Moment of inertia of a cone -- Why do we divide the equation dI = dm*r^2 by two?

[LCSphysicist]

Homework Statement

Moment of inertia of a cone

Relevant Equations

I = dmr2/2

Why we divide the equation dI = dm*r^2 by two when we calcule the moment of inertia of a cone relative to its symmetry axis?

 

[PeroK]

What's the moment of inertia of a disk?

 

[LCSphysicist]

What's the moment of inertia of a disk?

mr^2/2

but i tried a approach different, i will post:

 

[LCSphysicist]

I can't see where i am wrong :C

 

[haruspex]

View attachment 261891
I can't see where i am wrong :C

You used mr² instead of half that for the MoI of your disc element.

 

[archaic]

The equation $dI=r^2dm$ is the general one, where $r$ represents the perpendicular distance of a point to the axis of rotation. In this exercise, however, we notice that we can divide our cone into disks, find the moment of inertia of each disk, and then sum all of them, hence for each disk of thickness $dx\equiv\Delta x\to0$, its contribution to the total moment of inertia is $dI=\frac12r^2(x)dm$, where $r(x)$ represents the radius of a given disk.

 

[LCSphysicist]

I see, i tried to calc dividing the cone in disk, but i haven't see that i was summing the inertia moment of elemntary disks, i continued the sum like the disk were little points.