Altenbraknz
2022-09-25
Answered

Euler’s equations of motion for a rigid body can be interpreted as a rewriting of Newton’s second law for rotations in a rotating frame. They basically tell us the sum of the torques equals the rate of change of the body’s angular momentum, In the rotating frame. Do we then not need to take into account inertial forces when computing the torques in rotating coordinates?

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Lilliana Mason

Answered 2022-09-26
Author has **11** answers

No, but there is an inertial torque you have to worry about.

From the perspective of an inertial frame, the rotational analog of Newton's second law for rotation about the center of mass is

$\begin{array}{}\text{(1)}& \frac{d\mathit{L}}{dt}=\sum _{i}{\mathit{\tau}}_{\text{ext},i}\end{array}$

where L is the object's angular momentum with respect to inertial, τ,i is the i external torque, and the differentiation is from the perspective of the inertial frame. Note that this pertains to non-rigid objects as well as rigid bodies.

The relationship between the time derivatives of any vector quantity q from the perspectives of co-located inertial and rotating frames is

$\begin{array}{}\text{(2)}& {\left(\frac{d\mathit{q}}{dt}\right)}_{\text{inertial}}={\left(\frac{d\mathit{q}}{dt}\right)}_{\text{rotating}}+\mathbf{\Omega}\times \mathit{q}\end{array}$

where $\mathbf{\Omega}$ is the frame rotation rate with respect to inertial.

For a rigid body, the body's angular momentum with respect to inertial but expressed in body-fixed coordinates is $\mathit{L}=\mathbf{I}\phantom{\rule{thinmathspace}{0ex}}\mathit{\omega}$ where I is the body's moment of inertia tensor and $\mathit{\omega}$ is the body's rotation rate with respect to inertial but expressed in body-fixed coordinates. Since a rigid body's inertia tensor is constant in the body-fixed frame, we have

$\begin{array}{}\text{(3)}& {\left(\frac{d\mathit{L}}{dt}\right)}_{\text{body-fixed}}=\frac{d(\mathbf{I}\mathit{\omega})}{dt}=\mathbf{I}\frac{d\mathit{\omega}}{dt}\end{array}$

Combining equations (1), (2), and (3) yields

$\begin{array}{}\text{(4)}& \mathbf{I}\frac{d\mathit{\omega}}{dt}=\sum _{i}{\mathit{\tau}}_{\text{ext},i}-\mathit{\omega}\times (\mathbf{I}\times \mathit{\omega})\end{array}$

This is Euler's equations of motion for a rigid body. No inertial forces come into play. However, the term $-\mathit{\omega}\times (\mathbf{I}\times \mathit{\omega})$ is essentially an inertial torque. Just as inertial forces vanish in inertial frames, so does this inertial torque.

From the perspective of an inertial frame, the rotational analog of Newton's second law for rotation about the center of mass is

$\begin{array}{}\text{(1)}& \frac{d\mathit{L}}{dt}=\sum _{i}{\mathit{\tau}}_{\text{ext},i}\end{array}$

where L is the object's angular momentum with respect to inertial, τ,i is the i external torque, and the differentiation is from the perspective of the inertial frame. Note that this pertains to non-rigid objects as well as rigid bodies.

The relationship between the time derivatives of any vector quantity q from the perspectives of co-located inertial and rotating frames is

$\begin{array}{}\text{(2)}& {\left(\frac{d\mathit{q}}{dt}\right)}_{\text{inertial}}={\left(\frac{d\mathit{q}}{dt}\right)}_{\text{rotating}}+\mathbf{\Omega}\times \mathit{q}\end{array}$

where $\mathbf{\Omega}$ is the frame rotation rate with respect to inertial.

For a rigid body, the body's angular momentum with respect to inertial but expressed in body-fixed coordinates is $\mathit{L}=\mathbf{I}\phantom{\rule{thinmathspace}{0ex}}\mathit{\omega}$ where I is the body's moment of inertia tensor and $\mathit{\omega}$ is the body's rotation rate with respect to inertial but expressed in body-fixed coordinates. Since a rigid body's inertia tensor is constant in the body-fixed frame, we have

$\begin{array}{}\text{(3)}& {\left(\frac{d\mathit{L}}{dt}\right)}_{\text{body-fixed}}=\frac{d(\mathbf{I}\mathit{\omega})}{dt}=\mathbf{I}\frac{d\mathit{\omega}}{dt}\end{array}$

Combining equations (1), (2), and (3) yields

$\begin{array}{}\text{(4)}& \mathbf{I}\frac{d\mathit{\omega}}{dt}=\sum _{i}{\mathit{\tau}}_{\text{ext},i}-\mathit{\omega}\times (\mathbf{I}\times \mathit{\omega})\end{array}$

This is Euler's equations of motion for a rigid body. No inertial forces come into play. However, the term $-\mathit{\omega}\times (\mathbf{I}\times \mathit{\omega})$ is essentially an inertial torque. Just as inertial forces vanish in inertial frames, so does this inertial torque.

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