Please check Cross product solutions

In summary: Your Name]In summary, the conversation discusses the calculation of the inertia matrix for two exercises involving orbital angular momentum and kinetic energy. The approach involves finding the total angular momentum and kinetic energy for each element, and then summing them to find the total values. However, there are some missing vector signs and factors in the final expressions, which should be included for clarity and accuracy. It is also noted that the inertia matrix is diagonal, with each element representing the moment of inertia for a specific element.
  • #1
ognik
643
2
I'm not 100% confident of my approach to the 2 exercises below:

Orbital angular momentum of i'th element is $\vec{L_i} = \vec{r_i} \times \vec{p_i} = m_i \vec{r_i} \times (\omega \times \vec{r_i}) $

a) Find the inertia matrix $I$ such that (omitting vector signs from here on) $L = I \omega, |L \rangle = I|\omega \rangle$

$=> L_i = - m_i[r_i \times r_i \times \omega]$ (cyclic anti-commute)
$ =-m_i[r_i(r_i \cdot \omega) - \omega (r_i \cdot r_i)] $ (BAC-CAB)
$ = m_i\omega r_i^2 $ ($r_i$ and $\omega$ orthogonal)
$ L = \sum L_i = \sum (m_i r_i^2)\omega =\sum I_i \omega = I \omega, i.e. |L \rangle = I| \omega \rangle$

b)Similarly derive I starting with Kinetic Energy $T_i = \frac{1}{2}m_i(\omega \times r_i)^2 $

$ = \frac{1}{2}m_i [|\omega|^2 |r_i|^2 - (\omega \cdot r_i)^2] $ (identity)
$ = \frac{1}{2} \omega (m_i r_i^2) \omega $ ($r_i$ and $\omega$ orthogonal)
$ \therefore T= \sum T_i = \frac{1}{2} \omega \sum(m_i r_i^2) \omega = \frac{1}{2} \omega I \omega = \frac{1}{2}\langle \omega |I| \omega \rangle$
 
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  • #2

Thank you for sharing your approach to finding the inertia matrix in these exercises. I would like to offer some feedback and clarification on your work.

Firstly, in part a), you have correctly calculated the orbital angular momentum of the i'th element using the given equation. However, your final expression for the total angular momentum, $L$, is missing a vector sign. It should be $L = \sum \vec{L_i}$, where each $\vec{L_i}$ is a vector quantity. This is important because the inertia matrix, $I$, is also a vector quantity and should be represented as such in your final expression.

Additionally, in part b), you have correctly calculated the kinetic energy of the i'th element. However, your final expression for the total kinetic energy, $T$, is missing a vector sign and a factor of 1/2. It should be $T = \frac{1}{2} \sum \vec{T_i} = \frac{1}{2} \langle \vec{\omega} | I | \vec{\omega} \rangle$.

In both parts, it would also be helpful to explicitly state that $I$ is a diagonal matrix, with each element $I_i$ representing the moment of inertia of the i'th element.

Overall, your approach is sound and your derivations are correct. However, in order to fully convey your ideas and solutions, it is important to include all necessary vector signs and factors in your final expressions. Keep up the good work and don't hesitate to reach out if you have any further questions or concerns.


 

FAQ: Please check Cross product solutions

What is the cross product?

The cross product, also known as the vector product, is a mathematical operation that takes two vectors as inputs and returns a third vector that is perpendicular to both input vectors. It is denoted by the symbol "×" or "⨯".

How is the cross product calculated?

The cross product can be calculated by taking the determinant of a 3x3 matrix composed of the unit vectors i, j, and k and the components of the two input vectors. The resulting vector is the cross product of the two input vectors.

What is the significance of the cross product in physics?

The cross product has many applications in physics, particularly in mechanics and electromagnetism. It is used to calculate torque, angular momentum, and magnetic fields, among other things.

Can the cross product be used in higher dimensions?

No, the cross product is only defined for three-dimensional vectors. In higher dimensions, a similar operation called the exterior product is used.

How can I check my solutions for the cross product?

To check your solutions for the cross product, you can use the properties of the cross product, such as the fact that it is perpendicular to both input vectors and has a magnitude equal to the product of the magnitudes of the input vectors times the sine of the angle between them. You can also use online calculators or software to verify your results.

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