The linear mass density of the 2 meter long thin bar extending along the x-axis in the figure is expressed as (PHYS101-105) Problem

Question

The linear mass density of the 2 meter long thin bar extending along the x-axis in the figure is expressed as 𝜆(𝑥) = 2𝑥 + 4 (x in meters; λ in kg/m).

a) (15 points) Find the total mass of the bar.

b) (15 points) Find the position of the center of mass of the bar (xkm).

------------------------

x=0 x=2

Answer 1

(a) The linear density λ = dm /dl, hence dm = λdl. Taking dl = dx, we can write

                                            dm = λ dx or dm = (2x + 4)dx

In order to find the mass of the beam, we need to integrate. So

                           M = ʃ (2x + 4) dx

We need to integrate from 0 to 2 (length of the rod), then

                                             M = 2 ʃ xdx + 4 ʃ dx = 2·x²/ 2 + 4x.

After putting the borders of integration and calculating, we get M = 4 + 8 = 12 (kg)

(b) For center of mas we will apply the formula

                                                                      x_cm = 1/M (ʃ xdm)        

                        

Because dm = λdx, and λ= 2x + 4, we will write that

dm = (2x + 4) dx

Let’s put it in equation for center of mass. Then

                                                                        x_cm = 1/M ʃ x(2x + 4) dx

Let’s at first take this integral separately.

                   ´      ʃ x(2x + 4) dx = ʃ(2x²+ 4x)dx = 2ʃx²dx + 4ʃxdx = 2·2/3 (x³) + 4·x²/2 = 4/3(x³) + 2x².

We integrate from 0 to 2, then we will have

                                                                           ʃ x(2x + 4) dx = 18.7 kg·m

Finally,

                                                                             x_cm = 18.7 / 12 = 1.6 m

                                                                               

                                                                  

                                                                      Answers: 12 kg, 1.6 m