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Author Topic: Moment of inertia of a cube  (Read 13045 times)
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« on: November 29, 2005, 12:35:34 AM »

In case you're interested, this is what we did in class today in finding a moment of inertia of a cube with respect to any axis passing through its center of mass, provided that the mass density,  \varrho , is constant.

Let  \vec{r} be the position vector of any infinitesimal mass form the centre of the cube,  \hat{u} be a unit vector pointing along an arbitrary axis, and  \rho be the distance from the axis to the infinitesimal mass.

 \begin{array}{rcl} \vec{r} &=& x \hat{i} + y \hat{j} + z \hat{k} \\[10pt] \hat{u} &=& \dfrac{a \hat{i} + b \hat{j} + c \hat{k}}{\sqrt{a^2 + b^2 + c^2}} \\[10pt] I &=& \int \varrho \; \mathrm{d} V \; \rho^2 \end{array}

 \begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left[ r^2 - (\vec{r} \cdot \hat{u})^2 \right] \\[10pt] &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left[ x^2 + y^2 + z^2 - \left( \frac{xa}{\sqrt{a^2 + b^2 + c^2}} + \frac{yb}{\sqrt{a^2 + b^2 + c^2}} + \frac{zc}{\sqrt{a^2 + b^2 + c^2}} \right)^2 \right] \\[10pt] &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left \{ x^2 + y^2 + z^2 - \frac{1}{a^2 + b^2 + c^2} \cdot \right. \\[10pt] && \qquad \qquad \left. \left[ x^2 a^2 + y^2 b^2 + z^2 c^2 + 2xyab + 2yzbc + 2xzac \right] \right \} \end{array}

It is clear that the integrals of the cross terms vanish.

 \begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \varrho \left[ \int_{- \frac{L}{2}}^{\frac{L}{2}} x^2 \left( 1 - \frac{a^2}{a^2 + b^2 + c^2} \right) \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} y \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} z \right. \\[10pt] && \left. + \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} y^2 \left( 1 - \frac{b^2}{b^2 + b^2 + c^2} \right) \mathrm{d} y \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} z \right. \\[10pt] && \left. + \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} y \int z^2 \left( 1 - \frac{c^2}{a^2 + b^2 + c^2} \right) \mathrm{d} z \right] \\[10pt] &=& \varrho \left[ \frac{2}{3} \frac{L^3}{8} L^2 \right] \left[ \left( 1 - \frac{a^2}{a^2 + b^2 + c^2} \right) + \left( 1 - \frac{b^2}{b^2 + b^2 + c^2} \right) + \left( 1 - \frac{c^2}{a^2 + b^2 + c^2} \right) \right] \end{array}

 \begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \frac{1}{6} \varrho L^5 \\[10pt] &=& \frac{1}{6} \frac{M}{L^3} L^5 \\[10pt] &=& \frac{1}{6} M L^2 \end{array}

Notice that this is independent of an axis chosen.

Wink
« Last Edit: October 19, 2009, 07:20:46 AM by ปิยพงษ์ - Head Admin » Logged

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« Reply #1 on: November 29, 2005, 07:30:05 AM »

Thank you.  Cheesy
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« Reply #2 on: November 29, 2005, 07:04:51 PM »

You're welcome. Cheesy
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« Reply #3 on: December 02, 2005, 08:17:28 PM »

..
It is clear that the integrals of the cross terms vanish.
..
เพราะว่าพออินติเกรตจะออกมาเป็นกำลัง2แล้วพอใส่ลิมิตแล้วมันจะลบกันหมดใช่หรือเปล่าครับ
« Last Edit: December 02, 2005, 08:20:49 PM by ccchhhaaammmppp » Logged

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« Reply #4 on: December 02, 2005, 09:02:22 PM »

ถูกต้องครับ Wink

สังเกตว่าไม่ว่าจะเลือก orientation ของแกน X, Y, Z แบบไหน อินทิกรัลเหล่านี้ก็ยังหายไปอยู่ดีด้วยครับ
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« Reply #5 on: December 06, 2005, 09:35:12 PM »

Some interesting remark from Marion/Thornton's.

Quote
In this regard, the cube is similar to a sphere as far as the inertia tensor is concerned (i.e., for an origin at the center of mass, the structure of the inertia tensor elements is not sufficiently detailed to discriminate between a cube and a sphere).
-- Thornton, Stephen T., Marion, Jerry B. Classical Dynamics of Particles and Systems, p. 432.
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