# In case of an axis where moment of inertia changes with time which of the following equations is valid? [closed]

$$T = I \alpha$$

$$L = I \omega$$

$$T d(\theta) = d(\tfrac12 I \omega^2)$$

If I differentiate the second and third one with respect to time… all three equations give a different expression for torque in terms of the $$dI/dt$$ term. Clearly only one of these can be correct. So which is it and why?

An example of a axis where Moment of inertia changes with time can for example be the point of contact of the ring with with the ground when a rotating ring has a mass attached at a point on its circumference. Since the distance of the mass from the point of contact can change as the ring rotates… the moment of inertia about this axis can change. Note: the axis is through the point of contact of the ring with the ground and perpendicular to the plane of the ring.

• why is the moment of inertia changing? Is the object losing mass by throwing it away, or is the object changing size and conserving its mass? Mar 25 at 16:06
• The question is unclear. This is because in general MMOI is fixed when viewed along the body-fixed coordinates and changing when viewed along the inertial coordinates. Are you adding an additional change in time, like the body is not a rigid body and it changes shape? Or there is mass added/subtracted with time? Please edit the question and clarify the situation more. Mar 25 at 16:17
• In the example stated the MMOI does not change, but the reference point chosen is not the center of mass, and it changes with time. Mar 25 at 16:21
• It will change about the point of contact of the ring with the ground because the distance of the mass from this point will change. Sorry if I was unclear in my question Mar 25 at 16:54

The relation $$L = I\omega$$ is always true because it is the definition of $$I$$. (Note that in general $$L$$ and $$\omega$$ might not be parallel, in which case $$I$$ will be a symmetric tensor rather than a scalar. To simplify the answer, I will only discuss the case where $$I$$ can be treated as a scalar.)

The correct general definition of the torque $$\tau$$ is $$\tau = \mathrm{d}L/\mathrm{d}t$$. It follows that

$$\tau = \frac{\mathrm{d}}{\mathrm{d}t} (I\omega) = \omega \frac{\mathrm{d}I}{\mathrm{d}t} + I\alpha$$

That means $$\tau = I\alpha$$ is valid only when $$I$$ is constant in time.

The third equation also can't be assumed to be valid when $$I$$ is time-varying. It helps to express the rotational kinetic energy in the equivalent form $$T = \frac{L^2}{2I}$$. ($$T$$ is the letter typically used for kinetic energy.) Torques will change the value of $$L$$ and thus also $$T$$, but another contribution to the change in $$T$$ comes from changes in $$I$$. The third equation as written assumes that all changes to $$T$$ are due to a torque being exerted over an angular displacement. (See related Where does a spinning figure skater's energy go when she slows down?)

• The kinetic energy isn't equal to $L^2/2I$ if the object is not rotating along a principal axis (which is one reason that you could have a time-varying moment of inertia in the space frame). The general expression would be $T = \frac{1}{2} \vec{\omega} \cdot (I \vec{\omega})$. Mar 25 at 17:38
• @MichaelSeifert Under conditions where $I$ can be treated as a scalar, that expression is, in fact, equal to $L^2/(2I)$ even if $I$ is time-varying. I'll edit the answer to clarify this. Mar 25 at 17:40

The angular momentum is:

$$\vec L= I\,\vec \omega$$

thus, in case that inertia tensor $$~I~$$ is constant , you obtain

$$\vec\tau =\frac {d}{dt} \vec L= I\,\frac {d}{dt}\vec\omega+\vec\omega\times I\,\vec \omega$$ and if inertia tensor $$~I=I(t)~$$ you obtain $$\vec\tau =\frac {d}{dt} \vec L= I\,\frac {d}{dt}\vec\omega+ \left(\frac {d}{dt}\,I\right)\,\vec\omega+\vec\omega\times I\,\vec \omega$$

the kinetic energy is:

$$T=\frac 12 \vec\omega^T\,I\,\vec\omega=\frac 12 \vec\omega\cdot\vec L\quad\Rightarrow\\ 2\,\left[\frac{\partial T}{\partial \vec\omega}\right]^T=\vec L$$

It seems you are trying to understand the relationship between torque and change in angular momentum for arbitrary points, not at the center of mass.

I am going to use vector equations because they are the more concise form of dynamics compared to treating the problem component by component, or in 2D which is just a projection of the general 3D case.

Consider the 3×3 rotation matrix $$R$$ describing the orientation of the body in the inertial coordinates. Each column of $$R$$ represents the unit vectors x,y, and z as they ride along the rotating body.

Now consider the mass moment of inertia (MMOI) of a rigid body that is fixed in the body-coordinates and changing in the inertial-coordinates as the body rotates about. This relationship is described by the following matrix equation that calculates $${\rm I}_{C}$$, the MMOI tensor (3×3 matrix) for use in the equations of motion summed up at the center of mass.

$${\rm I}_{C}=R\,I_{{\rm body}}R^{\intercal}\tag{1}$$

Additionally, since the body is continually rotating, the rotation matrix itself is changing, and this is described with the following equation. Given the rotational velocity $$\vec{\omega}$$ at some instant this is

$$\tfrac{{\rm d}}{{\rm d}t}R=[\vec{\omega}\times]R\tag{2}$$

where $$[\vec{\omega}\times ]$$ is a special 3×3 (skew-symmetric) matrix that allows the calculation of a cross product $$\times$$ using linear algebra (matrix operations).

