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

Answer

A, B, C

Explanation

Mass of element PQ of length dx situated at `x=x` is

The COM of the element has co-ordinates (x, 0, 0). Therefore, x-coordinate of COM of the rod will be
`x_(COM)=(int_0^Lxdm)/(int_0^Ldm)=(int_0^L(x)(a+bx)dx)/(int_0^L(a+bx)bx)`
`=[(ax^2)/2+(bx^3)/(3)]/[ax+(bx^2)/(2)]_0^L` or `x_(COM)=(3aL+2bL^2)/(6a+3bL)`
The y-coordinate of COM of the rod is
`y_(COM)=(intydm)/(intdm)=0` `(as y=0)`
Similarly, `z_(COM)=0``
Hence, the centre of mass of the rod lies at `[(3aL+2bL^2)/(6a+3aL),0,0]`