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Eli
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Lets look at the equations

$$\frac{d}{dt}\left(\mathbf I\,\vec\omega\right)=\vec\tau$$

if the torque $~\vec\tau~$ is equal zero then the momentum

$$\vec L=\mathbf I\,\vec\omega=\vec c$$ where $\vec c$ is constant

the angle between $~\vec L~$ and $~\vec\omega~$ is proportional to the $~\vec \omega ^T\,\vec L$ thus

$$\theta_1\propto \vec\omega^T\,\mathbf I\,\vec\omega\propto \vec\omega^T\vec c\tag 1$$

if the torque is not equal zero then

$$\vec L=\mathbf I\,\vec\omega=\int \vec\tau\,dt=\vec f(t)$$ and the angle is now

$$\theta_2\propto \vec\omega^T\vec f(t)\tag 2$$

thus angle $~\theta_1\ne \theta_2$


$$\cos(\theta)=\frac{\vec a^T\,\vec b}{a\,b}$$

Lets look at the equations

$$\frac{d}{dt}\left(\mathbf I\,\vec\omega\right)=\vec\tau$$

if the torque $~\vec\tau~$ is equal zero then the momentum

$$\vec L=\mathbf I\,\vec\omega=\vec c$$ where $\vec c$ is constant

the angle between $~\vec L~$ and $~\vec\omega~$ is proportional to the $~\vec \omega ^T\,\vec L$ thus

$$\theta_1\propto \vec\omega^T\,\mathbf I\,\vec\omega\propto \vec\omega^T\vec c\tag 1$$

if the torque is not equal zero then

$$\vec L=\mathbf I\,\vec\omega=\int \vec\tau\,dt=\vec f(t)$$ and the angle is now

$$\theta_2\propto \vec\omega^T\vec f(t)\tag 2$$

thus angle $~\theta_1\ne \theta_2$

Lets look at the equations

$$\frac{d}{dt}\left(\mathbf I\,\vec\omega\right)=\vec\tau$$

if the torque $~\vec\tau~$ is equal zero then the momentum

$$\vec L=\mathbf I\,\vec\omega=\vec c$$ where $\vec c$ is constant

the angle between $~\vec L~$ and $~\vec\omega~$ is proportional to the $~\vec \omega ^T\,\vec L$ thus

$$\theta_1\propto \vec\omega^T\,\mathbf I\,\vec\omega\propto \vec\omega^T\vec c\tag 1$$

if the torque is not equal zero then

$$\vec L=\mathbf I\,\vec\omega=\int \vec\tau\,dt=\vec f(t)$$ and the angle is now

$$\theta_2\propto \vec\omega^T\vec f(t)\tag 2$$

thus angle $~\theta_1\ne \theta_2$


$$\cos(\theta)=\frac{\vec a^T\,\vec b}{a\,b}$$

Source Link
Eli
  • 12.9k
  • 2
  • 11
  • 31

Lets look at the equations

$$\frac{d}{dt}\left(\mathbf I\,\vec\omega\right)=\vec\tau$$

if the torque $~\vec\tau~$ is equal zero then the momentum

$$\vec L=\mathbf I\,\vec\omega=\vec c$$ where $\vec c$ is constant

the angle between $~\vec L~$ and $~\vec\omega~$ is proportional to the $~\vec \omega ^T\,\vec L$ thus

$$\theta_1\propto \vec\omega^T\,\mathbf I\,\vec\omega\propto \vec\omega^T\vec c\tag 1$$

if the torque is not equal zero then

$$\vec L=\mathbf I\,\vec\omega=\int \vec\tau\,dt=\vec f(t)$$ and the angle is now

$$\theta_2\propto \vec\omega^T\vec f(t)\tag 2$$

thus angle $~\theta_1\ne \theta_2$