- #1

andyonassis

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- 1

- Homework Statement
- Find the moment of inertia of frustum perpendicular to axis of symmetry about the top with radius r and bottom with radius R

- Relevant Equations
- I = (3m/10)[(R^5-r^5)/(R^3-r^3)] (about axis of symmetry)

So i derived the moment of inertia about the axis of symmetry (with height h) and I am confused about the perpendicular axis theorem.

The problem ask to find the moment of inertia perpendicular to axis of symmetry

So the axis about h, i labelled as z, the two axis that are perpendicular to z, i labelled as x and y.

I already found $$I_{z} = (3m/10)\frac{(R^5-r^5)}{(R^3-r^3)]}$$

So according to perpendicular axis theorem,

$$I_z = I_x + I_y$$

and because of symmetry $$I_x = I_y$

So i derived a solution for Ix and Iy

$$Ix = I_y = (3m/20)\frac{(R^5-r^5)}{(R^3-r^3)]}$$

I'm confused as the theorem also says $$I_x = I_y+I_z$$ and $$I_y = I_x+I_z$$

The problem ask to find the moment of inertia perpendicular to axis of symmetry

So the axis about h, i labelled as z, the two axis that are perpendicular to z, i labelled as x and y.

I already found $$I_{z} = (3m/10)\frac{(R^5-r^5)}{(R^3-r^3)]}$$

So according to perpendicular axis theorem,

$$I_z = I_x + I_y$$

and because of symmetry $$I_x = I_y$

So i derived a solution for Ix and Iy

$$Ix = I_y = (3m/20)\frac{(R^5-r^5)}{(R^3-r^3)]}$$

I'm confused as the theorem also says $$I_x = I_y+I_z$$ and $$I_y = I_x+I_z$$

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