question_answer

                A rod is of length 3 m and its mass acting per unit length is directly proportional to distance x from its one end. The centre of gravity of the rod from that end will be at:                                                              

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.