Forest UCM RBM

Rigid Body Motion

Rigid Body

Rigidy Body

A Rigid Body is a system involving a large number of point masses, called particles, whose distances between pairs of point particles remains constant even when the body is in motion or being acted upon by external force.

Forces of Constraint

The internal forces that maintain the constant distances between the different pairs of point masses.

Total Angular Momentum of a Rigid Body

Consider a rigid body that rotates about a fixed z-axis with the origin at point O.

let

[math]\vec R[/math] point to the center of mass of the object

[math]\vec {r}_k[/math] points to a mass element [math]m_k[/math]

[math]\vec{r}_k^{\;\;\prime}[/math] points from the center of mass to the mass element [math]m_k[/math]

the angular momentum of mass element [math]m_k[/math] about the point O is given as

[math]\ell_k = \vec {r}_k \times \vec {p}_k = \vec {r}_k \times m \vec {\dot r}_k[/math]

The total angular momentum about the point O is given as

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

This can be cast in term of the angular momentum about the center of mass and the angular momentum of the CM motion

[math]\vec {r}_k = \vec R + \vec{r}_k^{\;\; \prime}[/math]

[math] \vec L = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

[math] = \sum (\vec R + \vec{r}_k^{\;\; \prime}) \times m_k (\vec \dot R + \vec{\dot r}_k^{\;\; \prime})[/math]

[math] = \sum \vec R \times m_k \vec \dot R + \sum \vec R \times m_k \vec{\dot r}_k^{\;\; \prime} + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R +\sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} [/math]

[math]\sum \vec R \times m_k \vec \dot R = \vec R \times \sum m_k \vec \dot R = \vec R \times M \vec \dot R = \vec R \times \vec P[/math]

[math]\vec P =[/math] momentum of the center of Mass

[math]\sum \vec R \times m_k \vec{\dot r}_k^{\;\; \prime} = \vec R \times \sum m_k \vec{\dot r}_k^{\;\; \prime} [/math]

[math]\sum m_k \vec{\dot r}_k^{\;\; \prime} = \sum m_k \left ( \vec {r}_k - \vec R\right ) = \sum m_k \vec {r}_k - \sum m_k \vec R = \vec {v}_{cm} - \vec{v}_{cm} = 0[/math]

The location of the center of mass is at [math]\vec{ r}_k^{\;\; \prime} = 0[/math] the derivative is also zero

[math]\sum \vec{r}_k^{\;\; \prime} \times m_k \vec \dot R = \sum m_k \vec{r}_k^{\;\; \prime} \times \vec \dot R =0 [/math] : The location of the CM is at 0

[math] \vec L = \vec R \times \vec P + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} [/math]

[math] = L_{\mbox{CM}} + L_{\mbox{about CM}} [/math]

The total angular momentum is the sum of the angular momentum of the center of mass of a rigid body [math] L_{\mbox{CM}} [/math] and the angular momentum of the rigid body about the center of mass [math] L_{\mbox{about CM}} [/math]

Planet example

What is the total angular momentum of the earth orbiting the sun?

There are two components

[math] \vec L_{\mbox{CM}} [/math] = angular momentum of the earth orbiting about the sun

[math] \vec L_{\mbox{about CM}} [/math] = angular momentum of the earth orbiting about the earth's center of mass (Spin)

[math]\vec L_{\mbox{tot}} = \vec L_{\mbox{CM}} + \vec L_{\mbox{about CM}}[/math]

[math] \vec L_{\mbox{CM}} [/math] is conserved and defined as Orbital angular momentum

[math]\vec \dot L_{\mbox{CM}} = \vec \dot R \times \vec P + \vec R \times \vec \dot P[/math]

[math]\vec \dot R \times \vec P = \vec V \times M \vec V = 0[/math]

[math]\Rightarrow \vec \dot L_{\mbox{orb}} = \vec R \times \vec \dot P=\vec R \times \vec {F}_{ext}[/math]

