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:             [AIPMT 2002]

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 \[(x,\text{ }0,\text{ }0)\] 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, \[L=3\text{ }m\]

\[\therefore \]      \[{{x}_{CG}}=\frac{3}{2}=1.5\,m\]

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.