Respuesta :
It is given that a straight rod has one end at the origin (that is (0,0)) and the other end at the point (L,0) and a linear density given by[tex]\lambda=ax^2[/tex], where a is a known constant and x is the x coordinate.
Therefore, the infinitesimal mass is given as:
[tex]dm=\lambda \times dx=\lambda dx[/tex]
Therefore, the total mass will be the integration of the above equation as:
[tex]\int\,dm= \int\limits^L_0 {ax^2} \, dx[/tex]
Therefore, [tex]m=a\int\limits^L_0 {x^2} \, dx=a[\frac{x^3}{3}]_{0}^{L}=\frac{a}{3}[L^3-0]= \frac{aL^3}{3}[/tex]
Now, we can find the center of mass, [tex]x_{cm}[/tex] of the rod as:
[tex]x_{cm}=\frac{1}{m} \int xdm[/tex]
[tex]x_{cm}=\frac{1}{m}\int_{0}^{L}x\times \lambda dx =\int_{0}^{L}x\times ax^2 dx=\int_{0}^{L}ax^3 dx[/tex]
Now, we have
x_{cm}=\frac{1}{\frac{aL^3}{3}}\int_{0}^{L}ax^3dx=\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}
Therefore, the center of mass, [tex]x_{cm}[/tex] is at:
[tex]\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}=\frac{3}{aL^3}\times \frac{aL^4}{4}=\frac{3}{4}L[/tex]
