Classical Rotation: Referencing Rotational Motion in Mechanics

[kelly0303]

Hello! I am a bit confused by the reference frames used in derivations for rotational motion in classical mechanics (assume that there is no translation and the body rotates around a fixed point). As far as I understand there are two main frames used in the analysis: a lab frame, which is fixed (and inertial) and a body frame, which rotates with the body and hence it is not inertial. In the lab frame the moment of inertia tensor is a function of time, but in the body frame it is constant and for a proper choice it can be made diagonal. In such a diagonal case, the kinetic energy is given by: $$T = \frac{1}{2}(I_x\omega_x^2+I_y\omega_y^2+I_z\omega_z^2)$$ So I am confused about this expression. Given that we have only the diagonal terms of the inertia tensor, it means that we are in the body frame, so I assume that this formula is expressed in the body frame. But I am not sure I understand how can you have kinetic energy in the body frame, if the frame is rotating at the same time with the object. Isn't the object fixed relative to the body frame, hence the kinetic energy of the body is zero? Also, $\omega_x, \omega_y, \omega_z$ should be also in the body frame (I guess it doesn't make sense to multiply terms from the body frame with terms from the lab frame in the same equation). But, again, how can you have an angular velocity in the body frame, given that the object is rotating at the same time with the frame. For example, assume that one particle of the object is located on the x-axis in the body frame. That particle will obviously rotate with respect to the lab frame, but in the object frame it will still be on the x axis. So according to the body frame that particle is not moving, hence it has no kinetic energy and no angular velocity (and the same logic can be applied for all the particles forming the body). So I am sure I am missing something here, probably having to do with the definitions of the frames or with associating the variables to the right frames, but I am really confused. Can someone clarify this for me? Thank you!

 

[wrobel]

There ate two very different cases. The first one is as follows.
You have two coordinate frames I and II. You can introduce the kinetic energy relative to the first coordinate frame
$$T_I=\frac{1}{2}\sum m_i|\boldsymbol v_i|^2,$$
$\boldsymbol v_i$ -- the velocities relative to the frame I;
and relative to the second coordinate frame:
$$T_{II}=\frac{1}{2}\sum m_i|\boldsymbol v'_i|^2,$$
$\boldsymbol v'_i$ -- the velocities relative to the frame II.

In general $T_I\ne T_{II}$

Another case is as follows. You have already decided to consider kinetic energy relative to the frame $I$. Velocities $\boldsymbol v_i$ are vectors their inner product is a scalar so $T_I$ is a scalar. You can expand the vectors $\boldsymbol v_i$ by any coordinate basis they remain the same vectors; the function $T_I$ remains the same scalar. Your question is about this last case.

 

[kelly0303]

There ate two very different cases. The first one is as follows.
You have two coordinate frames I and II. You can introduce the kinetic energy relative to the first coordinate frame
$$T_I=\frac{1}{2}\sum m_i|\boldsymbol v_i|^2,$$
$\boldsymbol v_i$ -- the velocities relative to the frame I;
and relative to the second coordinate frame:
$$T_{II}=\frac{1}{2}\sum m_i|\boldsymbol v'_i|^2,$$
$\boldsymbol v'_i$ -- the velocities relative to the frame II.

In general $T_I\ne T_{II}$

Another case is as follows. You have already decided to consider kinetic energy relative to the frame $I$. Velocities $\boldsymbol v_i$ are vectors their inner product is a scalar so $T_I$ is a scalar. You can expand the vectors $\boldsymbol v_i$ by any coordinate basis they remain the same vectors; the function $T_I$ remains the same scalar. Your question is about this last case.

Thank you for your reply! So if I understand what you said well, you mean that the kinetic energy I mentioned, is the energy in the LAB frame, but it is expanded in terms of moments of inertia and angular velocity written in the BODY frame (as in that frame they just have an easier expression). So, for example, the angular momentum in the body frame doesn't mean that the body rotates in the body frame, it is just writing a vector from the LAB frame (L) using BODY frame axis. Is that right?

 

[PeroK]

Thank you for your reply! So if I understand what you said well, you mean that the kinetic energy I mentioned, is the energy in the LAB frame, but it is expanded in terms of moments of inertia and angular velocity written in the BODY frame (as in that frame they just have an easier expression). So, for example, the angular momentum in the body frame doesn't mean that the body rotates in the body frame, it is just writing a vector from the LAB frame (L) using BODY frame axis. Is that right?

The idea is that the rotational KE does not depend on the orientation of the coordinate axes in the lab frame. At each point in its rotation, you could imagine a set of coordinates in the lab frame where the body has the same orientation that it had at $t = 0$. Think of these as a large number of fixed coordinate systems, not changing with time. They are all measuring the same KE. At each instant you choose the orientation that coincides with the motion of the body.

You have to be careful with this idea not to confuse it with a rotating coordinate system. To take a simple example, imagine a particle moving with constant linear motion:

1) Imagine a set of fixed coordinate systems with different origins along the line of motion. At time $t$ the particle is at the origin of one of these and has a KE of $\frac 1 2 mv^2$, where $v$ is measured in that fixed coordinate system. In that coordinate system in an interval $\Delta t$, the particle has moved $v\Delta t$.

You have, therefore, a whole sequence of coordinate systems with the particle moving at speed $v$ in all of them.

This is why, at some arbitrary time $t$ you can take the origin to be where the particle happens to be at time $t$. But, that's the origin of a fixed coordinate system.

This is different from.

2) Imagine an inertial frame/coordinate system moving with particle. In the frame the KE is $0$. In this case, you have a single moving coordinate system. The particle is always at the same origin here.

 

[wrobel]

Thank you for your reply! So if I understand what you said well, you mean that the kinetic energy I mentioned, is the energy in the LAB frame, but it is expanded in terms of moments of inertia and angular velocity written in the BODY frame (as in that frame they just have an easier expression). So, for example, the angular momentum in the body frame doesn't mean that the body rotates in the body frame, it is just writing a vector from the LAB frame (L) using BODY frame axis. Is that right?

Thats right

 

[A.T.]

So, for example, the angular momentum in the body frame doesn't mean that the body rotates in the body frame, it is just writing a vector from the LAB frame (L) using BODY frame axis. Is that right?

Note the difference between "frame of reference" and "coordinate system". You can have a coordinate system fixed relative to the lab, but instantaneously aligned with the local body axes.