Rotation: Instantaneous velocity and Instantaneous acceleration

[Win_odd Dhamnekar]

Rotation: Part 2

Rotation: Part 3

Now in the Rotation: Part 3



How to prove that \(\displaystyle \omega \times R\dot{\rho} + \dot{R}(R^TR)\dot{\rho} = 2\omega \times (R\dot{\rho})\)

I know both terms are equal. But what kind of vector triple product algebra was used to make both terms equal?



My other question is how to prove that the magnitude and the direction of the last term of the instantaneous acceleration are \(\displaystyle \omega r^2 \) and -r respectively.

 

[Win_odd Dhamnekar]

In the above question, please read the magnitude of the last term of instantaneous aceleration is \(\displaystyle r\omega^2\)

The magnitude of \(\displaystyle |\omega \times (\omega \times r )|= (\omega \cdot r)\omega - (\omega \cdot \omega)r =|-r\omega^2| =r\omega^2\)

Note: The angle between the \(\displaystyle \overrightarrow{\omega}\) and \(\displaystyle \overrightarrow{r}\) is \(\displaystyle \frac{\pi}{2}\)

How is the direction of \(\displaystyle \omega \times(\omega \times r)= -r?\)

 

[topsquark]

[imath]\omega \times R \dot{ \rho } + \dot{R} \dot{ \rho } = \omega \times (R \dot{ \rho }) + \dot{R} (R^T R) \dot{ \rho }[/imath]

[imath]= \omega ( R \dot{ \rho }) + ( \dot{R} R^T ) (R \dot{ \rho } )[/imath]

Remember your last post? [imath]\dot{R} R^T b = \omega \times b[/imath]

So [imath]\omega \times R \dot{ \rho } + \dot{R} \dot{ \rho } = = \omega ( R \dot{ \rho }) + ( \dot{R} R^T ) (R \dot{ \rho } ) = \omega \times (R \dot{ \rho }) + \omega \times (R \dot{ \rho } ) = 2 \omega \times (R \dot{ \rho )}[/imath]

As to your other question, the quick way is to use the right hand rule twice. But if you want the math let, for example, [imath]\omega = \hat{k}[/imath] and [imath]r = \hat{i}[/imath]. Then [imath]\omega \times ( \omega \times r ) = \omega \times ( \hat{j} ) = - \hat{i}[/imath].

-Dan

 

[Win_odd Dhamnekar]

[imath]\omega \times R \dot{ \rho } + \dot{R} \dot{ \rho } = \omega \times (R \dot{ \rho }) + \dot{R} (R^T R) \dot{ \rho }[/imath]

[imath]= \omega ( R \dot{ \rho }) + ( \dot{R} R^T ) (R \dot{ \rho } )[/imath]

Remember your last post? [imath]\dot{R} R^T b = \omega \times b[/imath]

So [imath]\omega \times R \dot{ \rho } + \dot{R} \dot{ \rho } = = \omega ( R \dot{ \rho }) + ( \dot{R} R^T ) (R \dot{ \rho } ) = \omega \times (R \dot{ \rho }) + \omega \times (R \dot{ \rho } ) = 2 \omega \times (R \dot{ \rho )}[/imath]

As to your other question, the quick way is to use the right hand rule twice. But if you want the math let, for example, [imath]\omega = \hat{k}[/imath] and [imath]r = \hat{i}[/imath]. Then [imath]\omega \times ( \omega \times r ) = \omega \times ( \hat{j} ) = - \hat{i}[/imath].

-Dan

What would you say about #2?

Is the angle between \(\displaystyle \overrightarrow{\omega}\) and \(\displaystyle \overrightarrow{r} \frac{\pi}{2}?\)

 

[topsquark]

What would you say about #2?

Is the angle between \(\displaystyle \overrightarrow{\omega}\) and \(\displaystyle \overrightarrow{r} \frac{\pi}{2}?\)

Technically r(t) points out from the origin and the axis of rotation is not the z-axis so in general r and [imath]\omega[/imath] are not perpendicular. Note, though, that [imath]\omega \times r[/imath] will have a magnitude of [imath]sin( \alpha ) \omega r[/imath], where [imath]\alpha[/imath] is the angle between r and [imath]\omega[/imath] and that [imath]\omega \times r[/imath] will be perpendicular to [imath]\omega[/imath] due to the nature of the cross product. That means [imath]\omega \times r[/imath] is in the plane of rotation so [imath]| \omega \times ( \omega \times r )| = sin( \alpha ) \omega ^2 r[/imath].

For simplicity we almost always do these problems in the plane of rotation and just let [imath]\alpha = \dfrac{ \pi }{2}[/imath] rad.

-Dan