Angular momentum and velocity about a point about which a rigid body is not rotating

Question

Suppose a rigid body of mass $m$ is rotating about its centre of mass with angular velocity $ω$, and the centre of mass is translating with linear velocity v, consider two cases:

(I) we want to calculate the angular momentum about a point on the rigid body other than the centre of mass at a distance $r$ from the centre of mass

(II) we want to calculate the angular momentum and velocity about a point outside the rigid body at a distance $d$ from the centre of mass

In case (I), we can use two approaches, either $L=mvr+Iω$, but here $ω$ is unknown, or $L=I'ω$, here $ω$ is unknown as well, where $I'$ is the moment of inertia about an axis passing through the given point and parallel to the axis through the centre of mass. My question is: (a) How can we find $ω$ in the first approach, and (b) is the second approach valid?

In case (II), usually the first approach is used, but would the second approach be valid? I think it will be valid if the axis is fixed with respect to the rigid body, i.e., the axis moves at the same speed as the centre of mass to ensure the moment of inertia doesn't change.

Answer 1

I)

given

• $~m~$ mass
• $~\mathbf I~$ center of mass inertia
• $~\mathbf v~$ center of mass velocity
• $~\mathbf r~$ distance from the CM to point P

the angular momentum L at point P is:

$$\mathbf L=\mathbf r\times m\,\mathbf v+\mathbf I\,\mathbf \omega$$

you obtain the angular velocity $~\mathbf\omega~$ from the rotation matrix $~\mathbf R(\alpha~,\beta~,\gamma)~$ where $~\alpha~,\beta~,\gamma~$ are Euler angles

II)

given

• $~m~$ mass
• $~\mathbf I~$ center of mass inertia
• $~\hat{\mathbf{d}}~$ rotation axis at point p ,inertial fixed
• $~\mathbf r~$ distance from the CM to point P
• $~\phi~$ rotation angle about the axis $~\hat{\mathbf{d}}~$

the angular momentum L at point P is:

$$\mathbf L=\mathbf I'\,\mathbf \omega$$

where $$\mathbf \omega=\hat{\mathbf{d}}\dot\phi$$ $$\mathbf I'=\mathbf I-m\, \left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}}\\ r_{ {z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array} \right] \left[ \begin {array}{ccc} 0&-r_{{z}}&r_{{y}}\\ r_{ {z}}&0&-r_{{x}}\\ -r_{{y}}&r_{{x}}&0\end {array} \right] $$

III)

the rotation axis $~\hat{\mathbf{d}}~$ is body fixed

assume the body rotation matrix $\mathbf R~$ is

$$\mathbf R= \mathbf R_z(\gamma)\,\mathbf R_y(\beta)~\,\mathbf R_x(\alpha)$$ from here the rotation axis e.g.

$$\hat{\mathbf{d}}=\mathbf R^T\,\mathbf e_z=\left[ \begin {array}{c} -\sin \left( \beta \right) \\ \cos \left( \beta \right) \sin \left( \alpha \right) \\ \cos \left( \beta \right) \cos \left( \alpha \right) \end {array} \right] $$

the rotation the body rotation matrix an point P is (Integrating Angular Velocity Vector using Rodrigues' Rotation Formula)

$$\mathbf R_R=\mathbf R_R(\hat{\mathbf{d}}~,\phi)=\mathbf R_R(\alpha,\beta,\phi)$$

Answer 2

A couple of points of clarification.

• Avoid doing 3D dynamics using components. Work out the quantities you need in vector form and use the rules of vector algebra to manipulate things. This requires all vector quantities to be expressed on the same basis-vectors, usually referred to as the inertial frame of reference.

• When a rigid body is rotating with rotational velocity $\boldsymbol{\omega}$, this velocity is shared among all points on the rigid body. Rotational velocity is the property of the entire body and not of an individual point. On the other hand, translational velocity $\boldsymbol{v}$ is a property that changes from point to point.

• Chasle's Theorem states that rigid body motion can be fully described by the rotational velocity $\boldsymbol{\omega}$ and the translational velocity $\boldsymbol{v}$ of some arbitrary point on the body. It does not have to be the center of mass, but doing so simplifies the equations of motion.

