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Question

# The density of a linear rod of length L varies as ρ=A+Bx where x is the distance from the left end. Its centre of mass is at a distance (from the left end) of

A
3AL2BL23(2ABL)
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B
3AL+2BL23(2A+BL)
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C
3AL+2BL22A+BL
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D
3AL+2BL22ABL
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Solution

## The correct option is B 3AL+2BL23(2A+BL)In this case, the rod is linear. So, its centre of mass will lie along x−axis. For the non-uniform body, the x−coordinate of COM will be, xcom=∫xdm∫dm Let us take a very small element of length dx from the left end of the rod as shown in the figure. ⇒xcom=∫L0xρdx∫L0ρdx [∵ρ=dmdx⇒dm=ρdx] ⇒xcom=∫L0x(A+Bx)dx∫L0(A+Bx)dx [ρ=A+Bx given] ⇒xcom=∫L0(Ax+Bx2)dx∫L0(A+Bx)dx ⇒xcom=[Ax22+Bx33]L0[Ax+Bx22]L0 ⇒xcom=[AL22+BL33−(0+0)][AL+BL22−(0+0)] ⇒xcom=3AL2+2BL362AL+BL22 ∴xcom=3AL+2BL23(2A+BL) Hence centre of mass of rod is at a distance of 3AL+2BL23(2A+BL) from the left end.

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