If there is only a central force

[math]\vec {F}(\mbox{ext}) = G \frac{Mm}{R^3} \vec R[/math]

Then

[math]\vec R \times \vec {F}(\mbox{ext}) = \vec R \times G \frac{Mm}{R^3} \vec R= G \frac{Mm}{R^3} \vec R \times \vec R =0 [/math]

Thus

[math]\vec \dot L_{\mbox{CM}} = \vec R \times \vec {F}(\mbox{ext}) = 0[/math]

[math]\vec L_{\mbox{CM}} \equiv \vec L_{\mbox{Orb}}[/math] = constant = Orbital angular momentum

The above is a good approximation even though the Sun's gravitational Field is not perfectly uniform

How about [math]\vec L_{\mbox{about CM}}[/math]?

Since

[math]\vec L_{\mbox{tot}} = \sum \vec {r}_k \times m_k \vec {\dot r}_k =\vec L_{\mbox{Orb}} + \vec L_{\mbox{about CM}}[/math]

as seen earlier

[math] \vec L = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

[math] = \sum (\vec R + \vec{r}_k^{\;\; \prime}) \times m_k (\vec \dot R + \vec{\dot r}_k^{\;\; \prime})[/math]

[math] = \vec R \times \vec P + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} [/math]

Then

[math]\dot \vec L = \vec \dot R \times \vec P +\vec R \times \vec \dot P + \sum \vec{\dot r}_k^{\;\; \prime} \times m_k \vec{\dot r}_k^{\;\; \prime} + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\ddot r}_k^{\;\; \prime} [/math]

[math] =\vec R \times \vec \dot P + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\ddot r}_k^{\;\; \prime} [/math]

[math] =\vec R \times \vec {F}(\mbox{ext}) + \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\ddot r}_k^{\;\; \prime} [/math]

[math] =\vec{\dot{ L}}_{\mbox{Orv}} + \vec{\dot {L}}_{\mbox{about CM}}[/math]

[math]\Rightarrow \vec{\dot {L}}_{\mbox{about CM}} = \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\ddot r}_k^{\;\; \prime} [/math]

[math] \vec{\dot {L}}_{\mbox{spin}}\equiv\vec{\dot {L}}_{\mbox{about CM}} = \sum \vec{r}_k^{\;\; \prime} \times m_k \vec{\ddot r}_k^{\;\; \prime} \equiv \tau(\mbox{ext about CM}) [/math]

The total angular momentum is the sum of the orbital angular momentum and the spin

[math]\vec L_{\mbox{tot}} = \vec L_{\mbox{orb}} + \vec L_{\mbox{spin}}[/math]

Precession of the Earth

http://courses.physics.northwestern.edu/Phyx125/Precession%20of%20the%20Earth.pdf

Total Kinetic energy of a Rigid Body

Using the same coordinate system as above

the kinetic energy of mass element [math]m_k[/math] about the point O is given as

[math]T_k = \frac{1}{2} m_k \left | \vec \dot{r}_k \right |^2[/math]

The total Kinetic about the point O is given as

[math] \vec T = \sum \frac{1}{2} m_k \left | \vec \dot{r}_k \right |^2[/math]

Rewriting this again in terms of the location of the CM of the body [math]\vec R[/math] and the location of a mass element from the CM [math]\vec{r}_k^{\;\; \prime}[/math]

[math]\vec {r}_k = \vec R + \vec{r}_k^{\;\; \prime}[/math]

[math]\vec {\dot r}_k = \vec \dot R + \vec{\dot r}_k^{\;\; \prime}[/math]

[math]\vec {\dot r}_k \cdot \vec {\dot r}_k = \left ( \vec \dot R + \vec{\dot r}_k^{\;\; \prime} \right ) \cdot \left ( \vec \dot R + \vec{\dot r}_k^{\;\; \prime} \right ) [/math]

[math] = \left | \vec \dot R \right |^2 + \left | \vec{\dot r}_k^{\;\; \prime} \right |^2 + 2 \vec \dot R \cdot \vec{\dot r}_k^{\;\; \prime} [/math]

