The Moments of Inertia - Revision history

Psaokar at 22:41, 11 November 2025

2025-11-11T22:41:57Z

← Older revision		Revision as of 18:41, 11 November 2025
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			<b> Claim By Prajwal Saokar Fall 2025 </b>

	claimed and written by san47		claimed and written by san47
	[http://hyperphysics.phy-astr.gsu.edu/hbase/images/inecon.gif][[File:Rlin.gif\|thumb\|300px\|Rotation-Linear Parallels]]		[http://hyperphysics.phy-astr.gsu.edu/hbase/images/inecon.gif][[File:Rlin.gif\|thumb\|300px\|Rotation-Linear Parallels]]

Seisner6: /* Examples */

2019-06-25T20:57:53Z

Examples

← Older revision		Revision as of 16:57, 25 June 2019
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	Now, substituting back into the integral and cancelling terms, we arrive at:		Now, substituting back into the integral and cancelling terms, we arrive at:

	<math> I_{sphere} = \frac[1}{2}\frac{m}{\frac{4}{3} \pi r^3}\pi \int_{-r}^{r} (r^2 - z^2)^2 dz </math>		<math> I_{sphere} = \frac{1}{2}\frac{m}{\frac{4}{3} \pi r^3}\pi \int_{-r}^{r} (r^2 - z^2)^2 dz </math>


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	Evaluating the integral, we get:		Evaluating the integral, we get:

	<math> I_{sphere} = \frac{m}{\frac{8}{3} r^3} (\frac{16r^5}{15} </math>		<math> I_{sphere} = \frac{m}{\frac{8}{3} r^3} (\frac{16r^5}{15}) </math>

	and finally, we arrive at:		and finally, we arrive at:

Seisner6: /* Examples */

2019-06-25T20:56:40Z

Examples

← Older revision		Revision as of 16:56, 25 June 2019
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	We then need to sum together the rotational inertia contributions for each of these spheres. Treating, the mass of each infintesimal disk as some <math> dm </math> we have:		We then need to sum together the rotational inertia contributions for each of these spheres. Treating, the mass of each infintesimal disk as some <math> dm </math> we have:

	<math> I_{sphere} = \int_{-r}^{r~~} \frac{1}{2~~}R_{disk}^2 dm </math>		<math> I_{sphere} = \frac{1}{2}\int_{-r}^{r} R_{disk}^2 dm </math>

	Now, we can express <math> dm </math> in terms of the mass and radius of the sphere by using the mass density of the sphere:		Now, we can express <math> dm </math> in terms of the mass and radius of the sphere by using the mass density of the sphere:
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	So therefore:		So therefore:

	<math> dm = \rho_{sphere}\pi(r^2 - z^2) dz		<math> dm = \rho_{sphere}\pi(r^2 - z^2) dz </math>

	We also know that since the sphere has uniform mass density, we have:		We also know that since the sphere has uniform mass density, we have:

Seisner6: /* Examples */

2019-06-25T20:54:33Z

Examples

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 For two masses, <math> I = m_1r^2 {_\perp,_1} + m_2r^2 {_\perp,_2}</math>. The distance between masses is d, so the distance of each object from the center of mass is <math>r{_\perp,1} = r{_\perp,2} = (d/2)</math>. Therefore
 :<math>I = M(d/2)^2 + M(d/2)^2 = 2M(d/2)^2</math>
 :<math>I = 2 \cdot (2.3 \times 10^{-26}  kg)(0.75 \times 10^{-10}  m)^2 </math>
 :<math>I = 2.6 \times 10^{-46} kg \cdot m^2</math>
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 Imagine a 0.002 kg ladybug resting on the edge of a spinning disk. If the mass of the disk is 2 kg and it has a radius of 0.15m, what is the combined rotational inertia of the disk and the ladybug about the center of the disk (hint: treat the ladybug as a point mass)?
-'''Solution'''
+'''''Solution'''''
 In order to solve this problem, we can write out the combined rotational inertia in terms of it's constituent parts:
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 '''Difficult'''
-Solve the Riemann Hypothesis
+Derive the equation for the rotational inertia of a sphere of mass <math> m </math> and radius <math> r </math> about one of its internal axes. Do this by using the definition of rotational inertia for continuous masses. (Also solve the Riemann Hypothesis)
 == See also ==

Seisner6: /* Examples */

2019-06-25T20:20:44Z

Examples

← Older revision		Revision as of 16:20, 25 June 2019
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	So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.		So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.



			'''Difficult'''

			Solve the Riemann Hypothesis

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

Seisner6: /* Examples */

2019-06-25T20:19:46Z

Examples

← Older revision		Revision as of 16:19, 25 June 2019
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	<math> I_L = 4.5 x 10^{-5} kg \cdot m^2 </math>		<math> I_L = 4.5 x 10^{-5} kg \cdot m^2 </math>


	Next we calculate the rotational inertia of the disk. As this is an often used object, we know that the rotational inertia of a disk is given by:		Next we calculate the rotational inertia of the disk. As this is an often used object, we know that the rotational inertia of a disk is given by:
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	<math> I_{disk} = 2.25 x 10^{-2} kg \cdot m^2 </math>		<math> I_{disk} = 2.25 x 10^{-2} kg \cdot m^2 </math>


	Finally, we have:		Finally, we have:


	<math> I_{tot} = 2.2545 x 10^{-2} kg \cdot m^2 </math>		<math> I_{tot} = 2.2545 x 10^{-2} kg \cdot m^2 </math>


	So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.		So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.

