Moment of Inertia for a cylinder: Difference between revisions

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#Calculate the kinetic energy of a rod rotating about its end given that the rod has mass 200 kg, rotational velocity of 12 m/s, and length 15m, and radius 0.5m:
#Calculate the kinetic energy of a rod rotating about its end given that the rod has mass 200 kg, rotational velocity of 12 m/s, and length 15m, and radius 0.5m:


First, the equation that equals kinetic energy in rotational motion is <math> KE = \frac{1}{2}*I*ω^2 </math> where <math> I = \frac{1}{4}*M*R^2 + \frac{1}{3}*M*L^2 </math>
; First, the equation that equals kinetic energy in rotational motion is <math> KE = \frac{1}{2}*I*ω^2 </math> where <math> I = \frac{1}{4}*M*R^2 + \frac{1}{3}*M*L^2 </math>


Thus, finding I: <math> I = \frac{1}{4}*200*0.5^2 + \frac{1}{3}*200*10^2 </math>
Thus, finding I: <math> I = \frac{1}{4}*200*0.5^2 + \frac{1}{3}*200*10^2 </math>

Revision as of 23:25, 9 April 2017

This page discusses the moment of inertia specifically in the case of a cylinder or rod.

Written by Jack Corelli Edited by: Sukanya Basu Spring 2017

A Brief Overview

The Moments of Inertia can be defined as the tendency bodies are resistant to angular motion. 

The moment of inertia of an object relates the mass of the object, the distance between the center of mass and the shape of rotation. In a ring, or uniform loop, the center of mass is exactly at the center of the ring, and the moment of inertia can be found by approximating the exterior ring as a single line, akin to a circle. The moment of inertia is dependent on the axis of rotation. In terms of a cylinder, the moment of inertia can be different even if the shape, volume, and mass of the cylinder are the same, given that each cylinder has different axes of revolution.

Governing Equations

Cylinders

As seen in the image above, in its simplest form, the moment of inertia can be found for a simple, constant density and shape cylinder by the equation:

[math]\displaystyle{ I = \frac{1}{2}MR^2 }[/math]

where:

  • I is the moment of inertia
  • M is the mass of the object being rotated
  • R is the radius of the rod

This describes the moment of inertia calculated when a rod is spun about its center axis as designated by the blue line.


In the picture above, when the axis of rotation is no longer the center of the cylinder, the equations for moment of inertia change distinctly. When both the radius of the cylinder and the length of the cylinder are significant, the following equation is used.

[math]\displaystyle{ I = \frac{1}{4}MR^2 + \frac{1}{12}ML^2 }[/math] where:

  • I is the moment of inertia
  • M is the mass of the object being rotated
  • R is the radius of the rod
  • L is the length of the rod


This describes the moment of inertia calculated when the rod is spun about an axis of revolution perpendicular to its length at the center of the cylinder.

Rods

As seen in the image above, when the axis of rotation is perpendicular to the length, but is touching at the middle of the rod, then the equation would be: [math]\displaystyle{ I = \frac{1}{12}ML^2 }[/math]

where:

  • I is the moment of inertia
  • M is the mass of the object being rotated
  • L is the length of the rod


As seen in the image above, when the axis of rotation is perpendicular to the length, but is touching at one end of the rod, then the equation would be: [math]\displaystyle{ I = \frac{1}{3}ML^2 }[/math]

where:

  • I is the moment of inertia
  • M is the mass of the object being rotated
  • L is the length of the rod

Examples

  1. Calculate the kinetic energy of a rod rotating about its end given that the rod has mass 200 kg, rotational velocity of 12 m/s, and length 15m, and radius 0.5m:
First, the equation that equals kinetic energy in rotational motion is [math]\displaystyle{ KE = \frac{1}{2}*I*ω^2 }[/math] where [math]\displaystyle{ I = \frac{1}{4}*M*R^2 + \frac{1}{3}*M*L^2 }[/math]

Thus, finding I: [math]\displaystyle{ I = \frac{1}{4}*200*0.5^2 + \frac{1}{3}*200*10^2 }[/math]

[math]\displaystyle{ I = 6679.166 kg*m^2 }[/math]

Thus [math]\displaystyle{ KE = \frac{1}{2}*6679.166 *kg*m^2*(12 \frac{m}{s})^2 }[/math]

[math]\displaystyle{ KE = 480900 J }[/math]


  1. Mounted on a low-mass rod of length 0.16 m are four balls (see figure below). Two balls (shown in red on the diagram), each of mass 0.85 kg, are mounted at opposite ends of the rod. Two other balls, each of mass 0.25 kg (shown in blue on the diagram), are each mounted a distance 0.04 m from the center of the rod. The rod rotates on an axle through the center of the rod (indicated by the "✕" in the diagram), perpendicular to the rod, and it takes 0.9 s to make one full rotation.

Connectedness

  1. How is this topic connected to something that you are interested in?

Moments of Inertia are very important in industrial processes, especially for engines. Think of a wind turbine. That is essentially the third equation above, and if an engineer wanted to figure out how much force would be needed to slow the blade, that is one of the components in that equation. But also, due to the torque applied to the blade, there would need to be a slow negative acceleration, so as to not break the wind turbine.


See also

You can find more information in the text book about moments of inertia for different sized objects.


External links

https://en.wikipedia.org/wiki/List_of_moments_of_inertia

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

https://www.quora.com/What-is-an-explanation-in-simple-words-of-the-moment-of-inertia

https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/a/rotational-inertia

http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html