[math] \vec T = \sum \frac{1}{2} m_k \left | \vec \dot{r}_k \right |^2[/math]

[math] = \sum \frac{1}{2} m_k \left | \vec \dot R \right |^2 + \left | \vec{\dot r}_k^{\;\; \prime} \right |^2 + 2 \vec \dot R \cdot \vec{\dot r}_k^{\;\; \prime}[/math]

[math] = \frac{1}{2} M \left | \vec \dot R \right |^2 + \frac{1}{2} \sum m_k \left | \vec{\dot r}_k^{\;\; \prime} \right |^2 + \vec \dot R \cdot \sum m_k \vec{\dot r}_k^{\;\; \prime}[/math]

For a Rigid body

[math] \sum m_k \vec{\dot r}_k^{\;\; \prime} =0 [/math] The internal kinetic energy is zero for a rigid body, otherwise it would be expanding

[math] \vec T = \frac{1}{2} M \left | \vec \dot R \right |^2 + \frac{1}{2} \sum m_k \left | \vec{\dot r}_k^{\;\; \prime} \right |^2 \cdot [/math]

[math] = T_{\mbox{CM}} + T_{\mbox{about CM}}[/math]

Total Potential energy of a Rigid Body

If all forces are conservative

[math]\vec \nabla \times \vec F_{ij} = 0[/math]

then a Potential Energy may be defined

[math]\Delta U_{ij}(r_{ij}) \equiv -\int_{r_o}^r \vec{F}_{ij}(r_{ij}) \cdot d\vec{r}_{ij} = - W_{cons}[/math]

Then the potential energy of a rigid body is given by

[math]\sum_{i\lt j} \Delta U_{ij}(r_{ij}) = U_{\mbox{int}} = [/math] constant

for a rigid body the inter-particle separation distance [math]r_{ij}[/math] is constant so the internal potential of a rigid body does not change

Thus the kinematics of a Rigid body only needs to consider the potential energy of external forces.

Rotation about a fixed axis

Consider a Rigid body rotating with a speed [math]\vec \omega[/math] about a fixed axis (z-axis) with its origin at the point O

[math]\vec \omega = \omega \hat k[/math]

The total angular momentum about the point O is given as

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

As seen in the non-inertial reference frame chapter

[math]\vec \dot r = \vec v = \vec \omega \times \vec r[/math]

thus

[math] \vec \omega \times \vec r_k = \left( \begin{array}{ccc} \hat i & \hat j & \hat k\\ 0 & 0 & \omega \\x_{k} & y_{k} & z_{k}\end{array} \right)= \omega \left ( - y_{k} \hat i + x_{k} \hat j \right ) [/math]

[math] \vec {r}_k \times \vec {\dot r}_k = \vec {r}_k \times \vec \omega \times \vec r = \vec {r}_k \times \omega \left ( - y_{k} \hat i + x_{k} \hat j \right ) [/math]

[math]= \left( \begin{array}{ccc} \hat i & \hat j & \hat k\\ x_{k} & y_{k} & z_{k} \\ -\omega y_{k} & \omega x_{k} & 0\end{array} \right)= \omega \left ( - z_{k}x_{k} \hat i - z_{k}y_{k} \hat j + \left (x_{k}^2+y_{k}^2 \right ) \hat k \right ) [/math]

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

[math] = \omega \sum m_k \left ( - z_{k}x_{k} \hat i - z_{k}y_{k} \hat j + \left (x_{k}^2+y_{k}^2 \right ) \hat k \right )[/math]

Moments (Products) of Inertia about the fixed axis

[math]L_z = \omega \sum m_k \left (x_{k}^2+y_{k}^2 \right )[/math]

Let

[math]I_{zz} \equiv \sum m_k \left (x_{k}^2+y_{k}^2 \right ) =[/math] moment of inertia about the z-axis in the "z" ([math]\hat k[/math]) direction.

