Thursday, 12 March 2020

homework and exercises - How can I relate linear and angular motion using a single formula?




I want to relate linear and angular motion using a single formula.



Assume I have a 10m rod, and I apply a force of 5N on it, 2.5m away from the axis of rotation for 1s. How can I determine the resultant translational and rotational velocities?





This question has been subject to a lot of change since the time of asking, as it is based on a great misconception in my interpretation of rotational dynamics.


My fallacy was in thinking that resultant translational motion varies by the distance between the point of force application and the center of rotation. The reality, however, is that if a constant force is applied at any point of a rigid body system, the resultant translational motion will remain constant, given that time of application is constant as well.


What will change, however, is the energy required to sustain this constant force over a constant duration. This energy varies by the distance between the point of force application and center of mass.



Answer



Use the spatial inertia to relate linear/angular momentum to changes in linear/angular speed



$$\begin{pmatrix} \vec{L}\\ \vec{H}_A \end{pmatrix} = \begin{bmatrix} m {\bf 1}_{3×3} & -m [\vec{c}\times] \\ m [\vec{c}\times] & I_{cm}-m [\vec{c}\times][\vec{c}\times] \end{bmatrix} \begin{pmatrix} \Delta\vec{v}_A\\ \Delta \vec{\omega} \end{pmatrix}$$


ref: Newton–Euler equations wikipedia.


where $\vec{L}$ is net linear momentum (impulse), $\vec{H}_A$ is the near angular momenum at a point A. $\Delta \vec{v}_A$ is the change in linear speed at the same point and $\Delta{\omega}$ the change in rotational speed.


The scalar mass is $m$ and $I_{cm}$ is the mass moment of inertia at the center of mass. Finally $[\vec{c}\times$] is the 3×3 skew symmetric matrix representing the vector cross product operator. See https://physics.stackexchange.com/a/111081/392. The vector $\vec{c}$ is the location of the center of mass here relative to A.


The combined impulse, as well as the resulting motion is a spatial screw of which the ratio of linear to angular properties is called a pitch. See https://physics.stackexchange.com/a/80552/392 on more details. Also look at the table below to see how to extract the screw properties (position, direction & pitch) from different types of screws.


$$ \begin{array}{cccc} \mbox{Property} & \mbox{Velocity (Twist)} & \mbox{Momentum (Wrench)} & \mbox{Force (Wrench)}\\ \hline \mbox{Screw} & \hat{v}_{A}=\begin{pmatrix}\vec{v}_{A}\\ \vec{\omega} \end{pmatrix} & \hat{\ell}_{A}=\begin{pmatrix}\vec{L}\\ \vec{H}_{A} \end{pmatrix} & \hat{f}_{A}=\begin{pmatrix}\vec{F}\\ \vec{\tau}_{A} \end{pmatrix}\\ \mbox{Direction} & \vec{e}=\frac{\vec{\omega}}{|\vec{\omega}|} & \vec{e}=\frac{\vec{L}}{|\vec{L}|} & \vec{e}=\frac{\vec{F}}{|\vec{F}|}\\ \mbox{Magnitude} & \omega=|\vec{\omega}| & L=|\vec{L}| & F=|\vec{F}|\\ \mbox{Position} & \vec{r}=\vec{r}_{A}+\frac{\vec{\omega}\times\vec{v}_{A}}{|\vec{\omega}|^{2}} & \vec{r}=\vec{r}_{A}+\frac{\vec{L}\times\vec{H}_{A}}{|\vec{L}|^{2}} & \vec{r}=\vec{r}_{A}+\frac{\vec{F}\times\vec{\tau}_{A}}{|\vec{F}|^{2}}\\ \mbox{Pitch} & h=\frac{\vec{\omega}\cdot\vec{v}_{A}}{|\vec{\omega}|^{2}} & h=\frac{\vec{L}\cdot\vec{H}_{A}}{|\vec{L}|^{2}} & h=\frac{\vec{F}\cdot\vec{\tau}_{A}}{|\vec{F}|^{2}} \end{array} $$


PS. If you provide more specific details I can provide with an exmple of how the above can me used to get what you want.


Edit 1


The inverse spatial inertia matrix is


$$\begin{pmatrix}\Delta\vec{v}_{A}\\ \Delta\vec{\omega} \end{pmatrix}=\begin{bmatrix}\frac{1}{m}-\vec{c}\times I_{cm}^{-1}\vec{c}\times & \vec{c}\times I_{cm}^{-1}\\ -I_{cm}^{-1}\vec{c}\times & I_{cm}^{-1} \end{bmatrix}\begin{pmatrix}\vec{L}\\ \vec{H}_{A} \end{pmatrix}$$



This is used in game programming to handle impulses (see equation 18-8 in http://www.cs.cmu.edu/~baraff/sigcourse/notesd2.pdf)


Edit 2


The kinetic energy of a rigid body with 6×6 spatial inertia matrix $\hat{I}_A$ is defined by


$$K=\frac{1}{2} \hat{v}_A^\top \hat{I}_A \hat{v}_A =\frac{1}{2} \hat{\ell}_A^\top \hat{I}_A^{-1} \hat{\ell}$$


where $\hat{v}_A = (\vec{v}_A,\vec{\omega})$ and $\hat{\ell}_A = (\vec{L},\vec{H}_A)$.


