Question

Question #3a15a

Answer 1

The centre-of-mass is at a distance of `\quad x_{cm} = (\frac{A+2/3BL}{2A+BL})L` from the less-heavy end.

Explanation:

Total Mass: We can write the total mass `M` of the rod in terms of the coefficients `A` and `B`, of the linear mass density, and its length `L`.

`M = \int_0^L \quad \lambda(x).dx = \int_0^L\quad (A+Bx).dx`
`M = A(L-0) + B/2(L^2-0) = AL + BL^2/2` ...... (Eq 1)

Centre-of-Mass:

`x_{cm} = 1/M\int\quad x.dm`

If `\lambda(x)` is the linear mass density (density per unit length) at a distance `x` from the less-heavy end,

`dm = \lambda(x)dx = (A+Bx).dx`

`x_{cm} = 1/M\int_0^Lx.(A+Bx).dx`
`x_{cm} = 1/M[A/2(L^2-0^2) + B/3(L^3-0^3)],`
`x_{cm}=1/M[A/2L^2+B/3L^3]`......(Eq 2)

Substituting for `M` using Eq 1 in Eq 2,

`x_{cm} = \frac{A/2L^2+B/3L^3}{AL+B/2L^2}=(\frac{A/2+B/3L}{A+B/2L})L`

`x_{cm} = (\frac{A+2/3BL}{2A+BL})L`