Then

[math]L_z = I_{zz} \omega[/math]

Similarly

Products of Inertia of a rigid body rotating about the z-axis for the angular momentum component along the x-axis

[math]I_{xz} \equiv \sum m_k(-x_{k} z_{k}) [/math]

Products of Inertia of a rigid body rotating about the z-axis for the angular momentum component along the y-axis

[math]I_{yz} \equiv \sum m_k \left (-y_{k} z_{k} )\right ) [/math]

Note the negative sign is indicative of the rigid body's resistance (inertia) to have an angular momentum component that is not in the same direction as its rotation.

[math] \vec L =\omega \left ( I_{xz} \hat i + I_{yz} \hat k + I_{zz} \hat k \right )[/math] = angular momentum about the z-axis

Example

Moment of Inertia of a sphere

Find the moment of inertia of a uniform solid sphere of Radius (R) and mass (M) about its diameter.

Using a coordinate system with its origin at the center of the sphere

For a set of discrete masses

[math]I_{zz} \equiv \sum m_k \left (x_{k}^2+y_{k}^2 \right ) =[/math] moment of inertia about the z-axis in the "z" ([math]\hat k[/math]) direction.

For a mass distribution the above summation is written in integral form as

[math]I_{zz} = \int dm_k \left (x_{k}^2+y_{k}^2 \right ) [/math]

one may define the sphere's density as

[math]\rho = \frac{M}{\frac{4}{3} \pi R^3}[/math]

A differential mass is written as

[math]dm_k = \rho dV_k = [/math] Imagine a differential circle in the x-y plane that you integrate along z

if using spherical coordinates

[math]x_k = r \sin \theta \cos \phi \;\;\;\; y_k = r \sin \theta \sin \phi[/math]

[math]x_k^2 + y_k^2 = r^2 \sin^2 \theta[/math]

[math]I_{zz} = \int dm_k \left (x_{k}^2+y_{k}^2 \right ) [/math]

[math]= \rho \int r^2 dV_k = \rho \int (r \sin \theta )^2 dV_k = \rho \int r^2 \sin^2 \theta (r^2 dr \sin \theta d \theta d \phi)[/math]

[math]= \rho \int_0^R r^4 dr \int_0^{\pi} \sin^3 \theta d \theta \int_0^{2\pi} d \phi[/math]

[math]= \frac{8 \pi}{15} \rho R^5 = \frac{2}{5} MR^2[/math]

[math]dV = r^2 dr \sin \theta d \theta d \phi[/math]

The moment of Inertia of a hollow sphere of outer radius [math]b[/math] and inner radius [math]a[/math]

Moments of Inertia, like masses, add and subtract like scalers.

[math]I(b,a) = I(b,0) - I(a,o) = \frac{8 \pi}{15} \rho (b^5 - a^5)[/math] = Moment of inertia of a sphere of inner radius [math]a[/math] and outer radius [math]b[/math].

[math]\rho = \frac{M}{V} = \frac{M}{\frac{4 \pi}{3} (b^3 - a^3)}[/math]

[math]I(b,a) = \frac{2}{5} M \frac{ (b^5 - a^5)}{ (b^3 - a^3)}[/math] = Moment of inertia of a sphere of inner radius [math]a[/math] and outer radius [math]b[/math].

Moment of inertia tensor

The angular momentum for a body spinning about an arbitrary axis.

let

[math]\vec \omega = \omega_x \hat i + \omega_y \hat j + \omega_z \hat k[/math]

Consider a Rigid body rotating with a speed [math]\vec \omega[/math] about a fixed axis (z-axis) with its origin at the point O

[math]\vec \omega = \omega \hat k[/math]

The total angular momentum about the point O is given as

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k= \sum m_k \left [ \vec {r}_k \times \vec {\dot r}_k \right ][/math]

