- #1
Master1022
- 611
- 117
- Homework Statement
- How to use these equations of motion
- Relevant Equations
- ## r = R + \rho ##
## \dot r = \dot R + (\Omega \times \rho) + \dot \rho ##
## \ddot r = \ddot R + \ddot \rho + 2(\Omega \times \dot \rho) + \dot \Omega \times \rho + \Omega \times (\Omega \times \rho) ##
Hi,
I am just writing a post to follow up on a previous thread I made which I don't think was very clear. The question is mainly about how to use the below equations when there is also a rotation of the body around the fixed reference point.
Please see the diagram here to see how the vectors have been defined:
We can then derive the following equations of motion:
$$ r = R + \rho $$
$$ \dot r = \dot R + (\Omega \times \rho) + \dot \rho $$
$$ \ddot r = \ddot R + \ddot \rho + 2(\Omega \times \dot \rho) + \dot \Omega \times \rho + \Omega \times (\Omega \times \rho) $$
However, what happens if there is also a rotation of A around O (lets say that the angular velocity is ## \omega ##)? How do we now use the equations to find the absolute velocity and acceleration of B?
My approach:
In an attempt to be clear and not keep re-using terms, let us define ## \alpha ## to be the rotation of the body (about A) relative to the the rotation of A around O and ## \omega ## is defined above.
Velocity:
The equation is:
$$ \dot r = \dot R + (\Omega \times \rho) + \dot \rho $$
Does ## \dot R ## include the rotation of point A around O and therefore become ## \dot R = (\omega \times R)##? I originally thought that vector only included a linear translational velocity...
Now for the other two terms, one problem I was working seems to suggest that the rotation of B around point A is included in the ## \dot \rho ## term and not the ## (\Omega \times \rho) ## term. Therefore, it suggests that ## \dot \rho = (\alpha \times \rho) ## where ## \alpha ## is defined as above. It also suggests that ## (\Omega \times \rho) = (\omega \times \rho) ##. I am struggling to understand why this is the case.
Acceleration:
The equation is:
$$ \ddot r = \ddot R + \ddot \rho + 2(\Omega \times \dot \rho) + \dot \Omega \times \rho + \Omega \times (\Omega \times \rho) $$
the ## \ddot R ## term contains ## \omega \times (\omega \times R) + \dot \omega \times R ## (it would also contain any linear relative acceleration but there is none).
the ## 2(\Omega \times \dot \rho) ## is the Coriolis term which is ## 2(\omega \times \dot \rho) ## where ## \dot \rho ## has been calculated above. This term makes sense to me.
the ## \ddot \rho ## term contains the ## \alpha \times (\alpha \times \rho) + \dot \alpha \times \rho ##
the ## \dot \Omega \times \rho ## term is ## \dot \omega \times \rho ##
the ## \Omega \times (\Omega \times \rho) = \omega \times (\omega \times \rho) ##
I am also struggling to see why the last two terms relate ## \omega ## and ## \rho ##.
Any correction/ clarification is greatly appreciated. In the lecture notes, we only saw situations with ## R = 0 ## and body just rotating around a fixed point, so this is new to me.
Thanks
I am just writing a post to follow up on a previous thread I made which I don't think was very clear. The question is mainly about how to use the below equations when there is also a rotation of the body around the fixed reference point.
Please see the diagram here to see how the vectors have been defined:
We can then derive the following equations of motion:
$$ r = R + \rho $$
$$ \dot r = \dot R + (\Omega \times \rho) + \dot \rho $$
$$ \ddot r = \ddot R + \ddot \rho + 2(\Omega \times \dot \rho) + \dot \Omega \times \rho + \Omega \times (\Omega \times \rho) $$
However, what happens if there is also a rotation of A around O (lets say that the angular velocity is ## \omega ##)? How do we now use the equations to find the absolute velocity and acceleration of B?
My approach:
In an attempt to be clear and not keep re-using terms, let us define ## \alpha ## to be the rotation of the body (about A) relative to the the rotation of A around O and ## \omega ## is defined above.
Velocity:
The equation is:
$$ \dot r = \dot R + (\Omega \times \rho) + \dot \rho $$
Does ## \dot R ## include the rotation of point A around O and therefore become ## \dot R = (\omega \times R)##? I originally thought that vector only included a linear translational velocity...
Now for the other two terms, one problem I was working seems to suggest that the rotation of B around point A is included in the ## \dot \rho ## term and not the ## (\Omega \times \rho) ## term. Therefore, it suggests that ## \dot \rho = (\alpha \times \rho) ## where ## \alpha ## is defined as above. It also suggests that ## (\Omega \times \rho) = (\omega \times \rho) ##. I am struggling to understand why this is the case.
Acceleration:
The equation is:
$$ \ddot r = \ddot R + \ddot \rho + 2(\Omega \times \dot \rho) + \dot \Omega \times \rho + \Omega \times (\Omega \times \rho) $$
the ## \ddot R ## term contains ## \omega \times (\omega \times R) + \dot \omega \times R ## (it would also contain any linear relative acceleration but there is none).
the ## 2(\Omega \times \dot \rho) ## is the Coriolis term which is ## 2(\omega \times \dot \rho) ## where ## \dot \rho ## has been calculated above. This term makes sense to me.
the ## \ddot \rho ## term contains the ## \alpha \times (\alpha \times \rho) + \dot \alpha \times \rho ##
the ## \dot \Omega \times \rho ## term is ## \dot \omega \times \rho ##
the ## \Omega \times (\Omega \times \rho) = \omega \times (\omega \times \rho) ##
I am also struggling to see why the last two terms relate ## \omega ## and ## \rho ##.
Any correction/ clarification is greatly appreciated. In the lecture notes, we only saw situations with ## R = 0 ## and body just rotating around a fixed point, so this is new to me.
Thanks