The density of a 5.0-m long rod can be described by the linear density function λ(x) = 140 g/m + 14.8x g/m2. One end of the rod is positioned at x = 0 and the other at x = 5 m.
(a) Determine the total mass of the rod.
(b) Determine the center-of-mass coordinate.
I found the total mass of the rod to be 885 g, how does one find the center of mass coordinate in the x-direction given this information?
Thank you in advance.
m=∫ρ•dx =∫(140+14.8x) •dx =
= ∫140•dx+∫14.8•x•dx =
=140•x + 14.8x²/2=
=140•5 + 14.8•25/2 =885 g.
Calculate the integral
∫ρ•x•dx =
=∫(140+14.8x) •x •dx =
= ∫140•x•dx+∫14.8•x²•dx =
=140•x²/2 + 14.8x³/3=
=140•25/2 + 14.8•125/3=2367 kg.
x(c/m/) =∫ρ•x•dx/∫ρ•dx =2367/885=2.67 m.
C.M. (2.67 m; 0)