The density of a 5.0-m long rod can be described by the linear density function λ(x) = 145 g/m + 14.2x g/m2. One end of the rod is positioned at x = 0 and the other at x = 5 m.

Question

a) Determine the total mass of the rod.
b) Determine the center-of-mass coordinate.

Answer 1

m=∫ρ•dx =∫(145+14.2x) •dx =
= ∫145•dx+∫14.2•x•dx =
=145•x + 14.2x²/2=
=145•5 + 14.2•25/2 =902.5 g.

Calculate the integral
∫ρ•x•dx =
=∫(145+14.2x) •x •dx =
= ∫145•x•dx+∫14.2•x²•dx =
=145•x²/2 + 14.2x³/3=
=145•25/2 + 14.2•125/3=2404.2 kg.

x(c/m/) =∫ρ•x•dx/∫ρ•dx =2404.2/902.5=2.66 m.

C.M. (2.66 m; 0)

Answer 2

Question: Shouldn't the 2404.2 be in grams and not Kg?