Combine the two above equations to qualify how the MMOI tensor changes with time

$$\tfrac{{\rm d}}{{\rm d}t}{\rm I}_{C}=[\vec{\omega}\times]{\rm I}_{C}-{\rm I}_{C}[\vec{\omega}\times]\tag{3}$$

Here is a general situation where the center of mass (point C) and some other arbitrary location (point A) exist and quantities such as motion, momentum, and torques can be expressed on either one.

Start from the center of mass with linear velocity $$\vec{v}_C$$ and some body rotation $$\vec{\omega}$$ about the center of mass.

Linear momentum vector $$\vec{p}$$ is

$$\vec{p}=m\,\vec{v}_{C}\tag{4}$$

and angular momentum about the center of mass is

$$\vec{L}_{C}={\rm I}_{C}\vec{\omega}\tag{5}$$

The above is the definition of the MMOI tensor. This means that we define the MMOI about the arbitrary point based on $$\vec{L}_A = {\rm I}_A \vec{\omega}$$. The mass moment of inertia tensor at some point is always exactly was is needed to transform the rotation of the body $$\vec{\omega}$$ into angular momentum at the same point.

Now the standard equations of motion, as derived from Newton's 2nd law for the center of mass.

$$\vec{F}=\tfrac{{\rm d}}{{\rm d}t}\vec{p}=m\,\left(\tfrac{{\rm d}}{{\rm d}t}\vec{v}_{C}\right)\tag{6}$$

$$\vec{\tau}_{C}=\tfrac{{\rm d}}{{\rm d}t}\vec{L}_{C}={\rm I}_{C}\left(\tfrac{{\rm d}}{{\rm d}t}\vec{\omega}\right)+[\vec{\omega}\times]{\rm I}_{C}\vec{\omega}\tag{7}$$

Now use the following transformation rules to express momentum and the equations of motion about some arbitrary point

$$\vec{v}_{C}=\vec{v}_{A}+\vec{\omega}\times\left(\vec{r}_{C}-\vec{r}_{A}\right)\tag{8}$$

$$\vec{L}_{A}=\vec{L}_{C}+\left(\vec{r}_{C}-\vec{r}_{A}\right)\times\vec{p}\tag{9}$$

I am going to be using a shorthand notation of $$\vec{c} = \vec{r}_{C}-\vec{r}_{A}$$ to designate the location of the center of mass relative to the reference point.

Linear momentum is

$$\vec{p}=m\,\left(\vec{v}_{A}+\vec{\omega}\times\left(\vec{r}_{C}-\vec{r}_{A}\right)\right)\tag{10}$$

and angular momentum, which depends on both the angular velocity of the body and the linear velocity of the point.

$$\vec{L}_{A}={\rm I}_{C}\vec{\omega}+\vec{c}\times\vec{p}=\vec{c}\times m\,\vec{v}_{A}+\underbrace{{\rm I}_{C}\vec{\omega}-\vec{c}\times m\,\left(\vec{c}\times\vec{\omega}\right)}_{{\rm I}_{A}\vec{\omega}}\tag{11}$$

the above gives rise to the definition of $${\rm I}_A$$ the MMOI about point A.

$${\rm I}_{A}={\rm I}_{C}-m[\vec{c}\times][\vec{c}\times]\tag{12}$$

Now consider the arbitrary point A is riding on the body and from (8) we have the derivative $$\tfrac{{\rm d}}{{\rm d}t}\vec{c}=\vec{v}_{C}-\vec{v}_{A}=\vec{\omega}\times\vec{c}$$

The net force is still given by Newton's 2nd law

$$\vec{F}=\tfrac{{\rm d}}{{\rm d}t}\vec{p}=m\,\left(\left(\tfrac{{\rm d}}{{\rm d}t}\vec{v}_{A}\right)+\left(\tfrac{{\rm d}}{{\rm d}t}\vec{\omega}\right)\times\vec{c}+\vec{\omega}\times\left(\vec{\omega}\times\vec{c}\right)\right)\tag{13}$$

The net torque about A is given by the transformation $$\vec{\tau}_{A}=\vec{\tau}_{C}+\vec{c}\times\vec{F}$$ and the law of rotation (7)

$$\vec{\tau}_{A}={\rm I}_{C}\left(\tfrac{{\rm d}}{{\rm d}t}\vec{\omega}\right)+[\vec{\omega}\times]{\rm I}_{C}\vec{\omega}+\vec{c}\times\vec{F}$$

$$\vec{\tau}_{A}=\vec{c}\times m\,\left(\tfrac{{\rm d}}{{\rm d}t}\vec{v}_{A}\right)+{\rm I}_{A}\left(\tfrac{{\rm d}}{{\rm d}t}\vec{\omega}\right)+[\vec{\omega}\times]{\rm I}_{A}\vec{\omega}\tag{14}$$

As you can see (13), (14) are far more complex than (6) and (7) which is why we usually try to resolve everything on the center of mass in dynamics. But you don't have to do so, and the above is equally as valid.

Finally, the question about kinetic energy, you have to include both translational and rotational motion.

$$KE=\tfrac{1}{2}\vec{v}_{C}\cdot\vec{p}+\tfrac{1}{2}\vec{\omega}\cdot\vec{L}_{C}=\tfrac{1}{2}\vec{v}_{A}\cdot\vec{p}+\tfrac{1}{2}\vec{\omega}\cdot\vec{L}_{A}\tag{15}$$

which you are welcome to take a derivative of