Moment Of Inertia Of Cone

View Notes

Introduction: Moment of Inertia

The moment of inertia of a rigid object, commonly known as the mass moment of inertia, angular momentum, or rotational inertia, is a quantity that specifies the torque required by a rotational axis for a desired angular acceleration; analogous to how mass calculates the force needed for the desired acceleration.

Consider a solid uniform cone with mass M, radius R, and height h. Then density is

p= $\frac{M}{\frac{1}{3}πR^2h}$,

And the moment of inertia tensor over the base center is

$\int_{V}p(x,y,z)\begin{bmatrix} y^2 + z^2 & -xy & -xz \\ -xy & z^2 + x^2 & -yz \\ -xz & -yz & x^2 + y^2\end{bmatrix}dxdydz$

$\begin{bmatrix} \frac{1}{10}Mh^2 + \frac{3}{20}MR^2 & 0 & 0 \\ 0 & \frac{1}{10}Mh^2 + \frac{3}{20}MR^2 & 0 \\ 0 & 0 & \frac{3}{10}MR^2 \end{bmatrix}$,

$p\int_{h}^{0}\int_{-R(h-z)/h}^{R(h-z)/h}\int_{-\sqrt{[R(h-z)/h]^2-y^2}}^{\sqrt{[R(h-z)/h]^2-y^2}}\begin{bmatrix} y^2 + z^2 & -xy & -xz \\ -xy & z^2 + x^2 & -yz \\ -xz & -yz & x^2 + y^2\end{bmatrix}dxdydz$

Moment of Inertia of Circular Cone Derivation

Here we'll look at the derivation as well as the approximation to find the moment of inertia about an axis of a standard right circular cone.

We divide the cone into a small elementary disk where we consider the radius of the cone to be r at a distance x. Breakthrough. We'll need to find the disk mass, and it's given as;

dm = M / ⅓ π R2H (πr2dx)

dm = 3 M / π R2H (R2 / H2 X x2) dx

dm = 3Mx2dx / H3

Now if we consider the triangle's similarity we obtain;

r / x = R / H

r = R / H X x

We also need to remember the moment of elementary disk inertia which is;

½ dmr2

Meanwhile,

dI = ½ (3Mx2dx / H3) . R2 / H2 X x2

I = ∫dI = 3/2 MR2 / H5 . H5 / 5 = 3MR2 / 10