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Magazine Chapter 15: Problem 38
A slender rod of constant density lies along the line segment \(\mathbf{r}(t)=t \mathbf{j}+(2-2 t) \mathbf{k}, 0 \leq t \leq 1,\) in the yz-plane. Find the moments of inertia of the rod about the three coordinate axes.
Short Answer
Expert verified
The moments of inertia are \(I_x = \frac{\rho\sqrt{5}}{3}\), \(I_y = \frac{4\rho\sqrt{5}}{3}\), and \(I_z = \frac{\rho\sqrt{5}}{3}\).
Step by step solution
01
Understand the Problem
We need to calculate the moments of inertia of a slender rod about the x-axis, y-axis, and z-axis. The rod is described by the vector equation \( \mathbf{r}(t) = t \mathbf{j} + (2 - 2t) \mathbf{k} \), and is located in the yz-plane from \( t = 0 \) to \( t = 1 \).
02
Set up the Moment of Inertia about the x-axis (\(I_x\))
The moment of inertia about the x-axis is given by \[I_x = \int y^2 \, dm\]where \( y = t \) and \( dm = \rho \, ds \). Calculate \( ds \) and find the expression for \( dm.\)
03
Calculate the Element Length (ds)
The differential length element is:\[ ds = \sqrt{\left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \]For \( y = t \) and \( z = 2 - 2t \), compute:\[ \frac{dy}{dt} = 1, \quad \frac{dz}{dt} = -2 \]Thus,\[ ds = \sqrt{1^2 + (-2)^2} \, dt = \sqrt{5} \, dt \]
04
Calculate dm
The mass element \( dm \) is given by:\[ dm = \rho \, ds = \rho \, \sqrt{5} \, dt \]Here, \( \rho \) is the linear density of the rod.
05
Calculate the Moment of Inertia about the x-axis
Integrate to find \( I_x \):\[ I_x = \int_0^1 y^2 \, dm = \int_0^1 t^2 \, (\rho \, \sqrt{5} \, dt) = \rho \, \sqrt{5} \int_0^1 t^2 \, dt = \rho \, \sqrt{5} \left[ \frac{t^3}{3} \right]_0^1 = \frac{\rho \, \sqrt{5}}{3} \]
06
Set up the Moment of Inertia about the y-axis (\(I_y\))
The moment of inertia about the y-axis is given by \[I_y = \int z^2 \, dm = \int_0^1 (2-2t)^2 \, \rho \, \sqrt{5} \, dt.\]
07
Calculate the Moment of Inertia about the y-axis
Substitute into the integral:\[ I_y = \rho \, \sqrt{5} \int_0^1 (2-2t)^2 \, dt = \rho \, \sqrt{5} \int_0^1 (4 - 8t + 4t^2) \, dt \]Solve the integral:\[ I_y = \rho \, \sqrt{5} \left[ 4t - 4t^2 + \frac{4t^3}{3} \right]_0^1 = \rho \, \sqrt{5} \left( 4 - 4 + \frac{4}{3} \right) = \frac{4\rho \, \sqrt{5}}{3} \]
08
Set up and Calculate the Moment of Inertia about the z-axis (\(I_z\))
The moment of inertia about the z-axis:\[ I_z = \int (x^2 + y^2) \, dm \]Since the rod lies in the yz-plane, \( x = 0 \), so \[ I_z = \int y^2 \, dm = I_x \].
09
Moment of Inertia about the z-axis Result
Since the expression for \( I_z \) is the same as \( I_x \), the calculation yields the same result.\[ I_z = \frac{\rho \, \sqrt{5}}{3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Axes
In three-dimensional space, the coordinate axes act as reference lines that help us to locate points, lines, or shapes. These axes are the x-axis, y-axis, and z-axis, which intersect at a common point known as the origin. Here are some key points about these axes:
- The x-axis runs horizontally, the y-axis runs vertically, and the z-axis comes out towards you in 3D.
- The intersection of these axes defines the origin, with coordinates (0, 0, 0).
- Each axis is orthogonal to the others, meaning they are positioned at right angles to each other.
- They create a 3-dimensional coordinate system used to describe the position of any point in space using coordinates (x, y, z).
yz-plane
A plane in three-dimensional coordinate systems is defined by two axes. The yz-plane is defined by the y-axis and z-axis, with the x-coordinate being zero on this plane. Here are some essential aspects of the yz-plane:
- It includes all points where the x-coordinate is 0, expressed as (0, y, z).
- The yz-plane is coplanar, meaning every point on the plane lies in the same 2-dimensional flat surface.
- This plane is crucial in simplifying problems, as seen where the rod lies entirely along the yz-plane, freeing us from considering the x-direction.
- Introduces ease in computations, especially when dealing with symmetry around or on the plane.
Vector Equation
A vector equation represents a curve or a line in space using vectors. A vector provides direction and magnitude. In the context of the given exercise, the vector equation \( \mathbf{r}(t) = t \mathbf{j} + (2 - 2t) \mathbf{k} \)describes a line segment within the yz-plane. Here's what you should know:
- The vector equation shows us how the position of the rod changes with the parameter \(t\).
- Vector components, like \(t\mathbf{j}\) and \((2 - 2t)\mathbf{k}\), specify movement along the y and z directions, respectively.
- As \(t\) varies from 0 to 1, it traces the rod's entire length, showing the way it lies along the yz-plane.
- Using vector equations can simplify calculations of geometrical properties, as they make explicit the relationship between different components.
Linear Density
Linear density is a physical property that quantifies the mass distribution along a line or rod. It is expressed as mass per unit length. In the context of moments of inertia calculation:
- The linear density \(\rho\) is used to determine how mass is distributed along the rod, allowing for the integration in mass-related equations.
- In uniform rods, \(\rho\) is constant, simplifying computations as seen in this exercise.
- When deriving the moment of inertia, \(dm\) is expressed as \(\rho \, ds\), where \(ds\) is the differential length element.
- This linkage between density and differential elements allows us to perform integration over the rod's length, directly affecting the calculated moment of inertia.