As seen in the non-inertial reference frame chapter

[math]\vec \dot r = \vec v = \vec \omega \times \vec r[/math]

thus

[math] \vec \omega \times \vec r_k = \left( \begin{array}{ccc} \hat i & \hat j & \hat k\\ \omega_x & \omega_y & \omega_z \\x_{k} & y_{k} & z_{k}\end{array} \right)= \left ( \omega_y z_{k} - \omega_z y_k \right ) \hat i + \left ( \omega_z x_{k} - \omega_x z_k \right ) \hat j + \left ( \omega_x y_{k} - \omega_y x_k \right ) \hat k [/math]

[math] \vec {r}_k \times \vec {\dot r}_k = \vec {r}_k \times \vec \omega \times \vec r = \vec {r}_k \times \omega \left ( - y_{k} \hat i + x_{k} \hat j \right ) [/math]

[math]= \left( \begin{array}{ccc} \hat i & \hat j & \hat k\\ x_{k} & y_{k} & z_{k} \\ \left ( \omega_y z_{k} - \omega_z y_k \right ) & \left ( \omega_z x_{k} - \omega_x z_k \right ) & \left ( \omega_x y_{k} - \omega_y x_k \right ) \end{array} \right) [/math]

[math]L_x = \sum m_k \left [ \left . \vec {r}_k \times m \vec {\dot r}_k \right |_x \right ][/math]

[math]=\sum m_k \left [ y_k \left ( \omega_x y_{k} - \omega_y x_k \right )-z_k \left ( \omega_z x_{k} - \omega_x z_k \right ) \right ] [/math]

[math]= \sum m_k \left [ \left ( y_{k}^2 + z_k ^2 \right )\omega_x - y_k x_k \omega_y - z_k x_{k} \omega_z \right ][/math]

[math]L_y = \sum m_k \left [ \left . \vec {r}_k \times m \vec {\dot r}_k \right |_y \right ][/math]

[math]= \sum m_k \left [ z_k \left ( \omega_y z_{k} - \omega_z y_k \right ) -x_k \left ( \omega_x y_{k} - \omega_y x_k \right ) \right ][/math]

[math]= \sum m_k \left [ -x_ky_k \omega_x + \left ( z_{k}^2 +x_k^2\right ) \omega_y - z_k y_k \omega_z \right ][/math]

[math]L_z = \sum m_k \left [ \left . \vec {r}_k \times m \vec {\dot r}_k \right |_z \right ][/math]

[math] =\sum m_k \left [ x_k \left ( \omega_z x_{k} - \omega_x z_k \right ) - y_k \left ( \omega_y z_{k} - \omega_z y_k \right) \right ][/math]

[math] =\sum m_k \left [ -x_k z_k \omega_x - y_k z_k \omega_y +\left ( x_{k}^2 + y_k^2 \right)\omega_z \right ][/math]

The Moment of Inertia tensor is defined such that

[math]\tilde{I} = \left( \begin{array}{ccc} \sum m_k\left ( y_{k}^2 + z_k ^2 \right ) & - \sum m_k y_k x_k & - \sum m_kz_k x_{k}\\ -\sum m_kx_ky_k & \sum m_k\left ( z_{k}^2 +x_k^2\right ) & - \sum m_kz_k y_k \\ -\sum m_kx_k z_k & - \sum m_ky_k z_k & \sum m_k\left ( x_{k}^2 + y_k^2 \right) \end{array} \right) [/math]

[math] \vec L = \tilde{I} \vec \omega \;\;\;\;\;\; \mbox {or} \;\;\;\;\;\; L_i = \sum_{j=1}^3 I_{ij} \omega_j [/math]

The diagonals of the moment of inertia tensor are the usual moments of inertia

The off-diagonals are the six products of inertia

[math]I_{ij} = - \sum_{\alpha} m_{\alpha} i_{\alpha} j_{\alpha}[/math] where [math]i[/math] & [math]j[/math] represent [math]x,y,z[/math] components and [math]i\ne j[/math]

[math]I_{ii} = \sum_{\alpha} m_{\alpha} j_{\alpha} k_{\alpha}[/math] where [math]j[/math] & [math] k[/math] represent [math]x,y,z[/math] components and [math]i\ne j \ne k [/math]

