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

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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.

Mathematics

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Svet_ta [12.7K] · Answer 1 · 2026-03-03T10:16:11+03:00

It is stated that a straight rod has one endpoint at the origin (0,0) and the opposite endpoint at (L,0), with a linear density defined by $\lambda=ax^2$ , where a is a constant and x is the x coordinate.

Thus, the infinitesimal mass is expressed as:

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

The total mass can be calculated by integrating the above expression as follows:

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

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

Now, we can calculate 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, it follows that

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 located at:

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