Moment of Inertia for a cylinder - Revision history

Mlandwermeyer3 at 03:11, 19 April 2018

2018-04-19T03:11:39Z

@@ Line 1: / Line 1: @@
 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
+Edited by: Micah Landwermeyer Spring 2018
 ==A Brief Overview==
-  [[The Moments of Inertia]] can be defined as the tendency bodies are resistant to angular motion.
+  [[The Moments of Inertia]] can be defined as the tendency of objects with mass to resist angular or rotational motion.
 ==Governing Equations==
@@ Line 23: / Line 70: @@
 *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.
+This describes the moment of inertia calculated when a rod is spun about its center axis as designated by the blue line. This equation is the same as the moment of inertia for a uniform density disk (a cylinder can be thought of as a tall disk).
@@ Line 38: / Line 85: @@
-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.
+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. The expression is identical to the expression for the inertia of a thin rod, except for the addition of a component that accounts for the radius of the cylinder.
 ===Rods===

Mlandwermeyer3 at 00:36, 19 April 2018

2018-04-19T00:36:21Z

← Older revision		Revision as of 20:36, 18 April 2018
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	This page discusses the moment of inertia specifically in the case of a cylinder or rod.		This page discusses the moment of inertia specifically in the case of a cylinder or rod.
			'''Claimed by: Micah Landwermeyer'''
	Written by Jack Corelli		Written by Jack Corelli
	Edited by: Sukanya Basu Spring 2017		Edited by: Sukanya Basu Spring 2017

Sbasu63: /* Examples */

2017-04-10T03:43:02Z

Examples

← Older revision		Revision as of 23:43, 9 April 2017
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	::<math> KE = \frac{1}{2}125 kg100^2(2 \frac{m}{s})^2 </math>		::<math> KE = \frac{1}{2}125 kg100^2(2 \frac{m}{s})^2 </math>
	::<math> KE = 25000 J </math>		::<math> KE = 25000 J </math>



	~~[[File:Screen.png]]~~

	==Connectedness==		==Connectedness==

Sbasu63: /* Examples */

2017-04-10T03:42:52Z

Examples

← Older revision		Revision as of 23:42, 9 April 2017
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	::<math> I = \frac{1}{2}1005^2 </math>		::<math> I = \frac{1}{2}1005^2 </math>
	::<math> I = 125 kg*m^2 </math>		::<math> I = 125 kg*m^2 </math>
	::<math> KE = \frac{1}{2}125 kgm^2(2 \frac{m}{s})^2 </math>		::<math> KE = \frac{1}{2}125 kg100^2(2 \frac{m}{s})^2 </math>
			::<math> KE = 25000 J </math>

Sbasu63: /* Examples */

2017-04-10T03:41:59Z

Examples

← Older revision		Revision as of 23:41, 9 April 2017
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	::<math> I = \frac{1}{2}1005^2 </math>		::<math> I = \frac{1}{2}1005^2 </math>
	::<math> I = 125 kg*m^2 </math>		::<math> I = 125 kg*m^2 </math>
	::<math> KE = \frac{1}{2}125 kgm^2(12 \frac{m}{s})^2 </math>		::<math> KE = \frac{1}{2}125 kgm^2(2 \frac{m}{s})^2 </math>

Sbasu63: /* Examples */

2017-04-10T03:41:30Z

Examples

← Older revision		Revision as of 23:41, 9 April 2017
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	::<math> KE = 480900 J </math>		::<math> KE = 480900 J </math>


			*Calculate the kinetic energy of a cylindrical piston rotating about its central axis given that the rod has mass 100 kg, rotational velocity of 2 m/s, and radius 5m

			::<math> KE = \frac{1}{2}Iω^2 </math> where <math> I = \frac{1}{2}MR^2 </math>
			::<math> I = \frac{1}{2}1005^2 </math>
			::<math> I = 125 kg*m^2 </math>
			::<math> KE = \frac{1}{2}125 kgm^2(12 \frac{m}{s})^2 </math>


	*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.

	[[File:Screen.png]]		[[File:Screen.png]]

Sbasu63: /* Connectedness */

2017-04-10T03:37:52Z

Connectedness

← Older revision		Revision as of 23:37, 9 April 2017
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	#How is this topic connected to something that you are interested in?		#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~~.		Moments of inertia are important because they can be used to calculate kinetic energy of solid objects. These can be used in real life scenarios like wind turbines. The calculation of the moment of inertia for the blades can be used to find the amount of force need to still or to start the motion of the blades.


	== See also ==		== See also ==

Sbasu63: /* Examples */

2017-04-10T03:27:11Z

Examples

← Older revision		Revision as of 23:27, 9 April 2017
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	==Examples==		==Examples==

	#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}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>		:: First, the equation that equals kinetic energy in rotational motion is <math> KE = \frac{1}{2}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>

	Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>		::Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>

	<math> I = 6679.166 kg*m^2 </math>		::<math> I = 6679.166 kg*m^2 </math>

	Thus <math> KE = \frac{1}{2}6679.166 kgm^2(12 \frac{m}{s})^2 </math>		::Thus <math> KE = \frac{1}{2}6679.166 kgm^2(12 \frac{m}{s})^2 </math>

	<math> KE = 480900 J </math>		::<math> KE = 480900 J </math>



	#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.		*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.

	[[File:Screen.png]]		[[File:Screen.png]]

Sbasu63: /* Examples */

2017-04-10T03:26:23Z

Examples

← Older revision		Revision as of 23:26, 9 April 2017
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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}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>		: First, the equation that equals kinetic energy in rotational motion is <math> KE = \frac{1}{2}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>

	Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>		Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>

Sbasu63: /* Examples */

2017-04-10T03:25:54Z

Examples

← Older revision		Revision as of 23:25, 9 April 2017
Line 67:		Line 67:
	#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}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>		; First, the equation that equals kinetic energy in rotational motion is <math> KE = \frac{1}{2}Iω^2 </math> where <math> I = \frac{1}{4}MR^2 + \frac{1}{3}ML^2 </math>

	Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>		Thus, finding I: <math> I = \frac{1}{4}2000.5^2 + \frac{1}{3}20010^2 </math>