# Variable MOI: Rotational 2nd Law

I'm having trouble finding any sources that discuss equations of motion that take into account a changing moment of inertia.

Just looking at the scaler case, angular momentum is $L = I\omega$

Then, according to the 2nd law, $\tau = \frac{d}{dt}(L) = \dot{I}\omega + I\alpha$

Is this correct? I understand the problem with doing this in the linear case (i.e. it does not respect Galilean invariance), but does the rotational case hold? I'm not sure it does...I mean, in the absence of torque ($\tau = 0$), the equation says that $\alpha \neq 0$ if $\dot{I} \neq 0$ which doesn't quite make sense. Does the $\dot{I}\omega$ term just get lumped into $\tau$? (like the linear case)

D H
Staff Emeritus
High school and freshman physics texts are very careful to avoid cases where the inertia tensor as expressed from the perspective of an inertial frame is not constant. They are also very careful to avoid cases where the inertia tensor needs to be treated as a tensor.

The inertia tensor for a rigid body is constant in a frame that rotates with the body. The angular momentum is still Iω. However, to be consistent with the fact that the inertia tensor is expressed in some body-fixed coordinate system, the angular velocity must be expressed in that same coordinate system. That means we're working in a rotating coordinate system. We're in a regime where Newton's laws don't exactly apply.

The transport theorem lets us relate the time derivatives in one frame to those in some other frame. For any vector quantity ##\vec q##, the time derivative of this quantity as observed in frame A and frame B are related by
$$\left(\frac {d\vec q}{dt}\right)_A = \left(\frac {d\vec q}{dt}\right)_B + \vec \omega_{A\to B} \times \vec q$$
The time derivative of angular momentum as expressed in the inertial frame is just the external torque. Applying the transport theorem to ##L=I\vec{\omega}## yields
$$\vec{\tau}_{\text{ext}} = \left(\frac {d\vec L}{dt}\right)_{I} = \left(\frac {d\vec L}{dt}\right)_B + \vec \omega \times \vec L = I\dot{\vec{\omega}} + \vec \omega \times (I\vec{\omega})$$
Note that the transport theorem also tells us that the angular acceleration is the same in the inertial and body-fixed frames because ##\vec{\omega}\times \vec{\omega}## is identically zero.

What if the inertia tensor is not constant in the body fixed frame? That means the object in question is not a rigid body. The mass might be changing, or the body might be a flexible body. Needless to say, things get a lot hairier when either of these is the case.

What if the inertia tensor is not constant in the body fixed frame? That means the object in question is not a rigid body. The mass might be changing, or the body might be a flexible body. Needless to say, things get a lot hairier when either of these is the case.

I guess this is the case I am after. Particularly a model of a rocket that is assumed to be a rigid body. That makes sense when you say the MOI (and mass) of a rigid body is constant, and if it isn't, then it's not technically a rigid body. So what assumptions can be made so we can model the changing MOI while still considering it a rigid body?

A model I have run across is one that Matlab uses for a 3 degree-of-freedom variable mass/MOI object: http://www.mathworks.com/help/aeroblks/simplevariablemass3dofbodyaxes.html

For the rotation, they use the same equation that I listed in my first post.

SteamKing
Staff Emeritus
Homework Helper
Rockets are tricky because the mass isn't constant and neither is the MOI. And for really large rockets, I don't think you can consider them to be rigid, either.

D H
Staff Emeritus