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Question

A rod of length L is placed along the x-axis between x=0 and x=L. The linear mass density (mass/length) ρ of the rod varies with the distance x from the origin as ρ=a+bx. Here, a and b are constants. Find the position of centre of mass of this rod.

A
[3aL+bL26bL, 0, 0]
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B
[3aL+5bL25a+2bL, 0, 0]
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C
[3aL+2bL26a+3bL, 0, 0]
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D
[aLbL, 0, 0]
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Solution

The correct option is C [3aL+2bL26a+3bL, 0, 0]
Let us consider an element of length PQ at distance x from the left end.


Mass of the considered element of the rod will be

dm=ρdx=(a+bx)dx

The COM of the element will have co-ordinates (x, 0, 0).

Therefore, xco-ordinate of COM of the rod is given by

xCOM=L0xdmL0dm=L0(x)(a+bx)dxL0(a+bx)dx

xCOM=[ax22+bx33]L0[ax+bx22]L0

xCOM=3aL2+2bL36aL+3bL2=3aL+2bL26a+3bL

Similarly, yco-ordinate of COM of the rod is

yCOM=ydmdm=0 [y=0]

And for z co-ordinate,

zCOM=0 (z=0)

Hence, the centre of mass of the rod lies at
[3aL+2bL26a+3bL, 0, 0]

So, option (a) is the correct answer.

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