4.2. Classical Mechanics

4.2.1. Rigid Body Rotation

In all the sections below, the rigid body is rotating around the \bomega axis, so:

{\bf v} = \bomega \times {\bf r}

Kinetic Energy

The kinetic energy is:

T = \int \half\rho({\bf r}) v^2 \d^3 r = = \int \half\rho({\bf r}) {\bf v}\cdot{\bf v} \d^3 r = = \int \half\rho({\bf r}) {\bf v}\cdot(\bomega \times {\bf r}) \d^3 r = = \int \half\rho({\bf r}) \bomega\cdot({\bf r}\times {\bf v}) \d^3 r = = \half \bomega \cdot \int\rho({\bf r}) ({\bf r}\times {\bf v}) \d^3 r = = \half \bomega \cdot {\bf L}

where {\bf L} is the total angular momentum:

{\bf L} = \int\rho({\bf r}) ({\bf r}\times {\bf v}) \d^3 r

Angular Momentum

Total angular momentum is:

{\bf L} = \int \rho({\bf r}) ({\bf r} \times {\bf v}) \d^3 r = = \int \rho({\bf r}) ({\bf r} \times (\bomega \times {\bf r})) \d^3 r= = \int \rho({\bf r}) (\bomega r^2 - {\bf r} ({\bf r} \cdot \bomega)) \d^3 r = = \int \rho({\bf r}) (\one r^2 - {\bf r} {\bf r}) \d^3 r \cdot \bomega = = {\bf I} \cdot \bomega

Where {\bf I} is the moment of inertia tensor:

{\bf I} = \int \rho({\bf r}) (\one r^2 - {\bf r} {\bf r}) \d^3 r

Moment of Inertia

The moment of inertia tensor and its components are:

{\bf I} = \int \rho({\bf r}) (\one r^2 - {\bf r} {\bf r}) \d^3 r I^{ij} = \int \rho({\bf r}) (\delta^{ij} r_k r^k - r^i r^j) \d^3 r

Let’s write \bomega=\omega {\bf n} (where {\bf n} is a unit vector), then the kinetic energy is:

T = \half \bomega \cdot {\bf L} = \half \bomega \cdot {\bf I} \cdot \bomega = \half {\bf n} \cdot {\bf I} \cdot {\bf n}\, \omega^2 = \half I \omega^2

where I is the moment of inertia about the axis of rotation:

I = {\bf n} \cdot {\bf I} \cdot {\bf n} = = {\bf n} \cdot \int \rho({\bf r}) (\one r^2 - {\bf r} {\bf r}) \d^3 r \cdot {\bf n} = = \int \rho({\bf r}) (r^2 - ({\bf r}\cdot {\bf n})^2) \d^3 r

Cylinder

Solid cylinder of radius R, height h and mass m. We’ll use cylindrical coordinates. First for rotation about the z axis:

V = \pi R^2 h {\bf n} = (0, 0, 1) {\bf r} = (\rho\cos\phi, \rho\sin\phi, z) {\bf r} \cdot {\bf n} = z r^2 = \rho^2 + z^2 I = \int \rho({\bf r}) (r^2 - ({\bf r}\cdot {\bf n})^2) \d^3 r = \int {m\over V} (\rho^2+z^2 - z^2) \d^3 r = = \int {m\over V} \rho^2 \d^3 r = {m\over V} \int_0^{2\pi}\d\phi \int_0^R\d R \int_{-{h\over2}}^{h\over2} \d z \rho^2 \rho = = {m\over V} 2\pi {R^4\over 4} h = {m\over \pi R^2 h} 2\pi {R^4\over 4} h = \half m R^2

Code:

>>> from sympy import var, integrate, pi
>>> var("m V R rho z phi h")
(m, V, R, rho, z, phi, h)
>>> I = m/V * integrate(rho**2 * rho, (rho, 0, R), (phi, 0, 2*pi), (z, -h/2, h/2))
>>> I.subs(V, pi * R**2 * h)
R**2*m/2

And about the x axis:

{\bf n} = (1, 0, 0) {\bf r} = (\rho\cos\phi, \rho\sin\phi, z) {\bf r} \cdot {\bf n} = \rho\cos\phi r^2 = \rho^2 + z^2 I = \int \rho({\bf r}) (r^2 - ({\bf r}\cdot {\bf n})^2) \d^3 r = \int {m\over V} (\rho^2+z^2 - \rho^2\cos^2\phi) \d^3 r = = {m\over V} \int_0^{2\pi}\d\phi \int_0^R\d R \int_{-{h\over2}}^{h\over2} \d z (\rho^2+z^2 - \rho^2\cos^2\phi)\rho = = {m\over V}\left({\pi R^4 h\over 2}+{\pi R^2 h^3\over 12} -{\pi R^4 h\over 4}\right) = = {m\over \pi R^2 h}\left({\pi R^4 h\over 2}+{\pi R^2 h^3\over 12} -{\pi R^4 h\over 4}\right) = = {m\over 12} (6R^2 + h^2 - 3R^2) = = {m\over 12} (3R^2 + h^2)

Code:

>>> from sympy import var, integrate, pi, cos
>>> var("m V R rho z phi h")
(m, V, R, rho, z, phi, h)
>>> I = m/V * integrate((rho**2+z**2-rho**2*cos(phi)**2) * rho, (rho, 0, R), (phi, 0, 2*pi), (z, -h/2, h/2))
>>> I.subs(V, pi * R**2 * h).simplify()
m*(3*R**2 + h**2)/12

Special cases are a rod of length h (set R=0 above) and a thin solid disk of radius R and mass m (set h=0 above).

Sphere

Solid sphere of radius R and mass m. We’ll use spherical coordinates. All axes are equivalent, so we use rotation about the z axis:

V = {4\over3} \pi R^3 {\bf n} = (0, 0, 1) {\bf r} = (\rho\cos\phi\sin\theta, \rho\sin\phi\sin\theta, \rho\cos\theta) {\bf r} \cdot {\bf n} = \rho\cos\theta r^2 = \rho^2 I = \int \rho({\bf r}) (r^2 - ({\bf r}\cdot {\bf n})^2) \d^3 r = \int {m\over V} (\rho^2 - \rho^2\cos^2\theta) \d^3 r = = {m\over V} \int_0^{2\pi}\d\phi \int_0^R\d R \int_0^\pi \d \theta \rho^2(1-\cos^2\theta) \rho^2\sin\theta = = {m\over V} \int_0^{2\pi}\d\phi \int_0^R\d R \int_0^\pi \d \theta \rho^4\sin^3\theta = = {m\over V} 2\pi {R^5\over 5} {4\over 3} = = {m\over V} {8\over 15}\pi R^5 = {m\over {4\over 3}\pi R^3} {8\over 15}\pi R^5 = {2\over 5} m R^2

Code:

>>> from sympy import var, integrate, pi, sin
>>> var("m V R rho theta phi")
(m, V, R, rho, theta, phi)
>>> I = m/V * integrate(rho**4 * sin(theta)**3, (rho, 0, R), (phi, 0, 2*pi), (theta, 0, pi))
>>> I
8*pi*R**5*m/(15*V)
>>> I.subs(V, 4*pi*R**3/3)
2*R**2*m/5