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Please help me check my understanding related to the rotational motion of a 3D rigid body after reading some Physics textbooks and googling for some more materials (e.g., Wikipedia's Torque, Wikipedia's Moment of Inertia).

The following points are to be checked against a situation where the rotational motion of a 3D rigid body happens about a single rotation axis in an inertial Cartesian coordinate system where the z-axis is taken to be the rotation axis but the origin $O$ of the coordinate system is not necessarily on the centroid of the 3D rigid body and can even be outside the 3D rigid body:

  1. Each particle in the rigid body has $\vec{v} = \vec{\omega} \times \vec{r}$ where $\vec{r}$ is the position vector of the particle with respect to the origin $O$, and therefore, the speed of each particle depends on the choice of origin $O$.
  2. With respect to $O$, the moment of force $\vec{\tau}$ and the angular momentum change $d\vec{L}$ about the z-axis have the same directions.
  3. With respect to $O$, the angular acceleration $\vec{\alpha}$ and the angular velocity change $d\vec{\omega}$ about the z-axis have the same directions.
  4. Moment of inertia $I$ is always calculated with respect to z-axis, not the centroid of the rigid body, such that the $s$ in $\int s^2 dm$ is always the distance between a point mass in the rigid body and z-axis.
  5. Angular momentum of each particle in the rigid body $\vec{L} = \vec{r}\times\vec{p}$ where $\vec{r}$ is the position vector of $\vec{p}$ with respect to the origin $O$ (i.e., $||\vec{r}||$ is not the distance between $\vec{p}$ and z-axis), and therefore, the direction of $\vec{L}$ may be different from $\vec{\omega}$.
  6. Total angular momentum of the rigid body is then $\sum \vec{L} = I \vec{\omega}$, and therefore, the direction of $\sum \vec{L}$ is the same as $\vec{\omega}$.
  7. The torque experienced by each particle in the rigid body $\vec{\tau} = \vec{r}\times\vec{F}$ where $\vec{r}$ is the position vector of $\vec{F}$ with respect to the origin $O$ (i.e., $||\vec{r}||$ is not the distance between $\vec{F}$ and z-axis), and therefore, the direction of $\vec{\tau}$ may be different from $\vec{\omega}$.
  8. The total torque experienced by the rigid body is then $\sum \vec{\tau} = I \vec{\alpha}$, and therefore, the direction of $\sum \vec{\tau}$ is the same as $\vec{\alpha}$.
  9. The torque experienced by each particle in the rigid body is related to each particle's angular momentum rate of change $\vec{r}\times\vec{F} = \vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d(\vec{r} \times \vec{p})}{dt} = m\frac{d(\vec{r} \times \vec{v})}{dt} = m\frac{d(\vec{r} \times (\vec{\omega}\times\vec{r}))}{dt} = m\frac{d((\vec{r}\cdot\vec{r}) \vec{\omega} - (\vec{r}\cdot\vec{\omega})\vec{r})}{dt}$, and therefore, for each particle the directions of $\vec{\tau}$ and $d\vec{L}$ are the same but may be different from the direction of $\vec{\omega}$.
  10. The total torque experienced by the rigid body is related to the total angular momentum's rate of change $I \vec{\alpha} = \sum \vec{\tau} = \frac{d\left(\sum\vec{L}\right)}{dt} = \frac{d\left(\sum I\;\vec{\omega}\right)}{dt} = \sum I \frac{d\vec{\omega}}{dt}$, and therefore, the directions of $\vec{\alpha}$, $\sum \vec{\tau}$, $\sum d\vec{L}$, and $d\vec{\omega}$ are the same.

Any mistake? Because the rotation axis can be outside the 3D rigid body, I feel uneasy especially for Point 5 to 10.

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Seems like googling using "moment of force" instead of "torque" is more useful. I found this slide from PSU that says that $\tau$ is calculated by projecting all the forces onto the plane perpendicular to the rotation axis. That is also what I feel being strongly suggested in introductory Physics textbooks although I found no explicit statement as the one in the slide. I wonder why that is the case. –  Tadeus Prastowo Apr 14 at 11:54
    
What I found in the introductory Physics textbooks is always along the line given by this lecture notes from University of Rochester: they define $\vec{\tau} = \vec{r} \times \vec{F}$ on a particle on a plane and then discuss 3D rigid bodies like a cylinder without explicitly relate how $\vec{r}$ now means with respect to the origin $O$ now that some particles in the rigid bodies are on planes that do not include the origin $O$. –  Tadeus Prastowo Apr 14 at 12:20

