Rotational Motion Question 49
Question: 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]
Options:
A) 1.5 m
B) 2.0 m
C) 2.5 m
D) 3.0 m
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Answer:
Correct Answer: A
Solution:
- 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 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) $ x-coordinate of centre of gravity of the rod will be
$ x _{CG}=\frac{\int\limits_O^{L}{xdm}}{\int{dm}} $
$ =\frac{\int\limits_O^{L}{(x)( \frac{M}{L} )dx}}{M} $
$ =\frac{1}{L}\int\limits_O^{L}{xdx=\frac{L}{2}} $
but as given, $ L=3m $
$ \therefore $ $ x _{CG}=\frac{3}{2}=1.5m $
The y-co-ordinate of centre of gravity
$ y _{CG}=\frac{\int{ydm}}{\int{dm}}=0 $
$ (asy=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 distance 1.5 m from one end.
BETA
