Question #3a15a

1 Answer
Dec 19, 2017

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#