Note

[math] \tilde I = \tilde I^{\;\;T}[/math] The inertia tensor is symmetric ( i.e. it is it's own transpose) or in other words

[math]I_{ij} = I_{ji}[/math]

In other words the product of inertia of a body rotating about the i axis for the angular momentum along the j axis is the same as

the product of inertia of a body rotating about the j axis for the angular momentum about the i axis

Inertia Tensor for a solid cone

Find the moment of inertia tensor for a uniform solid cone of mass [math]M[/math] , height [math]h[/math] , and base radius [math]R[/math] spinning about its tip.

Choose the z-axis so it is along the symmetry axis of the cone.

The mass density may be written as

[math]\rho = \frac{M}{V}[/math]

Using cylindrical coordinates ( [math]r, \phi, z[/math]) the volume of a cone is given by

[math]V = \int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr = \int_0^h dz \int _0^{2\pi}d\phi \frac{R^2z^2}{2h^2}= 2 \pi \frac{R^2}{2h^2}\int_0^h z^2 dz [/math]

[math] = \pi \frac{R^2}{h^2} \frac{h^3}{3} =\frac{1}{3} \pi R^2 h [/math]

the radius of the base of the cone may be written as a function such that [math] r=\frac{Rz}{h}[/math] .

When [math]z=0 \Rightarrow r=0[/math] and when [math]z=h \Rightarrow r = R[/math]

[math]\rho = \frac{M}{V} = \frac{M}{\frac{1}{3} \pi R^2 h} = \frac{ 3 M}{\pi R^2 h} [/math]

calculating the moment of inertia about the z-axis :

[math]I_{zz} = \int \rho \left (x^2+y^2 \right ) dV= \int_0^h \int _o^{2\pi} \int_0^{\frac{Rz}{h}} \rho r^2 r dr d\phi dz[/math]

[math]I_{zz} = \rho \int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r^2 r dr [/math]

[math] = \rho \int_0^h dz (2 \pi) \frac{R^4z^4}{4h^4} [/math]

[math] = 2 \pi \rho \frac{R^4}{4h^4}\int_0^h z^4 dz [/math]

[math] = 2 \pi \rho \frac{R^4}{4h^4} \frac{h^5}{5} [/math]

[math] = 2 \pi \left ( \frac{ 3 M}{\pi R^2 h}\right ) \frac{R^4}{4h^4} \frac{h^5}{5} [/math]

[math] = 3 M \frac{R^2}{2h^5} \frac{h^5}{5} = \frac{3}{10} M R^2 [/math]

[math]I_{xx} = \int \rho \left (z^2+y^2 \right ) dV= \int \rho z^2 dV +\int \rho y^2 dV[/math]

Since

[math]I_{zz} = \int \rho \left (x^2+y^2 \right ) dV[/math]

and the cone is symmetric in the x-y plane

[math]\int \rho y^2 dV = \frac{1}{2} I+{zz}[/math]

[math] \int \rho z^2 dV = \rho \int_0^h z^2 dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr = 2 \pi \rho \int_0^h z^2 dz \frac{R^2z^2}{2h^2} [/math]

[math] = \pi \rho \frac{R^2}{h^2}\int_0^h z^4 dz = \pi \rho \frac{R^2}{h^2} \frac{h^5}{5} [/math]

[math] = \pi \left ( \frac{ 3 M}{\pi R^2 h}\right ) frac{R^2h^3}{5} = \frac{3Mh^2}{5} [/math]

[math]I_{xx} =\frac{3}{20} M R^2 + \frac{3Mh^2}{5}[/math]

[math]I_{yy } = \int \rho \left (z^2+x^2 \right ) dV = \frac{3}{20} M R^2 + \frac{3Mh^2}{5}= \frac{3}{20} M \left ( R^2 + 4h^2 \right ) = I_{xx}[/math]

The products of inertia

Product of Inertia of a rigid body that relates the x(y) component of the angular momentum for rotations about the y(x)-axis

