Show that the moment of inertia of a tetrahedron moving about one of its exes is

ID: 1290465 • Letter: S

Question

Show that the moment of inertia of a tetrahedron moving about one of its exes is: I= Ma^2/20 where a is the side of the tetrahedron. Hint: In order to solve this problem first find the moment of inertia of an equilateral triangle. Show that it is: I triangle = sigma l^4/16 square root of 3 where l is the side of the triangle. It might be easier to first calculate the moment of inertia with respect to the top corner and then use the parallel axes theorem. Then, in order to find I of a tetrahedron sum over triangle slices along the z axis. Note that l for each slice varies linearly with z at a rate that is related to the height of the tetrahedron. Another hint: For a tetrahedron of side a the height is square root of 2/3 a. For a triangle of side l the center of mass is at height t/2 square root of 3 and the height is l square root of 3/2. The area of the triangle is l^2 square root of 3 /4 and the volume of the tetrahedron is a3 /6 square root of 2.

Explanation / Answer

Let s use the formula of moment of inertia of a triangular triangular homogenous prism

We have I = Ma^2/12 -----(1)

Also we have for a prism the product of inertia of atraingular prism wrt xz andyz plabe is

Ixy = ?a^4/24*(h) -------------- (2)

Let us take a traingular element of thickness dz and then claculte dIxy in terms of size and mass of the lement.

Let the density of the material is ?

We have dm = ?dV

= ?[1/2(a-z)(a-z)]dZ

So , m = ?/2 ??a*(a^2-2az+z^2)dZ = ?a^3/6 ---- (3)

Substituting the the present parameters in eqn (2)

We have dIxy = ?dZ/24(a-z)^4

Therefore

Ixy = ?/24 ??a*(a^4-4a^3z+6z^2a^2-4az^3+z^4)

= ?a^5/120

Substituting the value of ? from eqn (3) ie ?= 6m/a^3

We have Ixy = ma^2/20 Hence proved