3.3.O

Gegeben

Es geht um einen homogen gefüllten Kreiszylinder.

Symbole:

\[r, a, m\]

Der Kreiszylinder hat den Radius \(r\) und die Masse \(m\).
Der Trägheitstensor dieses Körper bezüglich O wird beschrieben über seine \((x,y,z)\)-Komponenten.

a) xx-Komponente

Für die gegebenen Symbole: Berechnen Sie die \(xx\)-Komponente des Trägheitstensors bzgl. O:

\[\Theta_{xx} = \int \rho \left( y^2 + z^2 \right) {\mathsf d}V\]

Verwenden Sie Zylinderkoordinaten \((\tilde r,\varphi, z)\) definiert wie üblich als:

\[\begin{split}(x, y, z) &= (\tilde r \cos\varphi, \tilde r \sin\varphi, z) \\ {\mathsf d}V = {\mathsf d}x\,{\mathsf d}y\,{\mathsf d}z &= \tilde r \, {\mathsf d}\tilde r\,{\mathsf d}\varphi\, {\mathsf d}z\end{split}\]

Hinweis

Verwenden Sie folgendes Integral mit Integrationskonstante \(C\):

\[\int\sin^2\!\varphi\,d\varphi = \tfrac{\varphi}{2} - \tfrac{1}{4} \sin (2\varphi) + C\]

Zeigen Sie, dass:

\[\Theta_{xx} = m\left( \tfrac{r^{2}}{4} + \tfrac{a^{2}}{3}\right )\]

Lösung

\[\begin{split}\Theta_{xx} &= \int \rho \left( y^2 + z^2 \right) {\mathsf d}V \\ \tfrac{\Theta_{xx}}{\rho}&= \underbrace{\int y^2 {\mathsf d}V}_{\Theta_1} + \underbrace{\int z^2 {\mathsf d}V}_{\Theta_2}\end{split}\]

Verwendung von Zylinderkoordinaten:

\[\begin{split}\Theta_1 &= \int_{z=0}^{a}\int_{\varphi=0}^{2\pi}\int_{\tilde r=0}^r \tilde r^2 \sin^2\!\varphi \, \tilde r \,{\mathsf d}\tilde r\,{\mathsf d}\varphi\, {\mathsf d}z \\ &= \tfrac 1 4 r^4 \int_{z=0}^{a}\int_{\varphi=0}^{2\pi} \sin^2\!\varphi \,{\mathsf d}\varphi\, {\mathsf d}z \\ &= \tfrac 1 4 r^4 a \int_{\varphi=0}^{2\pi} \sin^2\!\varphi \,{\mathsf d}\varphi\\ &= \tfrac 1 4 r^4 a \left[ \tfrac{\varphi}{2} - \tfrac{1}{4} \sin (2\varphi) \right]_{\varphi=0}^{2\pi} \\ &= \tfrac 1 4 \pi r^4 a\\ \\ \Theta_2 &= \int_{z=0}^{a}\int_{\varphi=0}^{2\pi}\int_{\tilde r=0}^r z^2 \tilde r {\mathsf d}\tilde r\,{\mathsf d}\varphi\, {\mathsf d}z \\ &= \tfrac13 a^3 \int_{\varphi=0}^{2\pi}\int_{\tilde r=0}^r \tilde r {\mathsf d}\tilde r\,{\mathsf d}\varphi\\ &= \tfrac23 \pi a^3 \int_{\tilde r=0}^r \tilde r {\mathsf d}\tilde r\\ &= \tfrac13 \pi a^3 r^2\\\end{split}\]

\(\Theta_1\) und \(\Theta_2\) einsetzen liefert:

\[\begin{split}\tfrac{\Theta_{xx}}{\rho} &= \tfrac 1 4 \pi r^4 a + \tfrac13 \pi a^3 r^2 \\ \Theta_{xx} &= \rho \tfrac 1 4 \pi r^4 a + \rho \tfrac13 \pi a^3 r^2 \\ &= \tfrac 1 4 r^2 m + \tfrac13 a^2 m \\ &= m\left( \tfrac{r^{2}}{4} + \tfrac{a^{2}}{3}\right )\end{split}\]

b) yy-Komponente

Zeigen Sie, dass:

\[\Theta_{yy} = \int \rho \left( x^2 + z^2 \right) {\mathsf d}V = \Theta_{xx}\]

Lösung

\(\Theta_{yy}=\Theta_{xx}\) gilt aufgrund der Symmetrie.

c) zz-Komponente

Zeigen Sie, dass:

\[\Theta_{zz} = \tfrac 1 2 m r^2\]

