Register | Login | Visitors: 182 / 24 h

Navigation

Export

CreateEditMediaDiscussionHistoryDelete
Welcome→ Course→ Mechanics→ Rotational Energy

Rotational Energy

Table of Contents

  1. Overview
  2. Derivation
  3. Examples
  4. Exercises

Overview

Derivation

For a rotating solid around an axis, the kinetic energy of all mass elements is given as $$E_\mathrm{rot}=\sum\frac{1}{2}\Delta m_iv_i^2$$ For distinguishing this energy form from kinetic energy, which was originally defined for translations, we call it rotational energy $E_\mathrm{rot}$. If we now insert $v=\omega r$m we obtain the rotational energy the following relation: $$E_\mathrm{rot}=\sum\frac{1}{2}\Delta m_i\omega^2r_i^2$$ Making the mass elements infinitesimally small, the sum has to be replaced by an integral: $$E_\mathrm{rot}=\int\frac{1}{2}\omega^2r^2\,\mathrm{d}m$$ The integral $${I = \int r^2\,\mathrm{d}m}$$ will now be defined as the so-called moment of inertia $I$. With this definition, the rotation energy of a body is then given as: $${E_\mathrm{rot} = \frac{1}{2}I\omega^2}$$ The resulting energy is now called rotational energy.
Rotational Energy

The energy of a rotating object is given as $$E_\mathrm{rot} = \frac{1}{2}I\omega^2$$ where $\omega$ is the angular speed and $$I = \int r^2\,\mathrm{d}m$$ the moment of inertia.

Examples

The following examples will illustrate how to solve this integral for different objects. For easy geometrical shapes, it can be solved analytically. However, for complicated structures a numerical computation is required.
Example

The moment of inertia for a mass $M$ rotating around a central point is simply given as $$I = MR^2$$ since the radius is constant the integral $\int \mathrm{d}m$ leads to the constant mass $M$.

Example

We want to calculate now the moment of inertia of a solid cylinder with the radius $r$ and the length $l$ assuming a constant density $\varrho$. The total mass of the cylinder is denoted with $M$. For that purpose, we replace $\mathrm{d}m$ with the product $\varrho \mathrm{d}V$ and the volume element with $\mathrm{d}V = r\,\mathrm{d}r\,\mathrm{d}\varphi\,\mathrm{d}z$ which leads to $$I = \varrho \int\!\!\!\int\!\!\!\int r^3 \,\mathrm{d}r\,\mathrm{d}\varphi\,\mathrm{d}z$$ The integral over $\mathrm{d}\varphi$ is the full angle $2\pi$ and the integral over $\mathrm{d}z$ results in the length $l$. After calculating the integral over $r^3$ we obtained $$I = \varrho \frac{1}{4}R^4 2\pi l$$ Now we can replace $\pi R^2 l$ with the volume $V$ and $\varrho V$ with the mass $M$ which results in $$I = \frac{1}{2}MR^2$$ When we compare this result with a point mass $M$ that orbits around the rotational axis with the distance $R$, we find out that the moment of inertia of a solid cylinder is by the factor 1/2 smaller.

Exercises

This page contains 597 words and 3519 characters.
Last modified: 2022-10-01 17:09:49 by mustafa
Administration