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Thin shell of constant density Find the center of mass and the moment of inertia about the \(z\) -axis of a thin shell of constant density \(\delta\) cut from the cone \(x^{2}+y^{2}-z^{2}=0\) by the planes \(z=1\) and \(z=2\) .

Short Answer

Expert verified
The center of mass is at \((0, 0, \frac{4}{3})\) and moment of inertia is \(\frac{15\pi\delta}{4}\).

Step by step solution

01

Understand the Geometry of the Problem

The thin shell is part of a cone given by the equation \(x^2 + y^2 = z^2\), cut between the planes \(z=1\) and \(z=2\). This implies the shell is bounded by circles of radii \(1\) and \(2\) in the \(xy\)-plane, at respective heights \(z=1\) and \(z=2\). Thus, it forms a truncated cone or frustum.
02

Express in Cylindrical Coordinates

In cylindrical coordinates, the transformations are \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The equation \(x^2 + y^2 = z^2\) becomes \(r^2 = z^2\), thus \(r = z\). The volume element on the surface of the shell is \(r \, dr \, d\theta\), with \(1 \leq z \leq 2\) and \(0 \leq \theta < 2\pi\).
03

Calculate the Mass of the Shell

For a surface of constant density \(\delta\), the differential element of mass \(dm\) is given by \(\delta \, dS\). Substituting \(dS=r \, d\theta \, dz\), the total mass \(M\) is \(M = \int_0^{2\pi} \int_1^2 \delta \, z \, d\theta \, dz = \delta \int_0^{2\pi} \int_1^2 z \, d\theta \, dz\). Evaluating the integrals, we get \(M = 3\pi\delta/2\).
04

Determine the Center of Mass

For symmetry about the \(z\)-axis, the \(x\) and \(y\) coordinates of the center of mass are zero. The \(z\)-coordinate of the center of mass is calculated as \(z_{cm} = \frac{1}{M} \int_0^{2\pi} \int_1^2 z \, \delta \, z \, d\theta \, dz\). After integration, \(z_{cm} = \frac{4}{3}\).
05

Compute the Moment of Inertia about the z-axis

The moment of inertia \(I_z\) about the \(z\)-axis is given by \(I_z = \int_0^{2\pi} \int_1^2 \delta \, r^2 \, z \, d\theta \, dz\). Since \(r = z\), this becomes \(\int_0^{2\pi} \int_1^2 \delta \, z^3 \, d\theta \, dz\). Evaluating the integrals yields \(I_z = \frac{\pi\delta(15)}{4}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Center of Mass
The center of mass is a crucial concept that helps you understand how mass is distributed within a three-dimensional object. For a symmetrical object like the cone mentioned, the center of mass can be determined due to the uniform distribution of the mass.

In our problem, the cone forms a thin shell of constant density \(\delta\). Given this symmetry, the center of mass in the x and y dimensions will be zero, as mass is evenly spread out around the z-axis. This simplifies the process, making it necessary to only find the position along the z-axis.

To find the z-coordinate of the center of mass, you use the formula: \( z_{cm} = \frac{1}{M} \int_0^{2\pi} \int_1^2 z \, \delta \, z \, d\theta \, dz = \frac{4}{3}\). This calculation shows that even in sections of varying heights, the mass is concentrated at \frac{4}{3}\ of the way from the base to the top of the shell.
Cylindrical Coordinates
Cylindrical coordinates provide a practical way to handle problems involving shapes like cones, as they align closely with the geometry of these objects. In this coordinate system, any point in space is defined by three values: radius \(r\), angle \(\theta\), and height \(z\).

This system simplifies handling the conical shape described by \(x^2 + y^2 = z^2\). When we convert to cylindrical coordinates, the equation becomes \(r^2 = z^2\), which simplifies to \(r = z\).

This relation highlights that the radius changes in line with height, perfectly capturing the way cones taper. Moreover, we integrate the surface differential element \(dS = r \, d\theta \, dz\) across the shell for calculations, making it easier to handle double integrals over a complex geometry.
Cone
A cone is a three-dimensional geometric shape that tapers smoothly from a flat, often circular base to a point called the apex. Understanding the properties of cones helps when dealing with problems involving mass, volume, and inertia.

