Moment of inertia of a cylinder

[gboff21]

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
$$\rho$$=density

To get to the correct I=1/2mr². you need to make dv=2$$\pi$$rh dr
why isn't it dv= dv=$$\pi$$r²h
as in volume=cross-sectional area*height

 

[Delphi51]

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.

 

[gboff21]

Thanks I've been struggling to get my head round that for a while!

 

[Delphi51]

Most welcome. Always draw a picture!

 

[rwisz]

?

Your derivation of the moment of inertia of a solid cylinder is correct. The reason why dv is not equal to \pi r^2 h is because dv represents an infinitesimal volume element, while \pi r^2 h represents the total volume of the cylinder. In other words, dv is a very small volume element, while \pi r^2 h is the total volume that is made up of many dv elements. When calculating the moment of inertia, we need to consider the distribution of mass throughout the entire volume, which is why we use dv instead of \pi r^2 h.