The simplest way to deal with your case is to consider the inertia at the center of mass and fix the angular momentum as $\vec{H}_C = -\vec{c}\times \vec{L}$ producing the spatial momentum screw $\hat{\ell}_C = (\vec{L}, -\vec{c} \times \vec{L})$ with the impulse $\vec{L} = \int \vec{F}\,{\rm d}t$. The inverse spatial inertia at the center of mass is a block diagonal 6×6 matrix $\hat{I}_C^{-1} = \begin{bmatrix} \frac{1}{m} & \\ & I_C^{-1} \end{bmatrix}$ producing the kinetic energy


$$ K = \frac{1}{2}\begin{pmatrix}\vec{L}\\ -\vec{c}\times\vec{L} \end{pmatrix}^{\top}\begin{bmatrix}\frac{1}{m}\\ & I_{C}^{-1} \end{bmatrix}\begin{pmatrix}\vec{L}\\ -\vec{c}\times\vec{L} \end{pmatrix} $$


The above can be expanded using vector identies into


$$\boxed{ K=\frac{1}{2}\vec{L}^{\top}\left(\frac{1}{m}{\bf 1}_{3×3}-[\vec{c}\times] I_{C}^{-1}[\vec{c}\times]\right)\vec{L} }$$


The part inside the parenthesis is the inverse effective mass of the rigid body about point A. The first part is due to the linear momentum and the second part due to the angular momentum (force offset). A rod of length $s=10 \rm m$ has mass moment of inertia $I_{cm} = {\rm diag}(\frac{1}{12}m s^2, \frac{1}{12}m s^2, 0)$ and point of application located at $a=2.5 \rm m$ to the right of the center of mass (with $\vec{c}=(-a,0,0)$).



If the applied force is along the vertical then $\vec{L} =(0,\int F{\rm d}t,0)$ and the kinetic energy is $$K = \frac{1}{2} \left( \frac{1}{m} + \frac{12 a^2}{m s^2} \right) L^2$$ where $L=\int F{\rm d}t$ is the impusle magnitude.


Edit 3


Of course you can directly arrive at the above by evaluating the inverse spatial inertia $$\hat{I}_{A}^{-1}=\begin{bmatrix}\frac{1}{m} & 0 & \ldots & 0\\ 0 & \frac{1}{m}+\frac{12a^{2}}{ms^{2}} & & \frac{12a}{ms^{2}}\\ \vdots & & \ddots\\ 0 & \frac{12a}{ms^{2}} & & \frac{12}{ms^{2}} \end{bmatrix}$$ and looking at the [2,2] element (along the translational y axis).


Given an impulse $\vec{L} = (0,J,0)$ the resulting motion at the force point A is $$ \begin{pmatrix} \Delta v_x \\ \Delta v_y \\ \Delta \omega \end{pmatrix} = \begin{bmatrix} \frac{1}{m} & 0 & 0 \\ 0 & \frac{1}{m}+\frac{12 a^2}{m s^2} & \frac{12 a}{m s^2} \\ 0 & \frac{12 a}{m s^2} & \frac{12}{m s^2} \end{bmatrix} \begin{pmatrix} 0 \\ J \\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ \left(\frac{1}{m}+\frac{12 a^2}{m s^2}\right) J\\ \frac{12 a}{m s^2} J \end{pmatrix} $$


Edit 4


The center of rotation of the rigid body (which gives you an idea on how much it rotates vs. how much it translates) is found by (position of screw axis)


$$ \vec{r}_{rot} = \vec{r}_A + \dfrac{\Delta \vec{\omega} \times \Delta \vec{v}}{|\Delta \vec{\omega}|^2} =\begin{pmatrix}a\\0\\0\end{pmatrix} + \frac{(0,0,\frac{12 a}{m s^2}J) \times (0, (\frac{1}{m}+\frac{12 a^2}{m s^2})J,0)}{(\frac{12 a}{m s^2}J)^2} =\begin{pmatrix} -\frac{s^2}{12 a} \\ 0 \\ 0 \end{pmatrix} $$


As you can see the amount of force (or impulse) is unimportant as the geometry of the problem is defined by the line of action of the force and the inertial properties of the rigid body. In this case the center of rotation is at $r=\frac{10}{3}\,{\rm m}$ to the left of the center of mass. The ratio of linear velocity of point A to linear velocity of center of mass is $$ \frac{v_A}{v_{cm}} = \dfrac{(a+r)\Delta \omega}{r \Delta \omega} = \dfrac{a+\frac{s^2}{12 a}}{\frac{s^2}{12 a}} = \frac{7}{4} $$


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