ランダウ・リフシッツ力学 §32 問題 2

ランダウ・リフシッツの力学(増補第3版) §32 問題2 のメモです。

a)

\begin{align} I_1 & = \sum mz^2 \\ & = \int_{-l/2}^{l/2} \rho z^2 dz \\ & = \cfrac{\mu}{l} \left. \cfrac{z^3}{3}\right|_{-l/2}^{l/2} \\ & = \cfrac{\mu}{l} \cfrac{2}{3} \cfrac{l^3}{8} \\ & = \cfrac{\mu l^2}{12} \\ & = I_2. \end{align}

$x_3$に関しては

\[ l_3 = \sum m(x^2 + y^2) = 0. \]

b)

\begin{align} I_1 + I_2 + I_3 & = 2\rho \int \left(x^2 + y^2 + z^2\right) dV \\ & = 2 \rho \int_0^{\pi}\int_0^{2\pi} \int_0^R r^2 r^2 dr d\varphi \sin\theta d\theta \\ & = 2 \rho \times 2\times 2\pi \cfrac{R^5}{5} \\ & = 2 \cfrac{\mu}{\cfrac{4\pi R^3}{3}} \times 2\times 2\pi \cfrac{R^5}{5} \\ & = \cfrac{6}{5} \mu R^2 \\ \end{align}

これから

\[ I_1 = I_2 = I_3 = \cfrac{2}{5}\mu R^2. \]

c)

$I_3$は簡単に計算できます:

\begin{align} I_3 & = \sum m(x^2 + y^2) \\ & = \rho \int_0^{2\pi} \int_{-h/2}^{h/2} \int_0^R r^2 rdr dz d\theta \\ & = \cfrac{\mu}{\pi R^2 h} 2 \pi h \cfrac{R^4}{4} \\ & = \cfrac{1}{2}\mu R^2 . \end{align} \begin{align} I_1 + I_2 & = \sum m\left(x^2 + y^2 + 2z^2\right) \\ \end{align}

まずは$ \sum m \left(x^2 + y^2 \right)$を計算します:

\begin{align} \sum m \left(x^2 + y^2 \right) & = \rho \int_0^{2\pi} \int_{-h/2}^{h/2} \int_0^R r^2 r dr dz d\theta \\ & = \cfrac{\mu}{\pi R^2 h} 2\pi h \cfrac{R^4}{4} \\ & = \cfrac{1}{2} \mu R^2\\ \end{align} \begin{align} \sum m z^2 & = \rho \int_0^{2\pi} \int_0^R \int_{-h/2}^{h/2} z^2 dz r dr d\theta \\ & = \cfrac{\mu}{\pi R^2 h} 2\pi \cfrac{R^2}{2} 2 \cfrac{1}{3} \left(\cfrac{h}{2}\right)^3 \\ & = \cfrac{1}{12}\mu h^2 \end{align}

これから

\[ I_1 = I_2 = \cfrac{1}{4} \mu R^2 + \cfrac{1}{12}\mu h^2 = \cfrac{1}{4}\mu \left(R^2 + \cfrac{1}{3} h^2\right) \]

$I_1$, $I_2$をまともに計算することもできます:

\begin{align} I_1 & = \sum m\left( y^2 + z^2\right) \\ \end{align} \begin{align} \sum m y^2 & = \rho \int_0^{2\pi} \int_0^R \int_{-h/2}^{h/2} r^2\sin^2\theta dzr dr d\theta \\ & = \cfrac{\mu}{\pi R^2 h} \cfrac{R^4}{4} h\int^{2\pi}_{0} \cfrac{1-\cos2\theta}{2} d\theta\\ & = \cfrac{\mu}{\pi R^2 h} \cfrac{R^4}{4} h\cfrac{2\pi}{2}\\ & = \cfrac{1}{4} \mu R^2 \end{align}

これと上の$\sum m z^2$の結果から

\[ I_1 = \cfrac{1}{4} \mu R^2+ \cfrac{1}{12}\mu h^2 . \]

$I_2$の場合は上の積分で$\sin\theta$を$\cos\theta$に変えます。積分は同様にできて同じ結果になります:

\[ I_2 = \cfrac{1}{4} \mu R^2+ \cfrac{1}{12}\mu h^2 . \]

d)