Lösung

\[\begin{split}\Theta_{zz} &= \int \rho \left( x^2 + y^2\right) {\mathsf d}V \\ &= \rho \int_{z=0}^{a}\int_{\varphi=0}^{2\pi}\int_{\tilde r=0}^r \tilde r^2 \tilde r {\mathsf d}\tilde r\,{\mathsf d}\varphi\, {\mathsf d}z \\ &= \rho \int_{z=0}^{a}\int_{\varphi=0}^{2\pi}\tfrac 14 r^4 \,{\mathsf d}\varphi\, {\mathsf d}z \\ &= 2 \rho \pi a\tfrac 14 r^4 \\ &= \tfrac 12 \rho \pi r^2 a r^2 \\ &= \tfrac 1 2 m r^2\end{split}\]

d) Komponenten für Größen

Für die gegebenen Größen:

\[\begin{split}r &= 1 \, \mathrm{m} \\ a &= \tfrac{\sqrt{3}}{2} \, \mathrm{m} \\ m &= 4 \, \mathrm{kg}\end{split}\]

Berechnen Sie \(\Theta_{xx}, \Theta_{yy}, \Theta_{zz}\). Und zeigen Sie, dass:

\[\Theta_{xx} = \Theta_{yy} = \Theta_{zz} = 2 \,\mathrm{kg\,m^2}\]

Lösung

Einsetzen von

\[\begin{split}r &= 1 \, \mathrm{m} \\ a &= \tfrac{\sqrt{3}}{2} \, \mathrm{m} \\ m &= 4 \, \mathrm{kg}\end{split}\]

in \(\Theta_{xx}=\Theta_{yy}= m\left( \tfrac{r^{2}}{4} + \tfrac{a^{2}}{3}\right )\) und in \(\Theta_{zz}=\tfrac 1 2 m r^2\) führt auf:

\[\begin{split}\Theta_{xx} = \Theta_{yy} &= 4 \, \mathrm{kg} \left( \tfrac{1}{4} + \tfrac{\tfrac{3}{4}}{3}\right ) \, \mathrm{m}^2 \\ &= 2 \,\mathrm{kg\,m^2} \\ \\ \Theta_{zz} &= \tfrac 1 2 m r^2\\ &= 2 \,\mathrm{kg\,m^2}\end{split}\]

SymPy

Source Code eines Programms dazu:

from sympy.physics.units import kg, m, s
from sympy import pi, S, pprint, symbols, cos, sin, Matrix, Eq, solve, sqrt, exp, simplify
from sympy import integrate

# Units:
N         =  kg*m/s**2
k, M, G   =  10**3, 10**6, 10**9
kN        =  k*N
Pa        =  N/m**2
MPa, GPa  =  M*Pa, G*Pa
deg       =  pi/180
half      =  S(1)/2
cm        =  m/100
mm        =  m/1000

# ---

rho, R, a = symbols("rho, R, a")

# Polar coordinates:
z, phi, r = symbols("z, phi, r")
x = r*cos(phi)
y = r*sin(phi)

# Mass:
M = symbols("M")

sub_list = [
    ( R,         1 *m  ),
    ( a, sqrt(3)/2 *m  ),
    ( M,         4 *kg ),
    ]

pprint("\nΘxx: ")
tmp = (y*y + z*z) *r
tmp = integrate( tmp, (r,   0, R   ) )
tmp = integrate( tmp, (phi, 0, 2*pi) )
tmp = integrate( tmp, (z,   0, a   ) )
tmp = rho*tmp
tmp = tmp.simplify()
tmp = tmp.subs(rho*pi*R*R*a, M)
tmp = tmp.simplify()
tmp = tmp.cancel()
pprint(tmp)

pprint("\nΘxx / (kg m²): ")
tmp = tmp.subs(sub_list)
tmp /= kg*m**2
pprint(tmp)


pprint("\nΘyy: ")
tmp = (x*x + z*z) *r
tmp = integrate( tmp, (r,   0, R   ) )
tmp = integrate( tmp, (phi, 0, 2*pi) )
tmp = integrate( tmp, (z,   0, a   ) )
tmp = rho*tmp
tmp = tmp.simplify()
tmp = tmp.subs(rho*pi*R*R*a, M)
tmp = tmp.simplify()
tmp = tmp.cancel()
pprint(tmp)

pprint("\nΘyy / (kg m²): ")
tmp = tmp.subs(sub_list)
tmp /= kg*m**2
pprint(tmp)

pprint("\nΘzz: ")
tmp = (x*x + y*y) *r
tmp = (x*x + z*z) *r
tmp = integrate( tmp, (r,   0, R   ) )
tmp = integrate( tmp, (phi, 0, 2*pi) )
tmp = integrate( tmp, (z,   0, a   ) )
tmp = rho*tmp
tmp = tmp.simplify()
tmp = tmp.subs(rho*pi*R*R*a, M)
tmp = tmp.simplify()
tmp = tmp.cancel()
pprint(tmp)

pprint("\nΘzz / (kg m²): ")
tmp = tmp.subs(sub_list)
tmp /= kg*m**2
pprint(tmp)

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