQ are (x, 0, 0). Therefore, x-coordinate of centre of gravity of the rod will be \[{{x}_{CG}}=\frac{\int\limits_{O}^{L}{x\,dm}}{\int{dm}}\] \[=\frac{\int\limits_{O}^{L}{(x)\,\left( \frac{M}{L} \right)dx}}{M}\] \[=\frac{1}{L}\int\limits_{O}^{L}{x\,\,dx=\frac{L}{2}}\] but as given, L = 3 m \[\therefore {{x}_{CG}}=\frac{3}{2}=1.5\,m\] The y-co-ordinate of centre of gravity \[{{y}_{CG}}=\frac{\int{ydm}}{\int{dm}}=0\] \[(as\,y=0)\] Similarly, \[{{Z}_{CG}}=0\] i.e., the co-ordinates of centre of gravity of n rod are (1.5, 0, 0) or it lies at the distance 1.5 m from one end.You need to login to perform this action.
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