# How can I relate linear and angular motion using a single formula? [closed]

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.

## closed as unclear what you're asking by David Z♦Sep 6 at 8:59

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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 $$\frac{v_A}{v_{cm}} = \dfrac{(a+r)\Delta \omega}{r \Delta \omega} = \dfrac{a+\frac{s^2}{12 a}}{\frac{s^2}{12 a}} = \frac{7}{4}$$

• Are you looking at the same thing exactly? The op question was not clear of what ratio was requested. – ja72 May 17 '15 at 22:49
• You suppose. I still don't know, velocity at which point? Is the velocity of the force application point important to the op? Is the center of rotation point important to the op. If you have a particular situation you have doubts about you can ask your own question so that others have a chance or respond. Note that you shouldn't ask a "Check my work" question, rather you should ask a "How do I approach this situation/concept". – ja72 May 18 '15 at 13:20
• Actually the op doesn't want a ratio of velocities. The op wants a single (fundamental) equation relating linear and angular quantities. – ja72 May 18 '15 at 15:32
• Whatever op wants (if he knows) this answer is wrong: your equation is just the ratio between m and your m' which relates 2 linear v (quantities?), virtual $v_{m'}$ (which op ignores) and real $v_{cm}$ – user77485 May 25 '15 at 6:32
• $m' = m_{rod} \frac{I_{rod}}{I_{rod}+m_{rod} r^2}$ => $m' = m \frac{ms^2/12}{ms^2/12+m a^2}$ => $m' = m \frac{s^2/12}{s^2/12+ a^2}$ => $m' = m \frac{s^2}{12a^2 +s^2}$ => $m'/m = \frac{100}{175}=\frac{4}{7} ; m/m'= \frac{12a^2 +s^2} {s^2}=\frac{7}{4}$; here you ignore own precepts, use lengthy derivation, get same simple result, and misinterpret it: $\dfrac{a+\frac{s^2}{12 a}}{\frac{s^2}{12 a}} = \frac{12a^2+s^2}{12a}* \frac{12 a}{s^2} = \frac{12a^2 +s^2} {s^2}= \frac{175}{100}= m/m' = v_{m'}/v_{cm}=\frac{7}{4}$ – user77485 May 25 '15 at 7:51

I'm going to clarify once again. Assume I have a 10m rod (moment of inertia is known), and I apply a force of 5N on it, 2.5m away from the axis of rotation for one second. Both the linear and angular velocities could be found out by plugging in 5N into both F⃗ = ma⃗ and τ⃗ = Iα⃗, right? It seems kinda counter intuitive to me though. Wouldn't the force be distributed by some ratio over both linear and angular motion?

The question has been radically changed several times (consequently Steeven had to delete his answer, and another answer has become obsolete and incomprehensible). This deplorable fact makes it almost impossible to give an adequate answer. Another glaring anomaly is that OP has accepted this answer for the second or third time, and hasn't un-accepted it (yet) although it is being heavily downvoted, and , lastly, I am not allowed to delete my own answer since it is the accepted one.

I'll try to address the issue from a general point of view in order to cover all possible future changes: I'll consider a rod of $l=1m$ (10 cm would be even better) since it is unrealistic that one can push a rod of 10m and $I = 1/12$. The figures are not relevant anyway, since OP is not looking for a solution to a specific problem, but is asking for a general rule/principle/formula and the ratios are always the same.

• 1) If the rod were fixed on a pivot the effective point-mass of the rod at the tip (50 cm) would be 1/3 and at 0.25 cm from the CM it would be $m° =4/3 Kg$

• 2) If the rod is free and can translate the point-mass at the tip is 1/4 and at 0.25 cm is $m°= 4/7 Kg$. **

• 3) if the the rod is free and we know the impulse it is rather easy to determine the ratio between linear and angular velocity.

• 4) if the rod is free and (as OP requests) you apply a force of K (= 5) N at 0.25 cm the effective mass of the rod of that point is, of course, still known but it becomes complex (or impossible) to determine. Besides that, you must also distinguish if:
• a) the force is applied in one direction (and this is, by default, the case here)
• b) if the force rotates with the body

In either case there are many complications because:

a) if the force acts in one direction, the normal component gradually decreases to zero (and there is also a chance that it cannot be exerted for 1 second)

b) if the force rotates with the body, the CM will not be pushed only forward but will translate in many directions.

The other answer is being upvoted although it does not address any of these problems, does not address the question and has not produced the general single formula requested by the question. This is, probably, another anomaly: probably, one should post and be upvoted only when one has already found an answer.

** I gave you the formula for (point-mass)linear velocity $v$: $\frac{I}{I+md^2}$ and the ratio linear/ rotational velocity $v/v°$: $\frac{I}{md^2}$ so, the ratio linear/angular velocity $v/\omega$ is: $v/\omega = \frac{I}{md}$. This, as I said, is valid only for impulse/directional force.You should specify what you meant by your question.

• I really don't understand where these numbers come from, or even what you mean by "point mass". – David Z Apr 5 '15 at 12:32
• @DavidZ, the point mass $m°$ of a rod at $d$ is given by its mass m times : $\frac{I}{I+md^2}$. One can check the figures considering any collision with a particle with any momentum – user76716 Apr 5 '15 at 17:25
• @DavidZ, you commented this question years ago, now you put it on hold?? – user157860 Sep 6 at 10:44
• Yeah, it's really odd , considering that he took a great interest in this question, even going to the trouble of suggesting OP a mean trick to unaccept the answer here – bobie Sep 26 at 15:18

The question is quite general, but let me try to give some answers.

how do I determine resultant translational and rotational velocities?

For translational velocity, use Newton's second law:

$$\vec F= m \vec a$$

For rotational (angular) velocity, use:

$$\vec \tau=I\vec \alpha$$

To find moment of inertia $I$ you need to know the object geometry. For torque $\vec \tau$ use the definition $\tau=Fr$, where $r$ is the distance to the rotational center (and $F$ must be the perpendicular component).

When you have done this, use the acceleration $\vec a$ and angular acceleration $\vec \alpha$, respectively, to find velocity $\vec v$ and angular velocity $\vec \omega$.

I'm also looking for equations connecting translational kinetic energy to rotational kinetic energy in the context of applying torque.

Well, those are two different portions of energy (and I don't know what you mean by in the context of applying torque). The object as a whole will contain the total energy (the sum):

$$K_{trans}+K_{rot}=½mv^2+½I\omega^2$$

To stop such an object, you would need to absorb both the translational and the rotational kinetic energy. You can consider those two portions seperately.

If you want equations connecting these two energies, it must be because you convert one into the other or do some other proces involving them both. This depends on what you do.

• What did I forget? That distance is $r$, as written. Included in the expression for $\tau$. To find the combined kinetic energy, simply find the $v$ and $\omega$ and use the lower part of the answer. What is unclear? – Steeven Apr 3 '15 at 17:41
• My bad. Just another example of the wondrous things that sleep-deprivation will do to you. StackExchange won't let me vote on your answer once again unless you edit it. Could you add a newline or two in order for me to upvote it? – Vatsal Manot Apr 3 '15 at 17:55
• Wait. Applying a force on an object would cause it to translate forwards and rotate at the same time, right? I'm talking about a single force applied a certain distance from the center of mass. – Vatsal Manot Apr 3 '15 at 18:15
• That is correct. – Steeven Apr 3 '15 at 18:33
• Since the question has been changed, my answer is not relevant anymore. I will delete it. – Steeven Apr 4 '15 at 21:50

## protected by Qmechanic♦May 26 '15 at 18:32

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