円柱の慣性モーメント

図のように回転軸を ${\displaystyle y}$ 軸にとり， ${\displaystyle y}$ 軸と直交する ${\displaystyle zx}$ 平面をとる．円柱の密度を ${\displaystyle \rho }$ とすると

${\displaystyle \rho =\frac{M}{\pi R^{2}L}}$

である．この円柱を ${\displaystyle z}$ 軸方向に微小厚み ${\displaystyle dz}$ でスライスし，この厚み ${\displaystyle dz}$ の円板部分をさらに ${\displaystyle x}$ 軸方向に幅 ${\displaystyle dx}$ で分割する．断面積 ${\displaystyle dxdz}$ の微小角柱の長さは ${\displaystyle 2\sqrt{R^2-x^2}}$ であるので，この部分の質量を ${\displaystyle dm}$ とおくと，

${\displaystyle dm=\rho \cdot 2\sqrt{R^2-x^2}\cdot dxdz}$ ${\displaystyle =\frac{M}{\pi R^{2}L}\cdot 2\sqrt{R^2-x^2}dxdz}$ ${\displaystyle =\frac{2M}{\pi R^{2}L}\sqrt{R^2-x^2}dxdz}$

となる．

${\displaystyle y}$ 軸の先から円柱を見た図

${\displaystyle y}$ 軸の先から円柱を見た図から分かるように，この微小角柱の回転軸までの距離(回転半径)は ${\displaystyle r=\sqrt{x^2+z^2}}$ であるので，微小角柱の回転軸周りの慣性モーメント ${\displaystyle dI}$ は，

${\displaystyle dI=r^{2}dm}$ ${\displaystyle =(x^2+z^2)\cdot \frac{2M}{\pi R^{2}L}\sqrt{R^2-x^2}dxdz}$

と表せる．したがって，求める慣性モーメント ${\displaystyle I}$ は，

${\displaystyle I={\displaystyle {\int }_{D}dI}={\displaystyle {\int }_D{r}^{2}dm}}$ ${\displaystyle ={\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}{\displaystyle {\int }_{-R}^R\left(x^2+z^2\right)\cdot \frac{M}{\pi R^{2}L}\cdot 2\sqrt{R^2-x^2}dx}dz}}$ ${\displaystyle =\frac{2M}{\pi R^{2}L}{\displaystyle {\int }_{-R}^R\sqrt{R^2-x^2}\left\{{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(x^2+z^2\right)dz}\right\}dx}}$

となる．ここで，

${\displaystyle {\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(x^2+z^2\right)dz}=2{\displaystyle {\int }_0^{\frac{L}{2}}\left(x^2+z^2\right)dz}}$ ${\displaystyle =2{\left[x^{2}z+\frac{1}{3}{z}^3\right]}_0^{\frac{L}{2}}}$ ${\displaystyle =2\left\{x^2\cdot \frac{L}{2}+\frac{1}{3}{\left(\frac{L}{2}\right)}^3\right\}}$ ${\displaystyle =L{x}^2+\frac{L^3}{12}}$

より，次式を得る．

${\displaystyle I=\frac{2M}{\pi R^{2}L}{\displaystyle {\int }_{-R}^R\sqrt{R^2-x^2}\left(L{x}^2+\frac{L^3}{12}\right)dx}}$ ${\displaystyle =\frac{2M}{\pi R^2}{\displaystyle {\int }_{-R}^R\sqrt{R^2-x^2}\left(x^2+\frac{L^2}{12}\right)dx}}$

次に， ${\displaystyle x=R\sin \theta }$ とおくような置換積分を考えると， ${\displaystyle \theta }$ についての積分範囲は

となり，形式的に

${\displaystyle \frac{dx}{d\theta }=\frac{d}{d\theta }R\sin \theta =R\cos \theta \Rightarrow dx=R\cos \theta d\theta }$

