Moment of inertia of a cylinder

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Homework Help Overview

The discussion revolves around deriving the moment of inertia of a solid cylinder, specifically addressing the integration process involved in the calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the moment of inertia using the integral form and questions the choice of volume differential, suggesting an alternative approach. Other participants clarify the volume differential and its relation to the cylinder's geometry.

Discussion Status

The discussion is ongoing, with participants providing clarifications and supportive comments. There is an exchange of ideas regarding the correct formulation of the volume differential, but no consensus has been reached on the derivation process.

Contextual Notes

Participants are working within the constraints of deriving the moment of inertia and are exploring different interpretations of the volume differential in the context of the cylinder's geometry.

gboff21
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I have been deriving the moment of inertia of a solid cylinder and have got this far:
I=∫ r2 dm
and dm=dv *density

h=height
r=radius
[tex]\rho[/tex]=density

To get to the correct I=1/2mr2. you need to make dv=2[tex]\pi[/tex]rh dr
why isn't it dv= dv=[tex]\pi[/tex]r2h
as in volume=cross-sectional area*height
 
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Consider a circle on the end of the cylinder at radius r. If it goes all the way through the circle, its area is 2πrh. Let dV be this area times an incremental radius, dr:
dV = 2πrh*dr
Imagine it "unrolled" into a rectangular solid to see it clearly.

Don't forget the density.
 
Thanks I've been struggling to get my head round that for a while!
 
Most welcome. Always draw a picture!
 

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