The specific cone in this exercise is bounded by the equation \(x^2 + y^2 = z^2\), an equation that delineates a right circular cone. In our case, the cone is further bounded, thus forming a truncated cone or frustum, by the planes \(z=1\) and \(z=2\).

These boundaries define the area and volume of the shell, and understanding them ensures that calculations of properties like inertia are accurate when integrated over this specified volume.
Constant Density
Constant density means that the mass per unit volume is uniform throughout the object. In this context, it simplifies calculations, as density \(\delta\) can be factored out of integrals when computing properties, such as mass and moment of inertia.

For our cone shell example, \(\delta\) determines the differential mass element as \(dm = \delta \, dS\), where \(dS = r \, d\theta \, dz\) represents a small element on the surface of the shell. This constant density approach streamlines setting up integrals by reducing the number of variables affecting the outcome, only depending on the geometry and limits of integration.

Constant density is key in achieving the integration solutions where the mass \(M\) and moment of inertia are calculated in steps, ultimately influenced only by the shape's geometry and bounds, not the distribution of mass within the shell.

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Most popular questions from this chapter

Let \(R\) be a region in the \(x y\) -plane that is bounded by a piecewise smooth simple closed curve \(C\) and suppose that the moments of inertia of \(R\) about the \(x\) - and \(y\) -axes are known to be \(I_{x}\) and \(I_{y}\) . Evaluate the integral $$\oint_{C} \nabla\left(r^{4}\right) \cdot \mathbf{n} d s$$ where \(r=\sqrt{x^{2}+y^{2}},\) in terms of \(I_{x}\) and \(I_{y}\)

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let n be the outward unit normal vector field on \(S .\) Show that it is not possible for \(F\) to be orthogonal to \(n\) at every point of \(S .\)

Conservation of mass Let \(\mathbf{v}(t, x, y, z)\) be a continuously differentiable vector field over the region \(D\) in space and let \(p(t, x)\) \(y, z )\) be a continuously differentiable scalar function. The variable \(t\) represents the time domain. The Law of Conservation of Mass asserts that $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=-\iint_{S} p \mathbf{v} \cdot \mathbf{n} d \sigma$$ where \(S\) is the surface enclosing \(D\) a. Give a physical interpretation of the conservation of mass law if \(\mathbf{v}\) is a velocity flow field and \(p\) represents the density of the fluid at point \((x, y, z)\) at time \(t\) b. Use the Divergence Theorem and Leibniz's Rule, $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=\iiint_{D} \frac{\partial p}{\partial t} d V$$ to show that the Law of Conservation of Mass is equivalent to the continuity equation, $$\nabla \cdot p \mathbf{v}+\frac{\partial p}{\partial t}=0$$ (In the first term \(\nabla \cdot p \mathbf{v},\) the variable \(t\) is held fixed, and in the second term \(\partial p / \partial t,\) it is assumed that the point \((x, y, z)\) in \(D\) is held fixed.)

Let \(\mathbf{F}\) be a differentiable vector field and let \(g(x, y, z)\) be a differentiable scalar function. Verify the following identities. $$\begin{array}{ll}{\text { a. }} & {\nabla \cdot(g \mathbf{F})=g \nabla \cdot \mathbf{F}+\nabla g \cdot \mathbf{F}} \\ {\text { b. }} & {\nabla \times(g \mathbf{F})=g \nabla \times \mathbf{F}+\nabla g \times \mathbf{F}}\end{array}$$

Find the work done by \(\mathbf{F}\) in moving a particle once counterclockwise around the given curve. \(\mathbf{F}=2 x y^{3} \mathbf{i}+4 x^{2} y^{2} \mathbf{j}\) C: The boundary of the "triangular" region in the first quadrant enclosed by the \(x\) -axis, the line \(x=1,\) and the curve \(y=x^{3}\)

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