[math]I_{xy} \equiv \sum m_k(-x_{k} y_{k}) =\int \rho xy dV[/math]

From the viewpoint of the summation equation one can see that for every mass at point (x,z) there is an equal mass at the point (-x,z), Thus the sum will add to zero.

or mathematically

[math]I_{xy} =\int \rho xy dV =\int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr \rho xy =\int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr \rho (r \cos \phi ) ( r \sin \phi) [/math]

[math]=\rho \int_0^h dz \int _0^{2\pi}d\phi \cos \phi \sin \phi \int_0^{\frac{Rz}{h}} r^3 dr [/math]

[math]\int _0^{2\pi}d\phi \cos \phi \sin \phi =\int _0^{2\pi} \sin \phi d(\sin \phi) = \left . \frac{1}{2}\sin^2 \phi \right |_{0}^{2\pi} = 0 [/math]

Product of Inertia of a rigid body that relates the x(z) component of the angular momentum for rotations about the z(x)-axis

[math]I_{xz} \equiv \sum m_k(-x_{k} y_{k}) =-\int \rho xz dV [/math]

[math]= -\int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr \rho r \cos \phi z= -\rho \int_0^h z dz \int _0^{2\pi} \cos \phi d\phi \int_0^{\frac{Rz}{h}} r^2 dr [/math]

Product of Inertia of a rigid body that relates the y(z) component of the angular momentum for rotations about the y(x)-axis

[math]I_{yz} \equiv \sum m_k \left (-y_{k} z_{k} \right ) [/math]

[math]=-\int \rho yz dV = -\int_0^h dz \int _0^{2\pi}d\phi \int_0^{\frac{Rz}{h}} r dr \rho r \sin \phi z= -\rho \int_0^h z dz \int _0^{2\pi} \sin \phi d\phi \int_0^{\frac{Rz}{h}} r^2 dr [/math]

[math]\int _0^{2\pi} \sin \phi d\phi =\int _0^{2\pi} \cos \phi d\phi = 0[/math]

If the inertia tensor is diagonal (off diagnonal elements are zero)

Then The angular momentum points in the same direction as the angular velocity.

[math] I = \frac{3}{20} M \left( \begin{array}{ccc} R^2+4h^2 & 0 & 0\\ 0 & R^2+4h^2 & 0 \\0 & 0 & 2R^2 \end{array} \right) [/math]

http://scienceworld.wolfram.com/physics/MomentofInertiaSphere.html

http://www.phys.ufl.edu/~mueller/PHY2048/2048_Chapter10_F08_Part2_lecture.pdf

Principle Axis

as seen before, the angular momentum of a rigid body rotating about a fixed axis ( the z-axis in this case) is given by

[math] \vec L = \sum \ell_k = \sum \vec {r}_k \times m_k \vec {\dot r}_k[/math]

[math] = \omega \sum m_k \left ( - z_{k}x_{k} \hat i - z_{k}y_{k} \hat j + \left (x_{k}^2+y_{k}^2 \right ) \hat k \right )[/math]

Although the object was only rotating about the z-axis, there are angular momentum components along the other directions

In other words, unlike what you learned in introductory physics

[math]\vec L \ne I \vec \omega[/math] in general

instead the general expression for angular momentum is

[math]\vec L = \vec r \times \vec p[/math] is the more general statement

Consider the Kinetic energy of a rotating rigid body in terms of the sum of the kinetic energy of all of its fixed mass elements

[math]T = \frac{1}{2} \sum_k m_k v_k^2[/math]

If the body is only rotating about the z-axis then

[math]v_k = r_k \omega[/math]

[math]\Rightarrow T = \frac{1}{2} \sum_k m_k r_k^2 \omega = \frac{ \omega}{2} \sum_k m_k r_k^2[/math]

[math]= \omega I_z \equiv L_z[/math]

example of the tensor

Forest_Ugrad_ClassicalMechanics#Rigid_Body_Motion