/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Das Tr盲gheitsmoment eines Kreis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 10

Das Tr盲gheitsmoment eines Kreiskegels bez. seiner Achse ist zu berechnen (Dichte \(\rho=\) const.).

Short Answer

Expert verified
The moment of inertia of the cone is \( I = \frac{3}{10} M R^2 \).

Step by step solution

01

Understand the Problem

To calculate the moment of inertia of a cone with respect to its axis, we need to integrate over the volume of the cone taking into account the mass distribution with respect to the axis. We are given a constant density \( \rho \).
02

Express the Cone's Geometry

Consider a cone with a height \( h \) and a base radius \( R \). The axis of the cone is the line passing through the apex and the center of the base. Write the equation for the radius in terms of vertical position \( y \): \( r(y) = R \left( \frac{h-y}{h} \right) \).
03

Volume Element

A small disk inside the cone can be considered at a position \( y \) with thickness \( dy \). Its radius is \( r(y) \). The differential volume element \( dV \) would then be the volume of this disk: \( dV = \pi [r(y)]^2 \, dy = \pi \left( R \frac{h-y}{h} \right)^2 \, dy \).
04

Mass Element

The differential element of mass \( dm \) is given by \( \rho \, dV \). Substitute the expression for \( dV \) to get \( dm = \rho \pi \left( R \frac{h-y}{h} \right)^2 \, dy \).
05

Elemental Moment of Inertia

For a disk, the moment of inertia about its central axis is \( dI = \frac{1}{2} \, dm \, r(y)^2 \). Substitute \( dm \) and \( r(y) \) to get \( dI = \frac{1}{2} \rho \pi \left( R \frac{h-y}{h} \right)^4 \, dy \).
06

Integrate to Find Moment of Inertia

Integrate \( dI \) from \( y = 0 \) to \( y = h \) to find the total moment of inertia, \( I \). This leads to:\[ I = \frac{1}{2} \rho \pi R^4 \int_0^h \left( \frac{h-y}{h} \right)^4 \, dy \].Evaluate this integral.
07

Evaluate the Integral

The integration simplifies to:\[ I = \frac{1}{2} \rho \pi R^4 \frac{1}{5h^4} \cdot h \].Evaluating, we find:\[ I = \frac{1}{10} \rho \pi R^4 h \].
08

Substitute Mass

The mass \( M \) of the cone is given by \( M = \rho \frac{1}{3} \pi R^2 h \). Substitute \( \rho \) in terms of \( M \) to get:\[ I = \frac{3}{10} M R^2 \].
09

Final Answer

The moment of inertia of the cone about its axis is \( I = \frac{3}{10} M R^2 \) where \( M \) is the mass of the cone.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Solid Geometry
Solid Geometry is crucial when dealing with three-dimensional objects like cones. In our scenario, understanding the geometric structure of the cone is step one in computing its moment of inertia.
  • The cone has a specific height, denoted as \( h \), and a base radius, labeled as \( R \).
  • The axis of the cone is defined by the line running from the apex, through the center of the base.
  • To address this problem, it is necessary to understand how dimensions of the cone relate to the position inside it. This is where the radius as a function of height, or \( r(y) = R \left( \frac{h-y}{h} \right) \), becomes important.
Understanding these properties helps visualize the cone and is the foundation for further calculations. Solid Geometry allows us to translate these geometric descriptors into guiding equations.
Integral Calculus
Integral Calculus provides the tools needed to handle continuous distribution of mass in structures like cones through integration.
  • Firstly, identify the differential volume element, in this case a thin disk perpendicular to the axis at height \( y \) with thickness \( dy \).
  • The volume element \( dV \) is given as \( \pi [r(y)]^2 \, dy \). This expression accounts for the entire volume of that disk.
  • The next step is to compute the differential element of mass \( dm \), by substituting \( dV \) into \( dm = \rho \, dV \).
  • Using this distribution of mass, the differential moment of inertia \( dI \) can be written as \( \frac{1}{2} \, dm \, r(y)^2 \).
Finally, integrating these expressions from the base to the apex of the cone computes the total moment of inertia by summing up the contributions of all these differential elements. This technique shows how integral calculus simplifies understanding intricate systems by considering them incrementally.
Mass Distribution
Mass Distribution plays a key role in the calculation of moment of inertia. Knowing how mass is spread across the solid is vital.
  • A uniform density is an important assumption. Here, it is set constant as \( \rho \).
  • This constant density \( \rho \), combined with the geometry, determines how mass is distributed through the cone. This relationship needs to be factored into the integration to find inertia.
  • To find the total mass \( M \) of the cone, you can integrate the mass element over the entire volume: \( M = \rho \frac{1}{3} \pi R^2 h \).
In the end, Mass Distribution helps relate the cone's geometric and mass properties, seamlessly transitioning into the formula for the moment of inertia. This connection illustrates how mass distribution impacts rotational dynamics, affirming the tight link between physical properties and mathematical expressions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Die Funktion \(f\) habe in Polarkoordinaten die Gleichung \(z=u(r, \varphi)=r^{2}-8 \cdot \cos 2 \varphi\) (Lemniskate). Wie lauten ihre partiellen Ableitungen nach den kartesischen Koordinaten \(x\) und \(y\) ?

Untersuchen Sie, ob das Vektorfeld \(\vec{v}=(y, x, 0)\) ein Potentialfcld ist und bestimmen Sie ggf. das Potential. Ist \(y \mathrm{~d} x+x \mathrm{~d} y\) totales Differential einer Funktion \(f\) dreier Ver?nderlichen \((x, y, z)\) ? Wie lautet \(f\) gegebenenfalls?

Es sei \(K\) eine Kugel vom Radius \(R\) mit dem Mittelpunkt \((0,0,0)\) und \(f(x, y, z)=x^{2}+y^{2}\). Berechnen Sie \(^{K} \int f d k\).

Untersuchen Sie, ob die Differentialform $$ (2 x y+2 z \cdot \sin x \cdot \cos x) \mathrm{d} x+\left(x^{2}+z\right) \mathrm{d} y+\left(y+\sin ^{2} x\right) \mathrm{d} z $$ totales Differential einer Funktion \(f\) dreier Variablen ist und berechnen Sie ggf. \(f\). Vergleichen Sie auch mit Aufgabe \(12 .\)

Berechnen Sie den Schwerpunkt einer homogenen Halbkugel.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Dynamics

Read Explanation

Oscillations

Read Explanation

Measurements

Read Explanation

Fields in Physics

Read Explanation

Mechanics and Materials

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.