Principal Axes and Principal Moments of Inertia for a Rigid Body

                In this post, I will discuss the procedure for determining the principal axes and principal moments of inertia for a rigid body.  The principal axes of a body are a set of mutually perpendicular axes in which the resulting torques act independently of each other.  Furthermore, the principal moments of inertia are the respective inertial moments about these axes.

                Consider a cylinder of radius R and height h set such that the origin is at the center of the bottom base and the z axis is parallel with the height of the cylinder.  To begin, we will determine the center of mass of the cylinder.  This is done using the equation

where

Here, a represents the center of mass vector, r represents the unit vector pointing toward the center of mass, ρ is the mass density of the cylinder, V is the volume of the cylinder, and M is the total mass.  Additionally, as we are integrating over the volume of a cylinder, it is very helpful to convert to cylindrical coordinates.  For our cylinder, the center of mass is found to be

                The next step in our process is determining the inertial tensor for our rigid body.  As a result of a complex and lengthy derivation of which I will not be explaining in this post, we can construct this tensor using

In this equation J_ij represents our tensor with indices i and j, δ_ij is the kronecker delta operator, and x_k, x_i, and x_j represents our various axes with respect to the indices’ value.  Here, it is important to note that the kronecker delta operator has a value of 1 when i=j and a value of 0 when i≠j.  Upon computation of this tensor, we obtain

                Next, to determine our principal axes and principal moments of inertia, we must apply Steiner’s parallel-axis theorem to construct an inertial tensor relative to the center of mass of the cylinder.  This tensor can be solved for using the general equation

Here, I_ij represents the adjusted inertial tensor and a is our center of mass vector.  In application to our cylinder, we obtain

Finally, the principal axes of the body are the eigenvectors of tensor I_ij while the principal moments of inertia are the respective eigenvalues of I_ij.  For this body, the principal axes and principal moments of inertia are

Elastic Collisions in the CM Frame

                As mentioned in Elastic Collisions in the LAB Frame, there are two ways to solve elastic collision problems.  The first of which is in the LAB frame and was demonstrated in the previous post.  In this post, I will demonstrate how to avoid tedious algebra and the need to solve systems of equations by “boosting” into the center of mass (CM) frame.

                The main benefit of boosting into the CM frame is that contributions of linear momentum are eliminated.  When taking the CM frame approach, the process consists of four steps:

     1. Boost into CM frame

     2. Find velocities in CM frame

     3. Boost out of CM frame

     4. Solve for scattering angles in LAB frame using the determined final velocities

To begin, we will consider the same scenario as in the previous post.

As before, we will assume particles of initial masses m₁ and m₂ and initial velocities u₁ and u₂.  While these velocities can be vectors in any direction, we will assume u₁ and u₂ are only along the x axis.  Furthermore, as I previously mentioned, we must know at least one thing about what happens after the collision and will assume that particle 1 scatters at an angle of θ₁.

                First, we will boost into the CM frame.  To do this, we must determine the velocity of the system’s center of mass and then appropriately adjust the initial velocities from the LAB frame such that we observe them from the CM frame.  To determine the velocity of the system’s center of mass, we use

As we boost to the CM frame, the scenario will look a bit different.  Initially, it will appear as though the two particles are approaching the origin, or center of mass, from opposite sides.  Once they collide, the two particles will simply scatter in opposite directions. 

To properly adjust the initial velocities of the particles such that we observe them from the CM frame, we subtract the CM velocity V.  Thus,

and

where u’₁ and u’₂ are the initial velocities of the particles in the CM frame.

 As the particles are always moving in opposite directions, our linear momentum will always cancel, thus we must only worry about the conservation of kinetic energy.  Through algebraic manipulation of our energy and momentum relations, we determine that

and 

Here, v’₁ and v’₂ are the final velocities of the particles within the CM frame.

                Finally, we want to boost back out of the CM frame to determine our scattering angle θ₂.  To do this, we simply add back the velocity V we originally subtracted from each velocity and then use basic trigonometric relations.  Thus,

 and

are used to determine the particle velocities in the LAB frame.  Then, to determine θ₂, we rearrange the equation for the conservation of momentum, determining

Applying this technique for the case in which

 

and

we find that

and

These results are very close to those using the LAB frame procedure.  Most likely, this discrepancy is simply due to rounding error somewhere within our calculations.

Elastic Collisions in the LAB Frame

                While riddled with tedious algebra, elastic collision problems are quite simple to solve given the correct procedure.  In fact, there are two ways to solve them: the LAB frame and the CM frame.  The LAB frame refers to the point of view of an observer in a stationary reference frame while the CM frame, or center of mass frame, refers to the point of view of an observer from a frame moving with the center of mass of the colliding particles.  While it is necessary to know something about what happens after the collision in either case, approaching the scenario from the CM frame eliminates linear momentum and does not require solving a system of equations to determine the motion of the system.  In this blog, I will demonstrate the process for solving a system of colliding particles in the LAB frame.

                When taking the LAB frame approach, one must simply formulate and solve the system of equations describing the conservation of energy and linear momentum of the collision.  Consider the collision below.