と書けるので

${\displaystyle {\displaystyle {\int }_{-R}^R\sqrt{R^2-x^2}\left(x^2+\frac{L^2}{12}\right)dx}}$ ${\displaystyle ={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{R^2-R^2{\sin }^2\theta }\cdot \left(R^2{\sin }^2\theta +\frac{L^2}{12}\right)R\cos \theta d\theta }}$ ${\displaystyle =R^4{\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^2\theta \left(1-{\cos }^2\theta \right)d\theta }+\frac{R^2{L}^2}{12}{\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta }}$ ${\displaystyle =R^4\cdot 2{\displaystyle {\int }_0^{\frac{\pi }{2}}\left({\cos }^2\theta -{\cos }^4\theta \right)d\theta }+\frac{R^2{L}^2}{12}\cdot 2{\displaystyle {\int }_0^{\frac{\pi }{2}}{\cos }^2\theta d\theta }}$
↓ ここで，ウォリスの公式を利用
${\displaystyle =2R^4\left(\frac{1}{2}\cdot \frac{\pi }{2}-\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}\right)+\frac{R^2{L}^2}{6}\cdot \frac{1}{2}\cdot \frac{\pi }{2}}$ ${\displaystyle =\frac{\pi R^4}{8}+\frac{\pi R^2{L}^2}{24}}$

を得る．したがって，

${\displaystyle I=\frac{2M}{\pi R^2}\left(\frac{\pi R^2}{8}+\frac{\pi R^2{L}^2}{24}\right)=\frac{M{R}^2}{4}+\frac{M{L}^2}{12}}$

が求まる．

★ 極座標を用いて計算

円柱を ${\displaystyle z}$ 軸方向に微小厚み ${\displaystyle dz}$ でスライスしたときの円板について，極座標を用いて，微小部分に分割すると，その微小体積は

${\displaystyle dV=dSdz=rdrd\theta dz}$

となるので，この部分の微小質量は

${\displaystyle dm=\rho \cdot rdrd\theta dz}$ ${\displaystyle =\frac{M}{\pi R^{2}L}\cdot rdrd\theta dz}$

である．この微小部分の回転軸までの距離（回転半径）は

${\displaystyle u=\sqrt{x^2+z^2}}$ ${\displaystyle =\sqrt{{\left(r\cos \theta \right)}^2+z^2}}$

となるので，微小部分の，回転軸周りの慣性モーメント ${\displaystyle dI}$ は

${\displaystyle dI=u^{2}dm}$ ${\displaystyle =(r^2{\cos }^2\theta +z^2)\cdot \frac{M}{\pi R^{2}L}\cdot rdrd\theta dz}$

と表せる．したがって，求める慣性モーメント ${\displaystyle I}$ は，

${\displaystyle I={\displaystyle {\int }_{V}dI}={\displaystyle {\int }_V{u}^{2}dm}={\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}{\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^R\left(r^2{\cos }^2\theta +z^2\right)\frac{M}{\pi R^{2}L}rdr}d\theta }dz}}$ ${\displaystyle =\frac{M}{\pi R^{2}L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left\{{\displaystyle {\int }_0^{2\pi }\left({\displaystyle {\int }_0^R\left(r^3{\cos }^2\theta +r{z}^2\right)dr}\right)d\theta }\right\}dz}}$ ${\displaystyle =\frac{M}{\pi R^{2}L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left\{{\displaystyle {\int }_0^{2\pi }{\left[\frac{1}{4}{r}^4{\cos }^2\theta +\frac{1}{2}{r}^2{z}^2\right]}_0^{R}d\theta }\right\}dz}}$ ${\displaystyle =\frac{M}{\pi R^{2}L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left\{{\displaystyle {\int }_0^{2\pi }\left(\frac{1}{4}{R}^4{\cos }^2\theta +\frac{1}{2}{R}^2{z}^2\right)}d\theta \right\}dz}}$ ${\displaystyle =\frac{M}{2\pi L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left\{{\displaystyle {\int }_0^{2\pi }\left[\frac{1}{4}{R}^2\left(1+\cos 2\theta \right)+z^2\right]}d\theta \right\}dz}}$ ${\displaystyle =\frac{M}{2\pi L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}{\left[\frac{1}{4}{R}^2\left(\theta +\frac{\sin 2\theta }{2}\right)+\theta z^2\right]}_0^{2\pi }dz}}$ ${\displaystyle =\frac{M}{2\pi L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(\frac{\pi }{2}{R}^2+2\pi z^2\right)dz}}$ ${\displaystyle =\frac{M}{L}{\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(\frac{1}{4}{R}^2+z^2\right)dz}}$ ${\displaystyle =\frac{M}{L}{\left[\frac{1}{4}{R}^{2}z+\frac{1}{3}{z}^3\right]}_{-\frac{L}{2}}^{\frac{L}{2}}}$ ${\displaystyle =\frac{M}{L}\left(\frac{1}{4}{R}^{2}L+\frac{1}{12}{L}^3\right)}$ ${\displaystyle =\frac{M{R}^2}{4}+\frac{M{L}^2}{12}}$