\begin{align} I_1 & = \sum m \left( y^2 + z^2 \right) \\ & = \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} \left( y^2 + z^2 \right) dz dy dx \\ & = \cfrac{\mu}{a b c} a \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} \left( y^2 + z^2 \right) dz dy \\ & = \cfrac{\mu}{a b c} a \left( 2 c \cfrac{1}{3} \left(\cfrac{b}{2}\right)^3 + 2 b \cfrac{1}{3} \left(\cfrac{c}{2}\right)^3 \right) \\ & = \cfrac{\mu}{b c} \left( \cfrac{b^3c}{12} + \cfrac{bc^3}{12} \right) \\ & = \cfrac{\mu}{12} \left( b^2 + c^2 \right) \\ \end{align} \begin{align} I_2 & = \sum m \left( z^2 + x^2 \right) \\ & = \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} \left( z^2 + x^2 \right) dz dy dx \\ & = \cfrac{\mu}{a b c} b \int_{-a/2}^{a/2} \int_{-c/2}^{c/2} \left( z^2 + x^2 \right) dz dx \\ & = \cfrac{\mu}{a b c} b \left( 2 a \cfrac{1}{3} \left(\cfrac{c}{2}\right)^3 + 2 c \cfrac{1}{3} \left(\cfrac{a}{2}\right)^3 \right) \\ & = \cfrac{\mu}{a c} \left( \cfrac{ac^3}{12} + \cfrac{a^3c}{12} \right) \\ & = \cfrac{\mu}{12} \left( c^2 + a^2 \right) \\ \end{align} \begin{align} I_3 & = \sum m \left( x^2 + y^2 \right) \\ & = \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} \left( x^2 + y^2 \right) dz dy dx \\ & = \cfrac{\mu}{a b c} c \int_{-a/2}^{a/2} \int_{-c/2}^{c/2} \left( x^2 + y^2 \right) dx dy \\ & = \cfrac{\mu}{a b c} c \left( 2 b \cfrac{1}{3} \left(\cfrac{a}{2}\right)^3 + 2 a \cfrac{1}{3} \left(\cfrac{b}{2}\right)^3 \right) \\ & = \cfrac{\mu}{a b} \left( \cfrac{a^3b}{12} + \cfrac{ab^3}{12} \right) \\ & = \cfrac{\mu}{12} \left( a^2 + b^2 \right) \\ \end{align} \section{e)}

e)

まずは原点を円錐頂点にした場合を計算します。

ここの積分は円錐の体積を円柱座標系で真面目に計算してみたの円錐の計算を使います。

\begin{align} I^\prime_3 & = \sum m \left( x^2 + y^2 \right) \\ & = \rho \int_\mathrm{cone} \left( x^2 + y^2 \right) dV\\ & = \cfrac{\mu}{\cfrac{\pi R^2 h}{3}} \int_\mathrm{cone} r^2 dV\\ & = \cfrac{\mu}{\cfrac{\pi R^2 h}{3}} \int_0^h \int_0^{2\pi} \int_0^z \left(\cfrac{R}{h} z^\prime \right)^2 \left(\cfrac{R}{h}\right)^2 z^\prime dz^\prime d\theta dz \\ & = \cfrac{\mu}{\cfrac{\pi R^2 h}{3}} \cfrac{R^4}{h^4} 2\pi \int_0^h \cfrac{z^4}{4} dz \\ & = \cfrac{3 \mu}{ R^2 h } \cfrac{R^4}{h^4} \cfrac{1}{2} \cfrac{h^5}{5} \\ & = \cfrac{3 \mu R^2}{ 10 } . \end{align} \begin{align} I^\prime_1 & = \sum m \left(y^2 + z^2\right) \\ & = \rho \int_\mathrm{cone} \left( y^2 + z^2 \right) dV\\ & = \cfrac{\mu}{\cfrac{\pi R^2 h}{3}} \int_\mathrm{cone} \left( r^2 \sin^2\theta + z^2 \right) dV\\ & = \cfrac{3\mu}{\pi R^2 h} \int_0^h \int_0^{2\pi} \int_0^z \left( \left(\cfrac{R}{h} z^\prime \right)^2 \sin^2\theta + z^2 \right) \left(\cfrac{R}{h}\right)^2 z^\prime dz^\prime d\theta dz \\ & = \cfrac{3\mu}{\pi R^2 h} \int_0^h \int_0^{2\pi} \int_0^z \left( \left(\cfrac{R}{h} z^\prime \right)^2 \sin^2\theta + z^2 \right) \left(\cfrac{R}{h}\right)^2 z^\prime dz^\prime d\theta dz \\ & = \cfrac{3\mu}{\pi R^2 h} \cfrac{R^2}{h^2} \int_0^h \int_0^{2\pi} \left( \cfrac{R^2}{h^2} \cfrac{z^4}{4} \sin^2\theta + z^2 \cfrac{z^2}{2} \right) d\theta dz \\ & = \cfrac{3\mu}{\pi h^3} \int_0^h \left( \cfrac{R^2}{h^2} \cfrac{z^4}{4} \pi + z^2 \cfrac{z^2}{2} 2\pi \right) dz \\ & = \cfrac{3\mu}{\pi h^3} \left( \cfrac{R^2}{h^2} \cfrac{1}{4}\cfrac{h^5}{5} \pi + \pi \cfrac{h^5}{5} \right) \\ & = \cfrac{3\mu}{5} \left( \cfrac{R^2}{4} + h^2 \right) . \end{align} $I^\prime_2$の計算は上の計算で$\sin\theta$を$\cos \theta$に代えるだけで結果は同じです: \[ I^\prime_2 = \cfrac{3\mu}{5} \left( \cfrac{R^2}{4} + h^2 \right) . \]

