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 rCG=rdmM .

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 andtheotherat x=L $ .

Mass per unit length of the rod =ML

Hence, the mass of the element PQ of length dx==MLdx

The co-ordinates of the element PQ are (x,0,0) x-coordinate of centre of gravity of the rod will be

xCG=OLxdmdm

=OL(x)(ML)dxM

=1LOLxdx=L2

but as given, L=3m

xCG=32=1.5m

The y-co-ordinate of centre of gravity

yCG=ydmdm=0

(asy=0)

Similarly, zCG=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.



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