# Moment of inertia for a rhombus

OlderDan
Homework Helper
Hey I just did both integrals (mine and yours) and I get different answers. I have checked them and the same. Mine gives I = (8-3sqrt2)mc^2/(12sqrt2)
Yours is giving me a negative number which seems illogical. What is happening?
They are indeed the same.

$$I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c\sqrt 2 - x} {dydx} }$$

$$I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 dx} \left( {c\sqrt 2 - 2x} \right)$$

$$I = 2\sigma \int_0^{c/\sqrt 2 } {\left( {c\sqrt 2 x^2 - 2x^3 } \right)dx}$$

$$I = 2\sigma \left( {c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2}} \right) = 2\sigma \left( {\frac{2}{3} - \frac{1}{2}} \right)\left( {c/\sqrt 2 } \right)^4$$

$$I = 2\sigma \left( { - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2} + c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3}} \right) = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \frac{M}{{12}}c^2$$

My way:

$$I = 4\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c/\sqrt 2 } {dydx} }$$

$$I = 4\sigma \int_0^{c/\sqrt 2 } {\left( {c/\sqrt 2 x^2 - x^3 } \right)dx}$$

$$I = 4\sigma \left( {\frac{{\left( {c/\sqrt 2 } \right)^4 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{4}} \right) = 4\sigma \left( {\frac{1}{3} - \frac{1}{4}} \right)\left( {c/\sqrt 2 } \right)^4 = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \left( {\frac{M}{{12}}} \right)c^2$$

Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.

OlderDan
Homework Helper
Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.
You're welcome. Anything about that result look familiar?

It is identical to the case of a rod with the axis of rotation through the center. Other rotational lines for the same square also give similar rotational inertias to other cases of axis for a rod. I can speculate why this happens but I'm not sure. Why is it that this happens?

sorry the comparison should have used at least a rectangle (more general), not a square. However, the results for the rectangle are identical to those of a rod.

OlderDan