For this example, we will assume particles of initial masses m₁ and m₂ and initial velocities u₁ and u₂.  While these velocities can be vectors in any direction, we will assume u₁ and u₂ are only along the x axis.  Furthermore, as I previously mentioned, we must know at least one thing about what happens after the collision.  Thus, for this example, we will assume that particle 1 deflects at an angle of θ₁.  Next, we will construct our system of conservation equations.  As we know energy and linear momentum must be conserved, we know

and

where v₁ and v₂ represent the magnitude of the respective particles’ velocities after the collision and θ₂ is the scatter angle of the second particle.  The first equation is the conservation of kinetic energy while the second and third equations are the conservation of linear momentum in the x and y coordinates, respectively.  Finally, to determine the two particles’ velocities and particle 2’s scatter angle after the collision, we must simply solve the system of conservation energies.  Applying this technique for the case in which

and 

we find that

and

In the next post, I will demonstrate how to solve this same scenario from the CM frame.

Gravitational Forces

                As a young child, I often wondered what it would be like to dig a hole from one side of the Earth to the other.  While this task is clearly impossible for many reasons, the concept introduces an interesting modeling problem investigating the effects and interactions of gravitational force.  Newton’s law of universal gravitation states that each mass particle attracts every other particle in the universe.  Furthermore, Newton says that this force varies directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.  Mathematically, we write this as

where G is the gravitational constant, m is the mass of one particle, M is the mass of a second particle, r is the distance between the particles, and r hat tells us this force occurs along the radial axis.  While this equation works for a system of point particles, it needs some improvement for applications involving bodies of continuous mass distribution.  Thus, we can use an integral to sum up the forces due to each infinitesimally small distribution of mass.  This provides

In this variation, ρ represents the mass density of the distribution as a function of its radius.  Moreover, this volume integral allows for us to sum the entire mass of the object of interest given any unique mass distribution.

                Applying this to our problem of dropping an object down a hole spanning the diameter of the Earth, we can determine the behavior of that force over time and thus model the path of the object.  To begin, we will simplify our model by assuming the Earth is perfectly spherical and has a uniform mass distribution ρ.  Furthermore, we will designate the mass of our object the variable m.  Now, due to the symmetry of our simplified Earth, we can effectively apply a version of Gauss’s Law in determining our gravitational force.  In short, this means we only need to account for the volume of Earth’s mass within a radius r, or distance of the dropped object, to the center of our Earth.  Therefore, as our object gets closer to the center of the Earth the volume of mass integrated over will decrease and vice versa.  Upon evaluation of our gravitational force equation, we see

Interestingly, this force vector looks very similar to, and acts the same as, that of a simple harmonic oscillator.  The force for simple harmonic motion is

where k is the spring constant and x is the distance from equilibrium, or in our case the radius from the center of the Earth.  The motion of the object is modeled in the animation below.

        

Given that the object undergoes simple harmonic motion, we can determine the period of the objects flight.  The period of a simple harmonic oscillator is given by 

Substituting in our gravitational equivalence to k,

Finally, plugging in realistic values, we find that the period for the object to reach the other side of the world and return, neglecting air resistance, is approximately 84 minutes.

Hamiltonians

                While the formulation of a Lagrangian and Lagrange equations of motion can greatly simplify an otherwise complicated system, it is not the only way.  Similarly to the application of Lagrangians described in Hamilton’s Principle and Lagrangians and Lagrangians Pt. 2, the formulation of a system’s Hamiltonian provides easy access to the time depended equations of motion for each generalized coordinate necessary to describe the system.  To begin, we determine the Lagrangian

as before, where q represents the chosen general coordinates and T and U represent the kinetic and potential energies of the system, respectively.  Next, we construct our Hamiltonian equation given by

While I will not be further discussing its derivation, this generalized Hamiltonian equation is the result of a Legendre transform of the generalized Lagrangian.  Here, p_k represents the momentum associated with the respective generalized coordinate q_k.  Furthermore, as the Hamiltionian is a function of q_k, p_k, and t instead of q_k, dq_k/dt, and t as is the Lagrangian, it is necessary that each velocity present within the equation be replaced with an equivalent expression in terms of momentum provided by the relation

Finally, the relations

and

known as the Hamilton equations of motion, provide the system’s 2n first order differential equations.  By applying initial conditions and solving this system of differential equations, time depended equations of motion for each generalized coordinate can be determined.  Below is an example of this process’s application to a simple pendulum.

Example:

(Based on Problem 7.24 from Classical Dynamics of Particles and Systems 5^th Edition by Thornton and Marion)

Consider a simple plane pendulum with a fixed suspension point consisting of a mass m attached to a string of length ℓ. After the pendulum is set into motion, the length of the string is shortened at a constant rate

Determine the Hamilton equations of motion for this system.

To begin, it will be useful to define the length of the pendulum ℓ by integrating over time the rate at which it is shortened.  Thus, we find

where L is the original length of the pendulum.  Next, we determine the pendulum’s Lagrangian.  As

we determine that

where g is gravitational acceleration.  Third, we construct our Hamiltonian equation

but as it must be in terms of momentum rather than velocity, we substitute the relation

Now, with our Hamiltonian

we can determine our Hamiltonian equations of motion

and

Below is an animation of this pendulum’s motion given that L=4.0m, α=0.5m/s, m=1.0kg, g=10 m/s^2, θ(0)=π/4 radians, and θ’(0)=0 radians/s.