次に重心を原点にしたときの計算をします。

重心$Z_0$は(対称性から$x_3$軸のみ考えます)

\begin{align} Z_0 &= \cfrac{\sum m z}{\mu}\\ & = \cfrac{\rho}{\mu} \int_\mathrm{cone} z dV\\ & = \cfrac{\mu}{\cfrac{\pi R^2 h}{3}} \cfrac{1}{\mu} \int_\mathrm{cone} z dV\\ & = \cfrac{3}{\pi R^2 h} \int_0^h \int_0^{2\pi} \int_0^z z \left(\cfrac{R}{h}\right)^2 z^\prime dz^\prime d\theta dz \\ & = \cfrac{3}{\pi R^2 h} \left(\cfrac{R}{h}\right)^2 \int_0^h \int_0^{2\pi} \int_0^z z z^\prime dz^\prime d\theta dz \\ & = \cfrac{3}{\pi R^2 h} \cfrac{R^2}{h^2} \int_0^h \int_0^{2\pi} \left( z \cfrac{z^2}{2} \right) d\theta dz \\ & = \cfrac{3}{\pi h^3} 2\pi \left( \cfrac{1}{2} \cfrac{h^4}{4} \right) \\ & = \cfrac{3}{4} h. \end{align}

式(32.12)を使って計算します:

\begin{align} I_1 & = I^\prime_1 – \mu a^2 = \cfrac{3\mu}{5} \left( \cfrac{R^2}{4} + h^2 \right) -\mu \cfrac{9}{16} h^2 \\ & = \cfrac{3\mu}{5} \cfrac{R^2}{4} + \mu \cfrac{3}{80} h^2 \\ & = \cfrac{3\mu}{20} \left( R^2 + \cfrac{h^2}{4} \right) \\ & = I_2. \end{align} \[ I_3 = I^\prime_3 = \cfrac{3 \mu R^2}{ 10 } . \]

f)

\begin{align} I_1 & = \sum m (y^2 + z^2 ) \\ & = \rho \int (y^2 + z^2 ) dV \\ & = \rho \int \int \int (y^2 + z^2 ) dx dy dz \\ \end{align}

$x = a\xi, y = b\eta, z = c\zeta$と置くと

\begin{align} I_1 & = \rho abc \int \int \int (y^2 + z^2 ) d\xi d\eta d\zeta \\ & = \rho abc \int \int \int (b^2 \eta^2 + c^2\zeta^2 ) d\xi d\eta d\zeta \\ \end{align}

単位半径の球の慣性モーメント$I^\prime$は

\begin{align} I^\prime & = \rho \int \int \int (y^2 + z^2) dx dy dz \end{align}

対称性から

\begin{align} \cfrac{I^\prime}{2} & = \rho \int \int \int x^2 dx dy dz \\ & = \rho \int \int \int y^2 dx dy dz\\ & = \rho \int \int \int z^2 dx dy dz \end{align}

$I_1$の計算に戻りましょう。$I^\prime$は問題2 b)で計算済みですね(単位半径の球の重さを$\mu^\prime$と書きます):

\[ I^\prime = \cfrac{2}{5}\mu^\prime =\cfrac{2}{5} \rho \cfrac{4\pi}{3}. \]

これを使って

\begin{align} I_1 & = abc (b^2 + c^2 )\cfrac{1}{2} I^\prime \\ & = abc (b^2 + c^2 )\cfrac{1}{2} \cfrac{2}{5} \rho \cfrac{4\pi}{3} \\ & = \rho \cfrac{4\pi}{3} abc \cfrac{1}{5} (b^2 + c^2 ) \\ & = \mu \cfrac{1}{5} (b^2 + c^2 ). \end{align}

同様に

\begin{align} I_2 & = \sum m (z^2 + x^2 ) \\ & = \rho \int (z^2 + x^2 ) dV \\ & = \rho \int \int \int (z^2 + x^2 ) dx dy dz \\ & = \rho abc \int \int \int (c^2 \zeta^2 + a^2\xi^2 ) d\xi d\eta d\zeta \\ & = \rho abc \cfrac{1}{2} (c^2 + a^2 ) I^\prime \\ & = \rho abc \cfrac{1}{2} (c^2 + a^2 ) \cfrac{2}{5} \rho \cfrac{4\pi}{3} \\ & = \mu \cfrac{1}{5} (c^2 + a^2 ). \end{align} \begin{align} I_3 & = \sum m (x^2 + y^2 ) \\ & = \rho \int (x^2 + y^2 ) dV \\ & = \rho \int \int \int (x^2 + y^2 ) dx dy dz \\ & = \rho abc \int \int \int (a^2\xi^2 + b^2 \eta^2 ) d\xi d\eta d\zeta \\ & = \rho abc \cfrac{1}{2} (a^2 + b^2 ) I^\prime \\ & = \rho abc \cfrac{1}{2} (a^2 + b^2 ) \cfrac{2}{5} \rho \cfrac{4\pi}{3} \\ & = \mu \cfrac{1}{5} (a^2 + b^2 ). \end{align}