4. A long thin rod of length L has a linear density 2(x) = Ax where x is the distance from the left end of the rod. How far is the center of mass of the rod from the left end of the rod?
4. A long thin rod of length L has a linear density 2(x) = Ax where x is the distance from the left end of the rod. How far is the center of mass of the rod from the left end of the rod?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement
4. A long thin rod of length \( L \) has a linear density \( \lambda(x) = Ax \) where \( x \) is the distance from the left end of the rod. How far is the center of mass of the rod from the left end of the rod?
### Explanation and Solution
This problem involves a rod of length \( L \) with a linear density that varies along its length according to the function \( \lambda(x) = Ax \). Here, \( \lambda(x) \) gives the mass per unit length at a distance \( x \) from the left end of the rod, and \( A \) is a constant.
To find the center of mass, we need to utilize the definition of the center of mass for a continuous mass distribution, which is given by:
\[ x_{\text{cm}} = \frac{\int_0^L x \lambda(x) \, dx}{\int_0^L \lambda(x) \, dx} \]
1. **Calculate the Mass of the Rod**:
The mass \( M \) of the rod can be found by integrating the linear density over the length of the rod:
\[
M = \int_0^L \lambda(x) \, dx = \int_0^L Ax \, dx
\]
Performing the integration:
\[
M = A \int_0^L x \, dx = A \left[ \frac{x^2}{2} \right]_0^L = A \cdot \frac{L^2}{2} = \frac{AL^2}{2}
\]
2. **Calculate the Moment**:
The moment is found by integrating \( x \) times the linear density over the length of the rod:
\[
\int_0^L x \lambda(x) \, dx = \int_0^L x \cdot Ax \, dx = A \int_0^L x^2 \, dx
\]
Performing the integration:
\[
A \int_0^L x^2 \, dx = A \left[ \frac{x^3}{3} \right]_0^L = A \cdot \frac{L^3}{3} = \frac{AL^3}{3}
\]
3. **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffddfa066-ae14-4710-b68e-f1412b1153a3%2F5ba000a5-c8f7-4e9c-87dc-2c5c0b158a46%2F7c9l91_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
4. A long thin rod of length \( L \) has a linear density \( \lambda(x) = Ax \) where \( x \) is the distance from the left end of the rod. How far is the center of mass of the rod from the left end of the rod?
### Explanation and Solution
This problem involves a rod of length \( L \) with a linear density that varies along its length according to the function \( \lambda(x) = Ax \). Here, \( \lambda(x) \) gives the mass per unit length at a distance \( x \) from the left end of the rod, and \( A \) is a constant.
To find the center of mass, we need to utilize the definition of the center of mass for a continuous mass distribution, which is given by:
\[ x_{\text{cm}} = \frac{\int_0^L x \lambda(x) \, dx}{\int_0^L \lambda(x) \, dx} \]
1. **Calculate the Mass of the Rod**:
The mass \( M \) of the rod can be found by integrating the linear density over the length of the rod:
\[
M = \int_0^L \lambda(x) \, dx = \int_0^L Ax \, dx
\]
Performing the integration:
\[
M = A \int_0^L x \, dx = A \left[ \frac{x^2}{2} \right]_0^L = A \cdot \frac{L^2}{2} = \frac{AL^2}{2}
\]
2. **Calculate the Moment**:
The moment is found by integrating \( x \) times the linear density over the length of the rod:
\[
\int_0^L x \lambda(x) \, dx = \int_0^L x \cdot Ax \, dx = A \int_0^L x^2 \, dx
\]
Performing the integration:
\[
A \int_0^L x^2 \, dx = A \left[ \frac{x^3}{3} \right]_0^L = A \cdot \frac{L^3}{3} = \frac{AL^3}{3}
\]
3. **
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