1 Answer 1

up vote 0 down vote accepted
  1. Check ✓ except the the speed depends on the center of rotation, not the choice of the origin. The general rule is that if the origin as velocity $\vec{v}_O$ then a point A located at $\vec{r}_A$ has speed $\vec{v}_A = \vec{v}_O + \vec\omega \times \vec{r}_A$. It happenstance that your origin does not move.
  2. Check ✓. In general, torques and angular momental are related with $\sum \vec{\tau}_A = \frac{{\rm d}}{{\rm d}t} \vec{L}_A$ where A is any point on the rotating frame.
  3. Check ✓. Angular velocity and acceleration is shared among all points fixed to the rotating frame. Acceleration is the time derivative of velocity $\frac{{\rm d}}{{\rm d}t}\vec{\omega} = \vec{\alpha}$.
  4. Incorrect ×. Mass moment of inertia $I$ is a tensor of 6 values represented by a 3×3 matrix $$I=\begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{pmatrix}$$ Some books have the cross terms negative and some positive. There is no convention of what is a positive $I_{xy} = \int x y {\rm d}m$ or $I_{xy} = \int -x y {\rm d}m$ (ref 1). Typically, this tensor (MMOI) is described on the center of mass first, along coordinates fixed to the body. So $I_{body}$ is constant. As the body rotates it has a 3×3 rotation matrix $R$ so the MMOI tensor in fixed coordinates at the center of mass is $$I_C = R I_{body} R^\top$$ with $^\top$ the matrix transpose operator. To move the MMOI tensor to a different location, like where the rotation center is (or the origin) one must apply the parallel axis theorem. This is done with $$I_O = I_C - m [\vec{r}_C] [\vec{r}_C]$$ where [vector] is the 3×3 skew symmetric cross product operator (ref 2) and $\vec{r}_C$ is the position of the center of mass relative to O in fixed coordinates.
  5. Incorrect ×. Angular momentum at any point not at the center of mass has two components. One is the intrinsic momentum at the center of mass $I_C \vec{\omega}$ and the second the moment of linear momentum $\vec{r}_C \times m \vec{v}_C$ where $\vec{v}_C$ is the linear velocity of the center of mass. $$\vec{L}_O = I_C \vec{\omega} + \vec{r}_C \times m \vec{v}_C = I_C \vec{\omega} + \vec{r}_C \times m ( \vec{v}_O + \vec\omega \times \vec{r}_C ) \\ = \left(I_C \vec{\omega} - m \vec{r}_C \times \vec{r}_C \times \vec\omega \right) + \vec{r}_C \times m \vec{v}_O = I_O \vec\omega + \vec{r}_C \times m \vec{v}_O$$ Do you see how the parallel axis theorem is derived from the angular momentum transformation laws.
  6. Incorrect ×. Because MMOI is a tensor and not a scalar the direction of $\vec{L}$ is generally not the same as $\vec\omega$.
  7. Check ✓. Torque is vector field (varying by location) with the general law of $$\vec{\tau}_O = \vec{\tau}_A + \vec{r}_A \times \vec{F}$$ This is dual to the velocity transformation law (above on 1.), if re-arranged like $\vec{\tau}_A = \vec{\tau}_O + \vec{F} \times \vec{r}_A $. The direction of $\vec\tau$ (at origin?) is different from $\vec\omega$ because $I_C$ is not constant (see below).
  8. The general equations of motion are $$\boxed{\begin{align} \sum \vec{F} & = \frac{\rm d}{{\rm d}t} \vec{p} = m \frac{\rm d}{{\rm d}t} \vec{v}_C = m \vec{a}_C \\ \sum \vec{\tau}_C & = \frac{\rm d}{{\rm d}t}L_C = I_C \frac{\rm d}{{\rm d}t} \vec\omega + \frac{\rm d}{{\rm d}t}\left( I_C \right) \vec\omega = I_C \vec\alpha + \vec\omega \times I_C \vec\omega \end{align} }$$ expressed at the center of mass always (ref 3). Then use the transformation laws to move the torque around (like knowing that the torque component along the rotation axis is zero). In your case $\hat{z}^\top \vec{\tau}_O =0$ and $\vec{\tau}_O =\sum \vec{\tau}_C + \vec{r}_C \times \sum \vec{F}$.
  9. You need to re-work the rest from the information given above.
share|improve this answer
    
See related post physics.stackexchange.com/a/95542/392 if interested in the subject. –  ja72 Apr 14 at 15:20
    
Thank you very much for your help! I really need it. So, introductory Physics textbooks simplify things in their discussions & exercises by always choosing a rotation axis that goes through the center of mass of a 3D rigid body? –  Tadeus Prastowo Apr 14 at 16:43
    
Yes, 3D dynamics and kinematics are straight forward if you get all your ducks in a row, but there are a lot of ducks to account for. –  ja72 Apr 14 at 16:47

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