# Forest UCM RBM

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# 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

point to the center of mass of the object
points to a mass element
points from the center of mass to the mass element

the angular momentum of mass element about the point O is given as

The total angular momentum about the point O is given as

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

momentum of the center of Mass
The location of the center of mass is at the derivative is also zero
: The location of the CM is at 0

The total angular momentum is the sum of the angular momentum of the center of mass of a rigid body and the angular momentum of the rigid body about the center of mass

#### Planet example

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

There are two components

= angular momentum of the earth orbiting about the sun
= angular momentum of the earth orbiting about the earth's center of mass (Spin)

is conserved and defined as Orbital angular momentum

If there is only a central force

Then

Thus

= constant = Orbital angular momentum

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

Since

as seen earlier

Then

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

Precession of the Earth

### Total Kinetic energy of a Rigid Body

Using the same coordinate system as above

the kinetic energy of mass element about the point O is given as

The total Kinetic about the point O is given as

Rewriting this again in terms of the location of the CM of the body and the location of a mass element from the CM

For a Rigid body

The internal kinetic energy is zero for a rigid body, otherwise it would be expanding

### Total Potential energy of a Rigid Body

If all forces are conservative

then a Potential Energy may be defined

Then the potential energy of a rigid body is given by

constant

for a rigid body the inter-particle separation distance 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 about a fixed axis (z-axis) with its origin at the point O

The total angular momentum about the point O is given as

As seen in the non-inertial reference frame chapter

thus

### Moments (Products) of Inertia about the fixed axis

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

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

In other words, unlike what you learned in introductory physics

in general

instead the general expression for angular momentum is

is the more general statement

The Moment of Inertia is the inertia of the rotating body for the angular momentum that is parallel to the angular velocity The other components of the angular momentum that are not along the direction of the angular velocity related to the angular velocity by the Product of Inertia

The Moment of Inertia for the above example is defined as

Let

moment of inertia about the z-axis in the "z" () direction.

Then

Similarly

The product of Inertia of a rigid body rotating about the z-axis for the angular momentum component along the x-axis is given by

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

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.

= 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

moment of inertia about the z-axis in the "z" () direction.

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

one may define the sphere's density as

A differential mass is written as

Imagine a differential circle in the x-y plane that you integrate along z

if using spherical coordinates

The moment of Inertia of a hollow sphere of outer radius and inner radius

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

= Moment of inertia of a sphere of inner radius and outer radius .
= Moment of inertia of a sphere of inner radius and outer radius .

## Moment of inertia tensor

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

let

Consider a Rigid body rotating with a speed about a fixed axis (z-axis) with its origin at the point O

The total angular momentum about the point O is given as

As seen in the non-inertial reference frame chapter

thus

The Moment of Inertia tensor is defined such that

The diagonals of the moment of inertia tensor are the usual moments of inertia
The off-diagonals are the six products of inertia

where & represent components and
where & represent components and
Note
The inertia tensor is symmetric ( i.e. it is it's own transpose) or in other words

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 , height , and base radius 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

Using cylindrical coordinates ( ) the volume of a cone is given by

the radius of the base of the cone may be written as a function such that .

When and when

calculating the moment of inertia about the z-axis :

Since

and the cone is symmetric in the x-y plane

#### 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

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

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

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

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

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

## Precession of a Top

As shown above,the angular momentum tensor of a top rotating about its axis of symmetry with respect to its tip is

Suppose the top make a small angle with respect to the normal of the surface it is spinning on top of.

The resulting torque from this gravitational force is

where is a vector pointing to top's center of mass from the principal axes that make up a coordinate system whose origin is at the tip of the top

( points normal to the surface that the top is spinning on)

Initially the angular velocity ( and angular momentum) are pointed along the principal axis.

Newton's second law for the angular motion

Let

= precession frequency of the top

The axis of the top rotates with an angular speed about the axis.

The direction of the external torque is in the direction at the instant shown in the above figure. As a result the top will rotate in the counter-clockwise direction about the z-axis as viewed from above the z-axis.

## Principal Axis

As seen in the case of a precessing top above, the physics of a rotating rigid body can be more easily described when the inertia tensor is diagonal. One need only find a coordinate system in which the tensor is diagonal in order to simplify the problem.

