Question

A rod of length 30.0 cm has linear density (mass per length) given by

$\mathbf{\lambda }\mathbf{=}\mathbf{50}\mathbf{+}\mathbf{20}\mathbf{x}$

where x is the distance from one end, measured in meters, and $\mathbf{\lambda }$ is in grams/meter. (a) What is the mass of the rod? (b) How far from the x = 0 end is its center of mass?

Short answer

1. The mass of rod is 15.9 g.
2. The center of mass of rod is 0.153 m from x=0.

Step-by-step solution

Identification of given data

The length of rod is L=30 cm

The density of rod is $\mathrm{\lambda }=50+20\mathrm{x}$

The distance between centers of Earth and Moon is $d=3.84\times {10}^8\mathrm{m}$

The center of mass of system of particles is the point at which total mass of the particle can be assumed to be act.

Determination of mass of rod

(a)

The mass of rod is given as:

$m={\int }_0^L\lambda dx$

Substitute all the values in the above equation.

$$\begin{gathered} m={\int }_0^{30cm}\left(50+20x\right)dx \\ m={\left(50x+10x^2\right)}_0^{30cm} \\ m=\left[50\left(30\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)+10{\left(\left(30\mathrm{cm}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2\right]-\left[50\left(0\mathrm{cm}\right)+10{\left(0\mathrm{cm}\right)}^2\right] \\ m=15.9\mathrm{g} \end{gathered}$$

Therefore, the mass of rod is 15.9 g.

Determination of center of mass of rod

(b)

The center of mass of rod is given as:

$$\begin{gathered} x_{cm}=\frac{{\int }_0^{L}xdm}{m} \\ {x}_{cm}=\frac{1}{m}{\int }_0^{L}x\left(\lambda dx\right) \\ {x}_{cm}=\frac{1}{m}{\int }_0^L\left(\lambda x\right)dx \end{gathered}$$

Substitute all the values in the above equation.

Therefore, the center of mass of rod is 0.153 m from x=0.