Question

A straight rod has one end at the origin and the other end at the point (l,0) and a linear density given by λ=ax2, where a is a known constant and x is the x coordinate. since this wire is not uniform, you will have to use integrtation to solve this part. use m=∫l0dm to find the total mass m. find xcm for this rod. express your answer in terms of one or both of a and l.

Answer 1

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$\lambda=ax^2$, where a is a known constant and x is the x coordinate.

Therefore, the infinitesimal mass is given as:

$dm=\lambda \times dx=\lambda dx$

Therefore, the total mass will be the integration of the above equation as:

$\int\,dm= \int\limits^L_0 {ax^2} \, dx$

Therefore, $m=a\int\limits^L_0 {x^2} \, dx=a[\frac{x^3}{3}]_{0}^{L}=\frac{a}{3}[L^3-0]= \frac{aL^3}{3}$

Now, we can find the center of mass, $x_{cm}$ of the rod as:

$x_{cm}=\frac{1}{m} \int xdm$

$x_{cm}=\frac{1}{m}\int_{0}^{L}x\times \lambda dx =\int_{0}^{L}x\times ax^2 dx=\int_{0}^{L}ax^3 dx$

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, $x_{cm}$ is at:

$\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}=\frac{3}{aL^3}\times \frac{aL^4}{4}=\frac{3}{4}L$