Seisner6: /* Examples */

2019-06-25T20:16:32Z

Examples

← Older revision		Revision as of 16:16, 25 June 2019
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	==Examples==		==Examples==
	'''Simple'''		'''Simple'''

	What is the moment of inertia of a diatomic nitrogen molecule <math>N_2</math> around its center of mass? The mass of a nitrogen atom is <math>2.3 \times 10^{-26} kg</math> and the average distance between nuclei is <math>1.5 \times 10^{-10} m.</math> Use the definition of moment of inertia carefully.		What is the moment of inertia of a diatomic nitrogen molecule <math>N_2</math> around its center of mass? The mass of a nitrogen atom is <math>2.3 \times 10^{-26} kg</math> and the average distance between nuclei is <math>1.5 \times 10^{-10} m.</math> Use the definition of moment of inertia carefully.

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	'''Medium'''		'''Medium'''

	Imagine a 0.002 kg ladybug resting on the edge of a spinning disk. If the mass of the disk is 2 kg and it has a radius of 0.15m, what is the combined rotational inertia of the disk and the ladybug about the center of the disk (hint: treat the ladybug as a point mass)?		Imagine a 0.002 kg ladybug resting on the edge of a spinning disk. If the mass of the disk is 2 kg and it has a radius of 0.15m, what is the combined rotational inertia of the disk and the ladybug about the center of the disk (hint: treat the ladybug as a point mass)?

	'''Solution'''		'''Solution'''

	In order to solve this problem, we can write out the combined rotational inertia in terms of it's constituent parts:		In order to solve this problem, we can write out the combined rotational inertia in terms of it's constituent parts:

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	So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.		So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.



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

Seisner6: /* Examples */

2019-06-25T20:15:52Z

Examples

← Older revision		Revision as of 16:15, 25 June 2019
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	:<math>I = 2 \cdot (2.3 \times 10^{-26} kg)(0.75 \times 10^{-10} m)^2 </math>		:<math>I = 2 \cdot (2.3 \times 10^{-26} kg)(0.75 \times 10^{-10} m)^2 </math>
	:<math>I = 2.6 \times 10^{-46} kg \cdot m^2</math>		:<math>I = 2.6 \times 10^{-46} kg \cdot m^2</math>



	'''Medium'''		'''Medium'''
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	Finally, we have:		Finally, we have:

	<math> I_{tot} = 2.2545 x 10^{-2} kg \cdot m^2		<math> I_{tot} = 2.2545 x 10^{-2} kg \cdot m^2 </math>

	So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.		So, we see that in this case the added rotational inertia of the ladybug is negligible compared to the rotational inertia of the disk.

Seisner6 at 20:14, 25 June 2019

2019-06-25T20:14:51Z

@@ Line 89: / Line 89: @@
 ==Examples==
-'''1. The Moment of Inertia of a Diatomic Molecule'''
+'''Simple'''
 What is the moment of inertia of a diatomic nitrogen molecule <math>N_2</math> around its center of mass? The mass of a nitrogen atom is <math>2.3 \times 10^{-26}  kg</math> and the average distance between nuclei is <math>1.5 \times 10^{-10} m.</math> Use the definition of moment of inertia carefully.
-:::'''''Solution'''''
+'''''Solution'''''
-:::::::For two masses, <math> I = m_1r^2 {_\perp,_1} + m_2r^2 {_\perp,_2}</math>. The distance between masses is d, so the distance of each object from the center of mass is <math>r{_\perp,1} = r{_\perp,2} = (d/2)</math>. Therefore
-::::::::<math>I = M(d/2)^2 + M(d/2)^2 = 2M(d/2)^2</math>
+For two masses, <math> I = m_1r^2 {_\perp,_1} + m_2r^2 {_\perp,_2}</math>. The distance between masses is d, so the distance of each object from the center of mass is <math>r{_\perp,1} = r{_\perp,2} = (d/2)</math>. Therefore
-::::::::<math>I = 2 \cdot (2.3 \times 10^{-26}  kg)(0.75 \times 10^{-10}  m)^2 </math>
+:<math>I = M(d/2)^2 + M(d/2)^2 = 2M(d/2)^2</math>
-::::::::<math>I = 2.6 \times 10^{-46} kg \cdot m^2</math>
+:<math>I = 2 \cdot (2.3 \times 10^{-26}  kg)(0.75 \times 10^{-10}  m)^2 </math>
 == See also ==

Seisner6: /* Main Idea */

2019-06-25T00:20:36Z

Main Idea

← Older revision		Revision as of 20:20, 24 June 2019
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	Finally, by noting that <math> \omega = \frac{v}{r} </math>, we can show:		Finally, by noting that <math> \omega = \frac{v}{r} </math>, we can show:

	<math> \vec{L} = mr^2\vec{omega} = mr^2\frac{v}{r} = mvr\hat{z} </math>		<math> \vec{L} = mr^2\vec{\omega} = mr^2\frac{v}{r} = mvr\hat{z} </math>

	In doing this, we have shown that the "spin" formulation of angular momentum often used for continuous masses is simply a reformulation of the translational formulation which is used to describe point particles.		In doing this, we have shown that the "spin" formulation of angular momentum often used for continuous masses is simply a reformulation of the translational formulation which is used to describe point particles.
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	<math> I = \iiint_V \rho(x, y, z) r^2 dV </math>		<math> I = \iiint_V \rho(x, y, z) r^2 dV </math>


	==Calculating Moment of Inertia==		==Calculating Moment of Inertia==