Principal axis
The axis of rotation for a rigid body in which the moment of inertia tensor is diagonal thereby the Angular momentum will be parallel to the angular velocity.

In the above example of the top ( solid cone), the x,y, and z axes of the chosen coordinate system are principal axes of inertia and the moments of inertia are the principal moments.

One can observe from the example of a spinning top, that a rotation axis around which the rigid body is symmetry will tend to be a principal axis.

Additionally, if the body has two perpendicular planes of reflection symmetry, then the three perpendicular axes defined by these two planes and a point of rotation are principal axes.

### Existence of Principal Axes

Although it is easy to see which are the principal axes when the rigid body is symmetric, it is true that any rigid body's rotating about a point has three principal axes.

### Finding the Principal axis of a Rigid body

Once the Inertia tensor is known, the principal axes of a rigid body may be found by "diagonalizing" the tensor matrix.

The process of "diagonalizing" the tensor matrix amounts to solving an eigenvalue problem.

The angular momentum of a rigid body can be given by

For the special case that the angular momentum (\vec L) is parallel to the angular velocity (\vec \omega)

Then

To determine the principal axes you determine the vector such that

In practice, one determines ( the eigen vectors) by solving for (the eigenvalues) such that

where

= unit matrix

In order for the above matrix problem (system of 3 equations and 3 unknowns) to have a non-zero solution the secular equation must hold ( the determinant must be non-zero)

One solves the above equation by first finding the three eigenvalues that will make the above determinant zero and then use Gauss-Jordan elimination of the augmented matrix using each of the eigenvalues separately to find the eigenvector for that particular eigenvalue

For example:

Suppose the momentum tensor is given by

3 solutions

You can find the eigenvector for each eigenvalue by performing Gauss-Jordan row-column reduction of the augmented matrix

for the eigenvalue

In augmented form:

Divide everything by 3

Add the second row to the first row

Add the first row to the second and third row

Divide the second row by 3

Subtract the first row by the second row

The above augmented matrix represents the following system of equations

row 1 :

row 2 :

row 1 : is arbitary

thus

but the above needs to be a unit vector such that

therefor

for the eigenvalue

All three components of add up to zero or in other words

The remaining two principal axis must be perpendicular to the principal axis with eigenvalue .

Otherwise the remaining two principal axes are arbitrary and indicate that the rigid body is symmetric about these two remaining axes.

BTW: The above moment of inertia represents the moment of inertia of a cube rotating about one of its corners. The principal axis is a vector that traverses the diagonal of the cube.

## Euler's Equations

### Space Frame and Body Frame

Space Frame
An inertial reference frame used to locate the center of mass of a rigid body
Body Frame
A non-inertial reference frame that uses the principal axes of a rotating rigid body

.

If a body has an angular momentum and principal moments and

Then the angular momentum in the Body reference frame is given by

The torque in the Space reference frame is given by

is moving as seen in the Space frame

remembering that

one can see that

: Euler's equation

#### Euler's equation for the spinning top

In the case of the spinning top:

The external torque applied by gravity is always normal to the z-axis () so value of the torque is zero in the direction

but

so

the tops angular velocity along the symmetry axis is constant.

### Euler's Equation for Zero Torque

#### Two principal moments are the same value (Free precession)

The motion of a top is an example of this situation. As seen above for the motion of the top,

The torque is alway zero in the z-direction
constant

The remaining Euler equations are

IF the top is aligned with the z-axis then there will be no torque. (But as soon as it is misaligned there will be a torque)

The above Euler equations become

where

let

The we as a coupled set of equations as we saw before

using the trick of defining a complex frequency such that

we have the first order differential equation

If we assume the initial conditions
at and

Then

Interpretation
the two components, and , of the rigid body's angular velocity are rotating with an angular velocity about the axis as viewed in the Body reference frame.
Body cone
The Body cone is defined as the orbital path taken by the top as it rotates about the axis with the angular velocity . The shape forms a cone relative to the rotation point of the top (the top's tip)

The angular momentum ( ) in the space frame is given by

Observation
The Body frame (the reference frame attached to the spinning top) sees and precess about at an angular velocity of