How Do You Calculate the Mass of a Rod with Non-Uniform Density?

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Homework Help Overview

The discussion revolves around calculating the mass of a rod with a non-uniform linear mass density defined by the function λ = 0.300x² + 0.500, where the rod extends from x=2.00 m to x=3.00 m. Participants are exploring the implications of this density function on both mass and center of mass calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster seeks assistance in calculating the mass of the rod using the provided density function. One participant suggests using integration to find the mass, while another raises a question about determining the center of mass.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem, including mass calculation and center of mass determination. Some guidance has been offered regarding the integration approach for mass, but there is no consensus on the methods for finding the center of mass.

Contextual Notes

Participants are working within the constraints of the problem as posed, focusing on the specific density function and the limits of integration defined by the endpoints of the rod.

AstroturfHead
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I'm having a hard time with this.
A long thin rod lies along the x-axis. One end is at x=2.00 m and the other at x=3.00 m. Its linear mass density λ = 0.300 x2+ 0.500, in kg/m. Calculate mass of the rod.

Can anybody help?
 
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I take it you mean [itex]\lambda = .300x^2 + .500[/itex]?

Note that

[tex]m = \int_{x_1}^{x_2}\lambda(x) dx[/tex].

--J
 
But then how would I find the center of mass?
 
The center of mass is just the weighted average of the density. You really should be able to look these formulas up on your own, or memorize them.

[tex]x_{CM}= \frac{1}{m}\int_{x_1}^{x_2} x \lambda(x) dx[/tex]

--J
 

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