We found that from the mechanics of system of particles, Total kinetic energy can be split into two parts as the sum of kinetic energy of the all system of particles with its total mass at the centre of mass position plus the kinetic energy of the all the system of particles about the the centre of mass point.
It is like splitting the motion into pure translational and pure rotational motion. Not only in mechanics, but in many places of physics, this kind of separation into purely translational and rotational is possible.
So, it is always possible in mechanics, choosing our coordinate system at the centre of mass or at the point that is stationary in a rotational motion.
About this stationary point, all the measurements will include only rotational quantities.
Let us determine the angular momentum of this system of particles about this stationary point,
$$\vec{L} = m_i \vec{r_i}\times \vec{v_i} $$ The summation convention is implied.
Applying the known relation of $$ \vec{v_i} = \vec{\omega}\times\vec{r_i} $$
Therefore, $$ \vec{L} = m_i [\vec{r_i} \times (\vec{\omega}\times \vec{r_i}) ] $$
Using the cross product rule, $$ \vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C} $$
$$ \vec{L} = m_i [r_i^2\vec{\omega} - (\vec{r_i}\cdot\vec{\omega})\vec{r_i}] \,\,\,...eq.(1)$$
Or, writing a general component of angular momentum vector in Einstein notation, $$ L_a = m_i [ \omega_a r_b^2 - (r_{ib}\omega_{ib})r_a] \,\,\,...eq.(2)$$ with Einstein summation convention.
If I consider only 3 dimensional space, then b - runs from 1 to 3 (or) x,y,z .
From the equation of angular momentum, we can know that the angular momentum does depend not only on the angular momentum about its axis lets say \omega_a but depends on all other axis where in \omega_b , b - runs from all a,b,c (if it is 3 dimension).
If I need to take out $\omega$ then eq.(2) becomes, $$ L_a = I_{ab} \omega_b $$
where $$ I_{ab} = m_i (r_i^2 \delta_{ab} - r_a r_b) $$ If we want to write the mass in terms of integration, then, $$ I_{ab} = \int_V \rho(\vec{r}) (r_i^2\delta_{ab} - r_ar_b) \,dV $$
Writing in terms of vectors, $$ \vec{L} = I\vec{\omega} $$
where "I" is called moment of Inertia Tensor.
It is like splitting the motion into pure translational and pure rotational motion. Not only in mechanics, but in many places of physics, this kind of separation into purely translational and rotational is possible.
So, it is always possible in mechanics, choosing our coordinate system at the centre of mass or at the point that is stationary in a rotational motion.
About this stationary point, all the measurements will include only rotational quantities.
Let us determine the angular momentum of this system of particles about this stationary point,
$$\vec{L} = m_i \vec{r_i}\times \vec{v_i} $$ The summation convention is implied.
Applying the known relation of $$ \vec{v_i} = \vec{\omega}\times\vec{r_i} $$
Therefore, $$ \vec{L} = m_i [\vec{r_i} \times (\vec{\omega}\times \vec{r_i}) ] $$
Using the cross product rule, $$ \vec{A}\times(\vec{B}\times\vec{C}) = (\vec{A}\cdot\vec{C})\vec{B} - (\vec{A}\cdot\vec{B})\vec{C} $$
$$ \vec{L} = m_i [r_i^2\vec{\omega} - (\vec{r_i}\cdot\vec{\omega})\vec{r_i}] \,\,\,...eq.(1)$$
Or, writing a general component of angular momentum vector in Einstein notation, $$ L_a = m_i [ \omega_a r_b^2 - (r_{ib}\omega_{ib})r_a] \,\,\,...eq.(2)$$ with Einstein summation convention.
If I consider only 3 dimensional space, then b - runs from 1 to 3 (or) x,y,z .
From the equation of angular momentum, we can know that the angular momentum does depend not only on the angular momentum about its axis lets say \omega_a but depends on all other axis where in \omega_b , b - runs from all a,b,c (if it is 3 dimension).
If I need to take out $\omega$ then eq.(2) becomes, $$ L_a = I_{ab} \omega_b $$
where $$ I_{ab} = m_i (r_i^2 \delta_{ab} - r_a r_b) $$ If we want to write the mass in terms of integration, then, $$ I_{ab} = \int_V \rho(\vec{r}) (r_i^2\delta_{ab} - r_ar_b) \,dV $$
Writing in terms of vectors, $$ \vec{L} = I\vec{\omega} $$
where "I" is called moment of Inertia Tensor.
No comments:
Post a Comment
Let everyone know what you think about this