となる．

★ 平行軸の定理を用いて計算

図に示すように，中心が原点となるように ${\displaystyle xy}$ 平面に置かれた厚さ ${\displaystyle dz}$ の一様な円板の， ${\displaystyle y}$ 軸のまわりの慣性モーメント ${\displaystyle d{I}_G}$ は，

${\displaystyle d{I}_G={\displaystyle {\int }_S{\left(r\cos \theta \right)}^2\rho dV}}$ ${\displaystyle ={\displaystyle {\int }_0^{2\pi }{\displaystyle {\int }_0^R{r}^2{\cos }^2\theta \frac{M}{\pi R^{2}L}rdrd\theta }dz}}$ ${\displaystyle =\frac{M}{\pi R^{2}L}{\displaystyle {\int }_0^{2\pi }\left\{{\displaystyle {\int }_0^R{r}^{3}dr}\right\}{\cos }^2\theta d\theta }dz}$ ${\displaystyle =\frac{M}{\pi R^{2}L}{\displaystyle {\int }_0^{2\pi }\left(\frac{1}{4}{R}^4\right){\cos }^2\theta d\theta }dz}$ ${\displaystyle =\frac{M{R}^2}{4\pi L}{\displaystyle {\int }_0^{2\pi }{\cos }^2\theta d\theta }dz}$ ${\displaystyle =\frac{M{R}^2}{4\pi L}{\displaystyle {\int }_0^{2\pi }\left(\frac{1+\cos 2\theta }{2}\right)d\theta }dz}$ ${\displaystyle =\frac{M{R}^2}{4\pi L}{\left[\frac{\theta }{2}+\frac{\sin 2\theta }{4}\right]}_0^{2\pi }dz}$ ${\displaystyle =\frac{M{R}^2}{4L}dz}$

である．原点 ${\displaystyle \text{O}}$ から ${\displaystyle z}$ 軸方向に ${\displaystyle z}$ だけ移動した厚さ ${\displaystyle dz}$ の円板の慣性モーメント ${\displaystyle d{I}_z}$ は，平行軸の定理により，

${\displaystyle d{I}_z=d{I}_G+\frac{M}{\pi R^{2}L}\cdot \pi R^{2}dz\cdot {\left|z\right|}^2}$ ${\displaystyle =\frac{M{R}^2}{4L}dz+\frac{M}{L}{z}^{2}dz}$

であるので，求める慣性モーメント ${\displaystyle I}$ は，

${\displaystyle I={\displaystyle {\int }_{z}d{I}_z}}$ ${\displaystyle ={\displaystyle {\int }_{-\frac{L}{2}}^{\frac{L}{2}}\left(\frac{M{R}^2}{4L}+\frac{M}{L}{z}^2\right)dz}}$ ${\displaystyle ={\left[\frac{M{R}^2}{4L}z+\frac{M}{3L}{z}^3\right]}_{-\frac{L}{2}}^{\frac{L}{2}}}$ ${\displaystyle =\frac{M{R}^2}{4}+\frac{M{L}^2}{12}}$

となる．