Moment of inertia of hollow cone can be determined using the following expression;
I = MR2 / 2
Here students will learn and understand how the formula is derived as well as its application.
Moment Of Inertia Of Hollow Cone Formula Derivation
For deriving the moment of inertia formula of a hollow cone we can basically follow some general guidelines which are;
- Determining the mass density in terms of per unit area.
- Defining parameters and finding the mass of small parts.
- Finding the moment of inertia of the small parts corresponding to a ring.
1. There is a hollow cone with radius R, height H and mass M.
Now, we will take the element disc at a slant height l and | radius r having thickness dl and mass dm.
Using the similarity of the triangle we get;
r / x = R / √ R2 + H2
r = R / R / √ R2 + H2 X x
The mass of the elemental disc mass is given by;
dm = M / π R√ R2 + H2 X 2 π rdx
dm = M / π R√ R2 + H2 X 2 X R / R√ R2 + H2 X xdx
dm = 2Mxdx / R2 + H2
2. Calculating the moment of inertia of the elemental disk. It will be given as;
dI = r2 dm
dI = 2Mxdx / R2 + H2 X R2 / R2 + H2X x2
3. Finding the moment of inertia through integration.
I = ∫ 2Mxdx / R2 + H2 X R2 / R2 + H2X x2dx
I = 2MR2 / (R2 + H2)2 ∫ x3dx
Using the limits of x where x usually varies from o to √ R2 + H2
We now get;
I = 2MR2 / (R2 + H2)2 X (R2 + H2)2 / 4
I = ½ MR2
Therefore, I = MR2 / 2