  To calculate the translational velocity at point A from the translational velocity at point B use the velocity kinematics

  $$ \boldsymbol{v}_A = \boldsymbol{v}_B + \boldsymbol{\omega} \times ( \boldsymbol{r}_A - \boldsymbol{r}_B) \tag{1}$$

  where $\boldsymbol{r}_A$ and $\boldsymbol{r}_B$ is the position vectors for each point and $\times$ is the vector cross-product.

• Linear and angular momentum are summed up over all the particles of a body with respect to a reference point (sometimes called the summation point). Linear momentum is a property of the entire body (just like angular velocity is) and angular momentum is specific to the summation point.

  You can transform angular momentum from one summation point to another using the following transformation

  $$ \boldsymbol{L}_A = \boldsymbol{L}_B + \boldsymbol{p} \times ( \boldsymbol{r}_A - \boldsymbol{r}_B ) \tag{2}$$

  where $\boldsymbol{p}$ is linear momentum of the body. If you notice a similarity with the velocity kinematics, it is not a coincidence.

  Interestingly, the motion of the summation point does not enter into the equation for angular momentum. So if you know angular momentum about the center of mass, it is trivial to find angular momentum about an arbitrary point riding on the body, or fixed to the ground.

For your situation, you have the following

Using the kinematics (1) and angular momentum transformation (2) you have the following expression for angular momentum summed at an arbitrary point A:

$$\begin{array}{r|c} & \text{Point A}\\ \hline \text{Realtive Center of Mass Position} & \boldsymbol{c}_{A}=\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\\ \text{Linear Velocity} & \boldsymbol{v}_{A}=\boldsymbol{v}_{C}+\boldsymbol{c}_{A}\times\boldsymbol{\omega}\\ \text{Linear Momentum} & \boldsymbol{p}=m\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{c}_{A}\right)\\ \text{Angular Momentum} & \boldsymbol{L}_{A}={\rm I}_{C}\boldsymbol{\omega}+\boldsymbol{c}_{A}\times\boldsymbol{p} \end{array} \tag{3}$$

In the case that point A rides with the body, linear momentum can be expressed in terms of its motion as seen in the second to last expression above. This allows the definition of mass movement of inertia summed at A and the simplification of angular momentum as follows:

$$\begin{array}{r|c} & \text{Point A}\\ \hline \text{Angular Momentum} & \boldsymbol{L}_{A}={\rm I}_{C}\boldsymbol{\omega}+\boldsymbol{c}_{A}\times m\left(\boldsymbol{v}_{A}+\boldsymbol{\omega}\times\boldsymbol{c}_{A}\right)\\ \text{Mass moment of Inertia} & {\rm I}_{A}={\rm I}_{C}+m\left((\boldsymbol{c}_{A}\cdot\boldsymbol{c}_{A}){\tt 1}-\boldsymbol{c}_{A}\odot\boldsymbol{c}_{A}\right)\\ \text{Angular Momentum} & \boldsymbol{L}_{A}={\rm I}_{A}\boldsymbol{\omega}+\boldsymbol{c}_{A}\times m\boldsymbol{v}_{A} \end{array} \tag{4}$$

where $\mathtt{1}$ is the identity matrix, $\cdot$ is the dot product and $\odot$ is the outer product.

Both (3) and (4) expressions for $\boldsymbol{L}_A$ are correct, but the second one is only in terms of the motion of A, translational velocity $\boldsymbol{v}_A$ and rotational velocity $\boldsymbol{\omega}$.

In summary both of $$\boldsymbol{L}_{A}={\rm I}_{C}\boldsymbol{\omega}+\boldsymbol{c}_{A}\times m\boldsymbol{v}_{C} \tag{3}$$ $$\boldsymbol{L}_{A}={\rm I}_{A}\boldsymbol{\omega}+\boldsymbol{c}_{A}\times m\boldsymbol{v}_{A}\tag{4}$$

are equally valid for a point riding on the body with the proper definition for $\boldsymbol{v}_C$, $\boldsymbol{v}_A$, $\mathrm{I}_C$ and $\mathrm{I}_A$.

In the case where point A is not part of the body then, only (3)is correct. This is because in this case $\boldsymbol{v}_A$ is independent of the motion of the body, and cannot be factored out.