# Term in definition of torque

1. Oct 19, 2009

### Kricket

Hello all,

I'm sure you all know the standard relation between torque and angular acceleration:

$$\tau = I\alpha$$

However, in reading a doc on kinematics this weekend, the author gave the following formula for torque:

$$\tau = I\alpha + \omega \times L$$

My question is, from whence cometh the omegaxL term? Something to do with kinetic energy, perhaps?

Thanks for your help

2. Oct 19, 2009

### jostpuur

Right now I don't remember how the definitions go, but anyway it is true that torque equals the time derivative of the angular momentum:

$$\tau = \dot{L}$$

I'll assume this as know now. It could be that that's a definition, or then it could be that it must be proven, I'm not sure.

Try to find introductory material to inertial tensor in three dimensions. For example: http://en.wikipedia.org/wiki/Moment_of_inertia I haven't read the Wikipedia page myself, so I cannot tell if it's good or bad. If you can find this book from somewhere https://www.amazon.com/Classical-Dynamics-Particles-Systems-Thornton/dp/0534408966/ref=pd_cp_b_0_img you might try it too.

Then go through some details of my post here: https://www.physicsforums.com/showthread.php?t=215122 so that you understand why $e^{\theta\times} = 1 + \theta\times + \cdots$ is a rotation operator.

Then consider a system of particles, which has a rigid structure. With respect to the center of mass, the locations and masses of the particles are originally $x_1(0), x_2(0), \ldots, x_N(0)$ and $m_1,m_2,\ldots, m_N$, but then the system may rotate so that at a given time $t$, the locations of the masses are

$$x_k(t) = e^{\theta(t)\times} x_k(0)$$

$\theta(t)$ is an angle which may change as a function of time. The velocities of the particles are

$$v_k(t) = \dot{x}_k(t) = \dot{\theta}(t)\times (e^{\theta(t)\times} x_k(0)) = \omega(t) \times x_k(t)$$

Here notation $$\omega(t) = \dot{\theta}(t)$$ was used. Also use notation $$\alpha(t) = \ddot{\theta}(t)$$. Then, by using the product rule $$D_t(fg) = \dot{f}g +f\dot{g}$$ we get

$$a_k(t) = \ddot{\theta}(t)\times (e^{\theta(t)\times} x_k(0)) \;+\; \dot{\theta}(t)\times\big(\dot{\theta}(t)\times (e^{\theta(t)\times} x_k(0))\big) = \alpha(t)\times x_k(t) \;+\; \omega(t)\times v_k(t)$$

The angular momentum is

$$L(t) = \sum_{k=1}^N m_k x_k(t)\times v_k(t)$$

Then compute its time derivative, notice $v_k\times v_k=0$ and use a vector identity $A\times (B\times C) = B(A\cdot C) - C(A\cdot B)$:

$$\dot{L}(t) = \sum_{k=1}^N m_k x_k(t)\times a_k(t) = \sum_{k=1}^N m_k x_k(t)\times\big(\alpha(t)\times x_k(t) \;+\; \omega(t)\times v_k(t)\big)$$
$$= \sum_{k=1}^N m_k\big(\alpha \|x_k(t)\|^2 \;-\; x_k(t)(\alpha(t)\cdot x_k(t))\big) \;+\; \sum_{k=1}^N m_k x_k(t)\times (\omega(t)\times v_k(t))$$

Now the first term can be recognized to be $I\alpha$.

In general $A\times (B\times C)\neq B\times (A\times C)$, but now we are in this situation:

$$x\times (\omega\times v) = x\times (\underbrace{\omega \times (\omega \times x)}_{=\omega(\omega\cdot x) - x|\omega|^2}) = (x\times\omega)(\omega\cdot x)$$

$$\omega \times (x\times v) = \omega \times (\underbrace{x\times (\omega\times x)}_{=\omega |x|^2 - x(\omega\cdot x)}) = -(\omega\times x)(\omega\cdot x)$$

so actually $x\times (\omega\times v) = \omega\times (x\times v)$. So the second term after $I\alpha$ is the seeked $\omega\times L$.

Last edited by a moderator: Apr 24, 2017
3. Oct 19, 2009

### D H

Staff Emeritus
It cometh from working in a rotating frame.

Typically, the inertia tensor, the angular velocity, and angular acceleration are represented in body or structural frame coordinates. The inertia tensor for a rigid body is constant in a frame fixed with respect to the body. If the body is rotating, the inertia tensor is anything but constant from the perspective of an inertial frame. Rate sensors such as gyros measure angular rate in the sensor's case frame, a fixed transformation from the body frame.

Given any vector quantity $\mathbf q$, the time derivative of this vector from the perspective of an inertial frame and from the perspective of a rotating frame are related via

$$\left(\frac{d\mathbf q}{dt}\right)_{\text{inertial}} = \left(\frac{d\mathbf q}{dt}\right)_{\text{rotating}} + \boldsymbol \omega \times \mathbf q$$

Now set $\mathbf q$ to the angular momentum vector $\mathbf L = \boldsymbol I \boldsymbol \omega$:

$$\left(\frac{d\mathbf L}{dt}\right)_{\text{inertial}} = \left(\frac{d}{dt}(\boldsymbol I \boldsymbol \omega)\right)_{\text{rotating}} + \boldsymbol \omega \times \mathbf L$$

From the rotational analog of Newton's second law, the left hand side is the external torque. Assuming the inertia tensor is constant in the body frame, the above becomes

$$\boldsymbol{\tau}_{\text{ext}} = \boldsymbol I \frac{d \omega}{dt} + \boldsymbol \omega \times \mathbf L$$

The frame of reference in the above is (implicitly) the rotating body frame.

4. Oct 19, 2009

### Kricket

jostpuur: thanks for your reply, although I think I'll have to look it over when I have more time!

D H: thanks also...so, if I understand correctly, the derivation would go like (for an object centered at P):

$$\textbf{q}_{obj} = \textbf{P}_{world} + \textbf{q}_{world}$$

$$\left(\frac{d\textbf{q}}{dt}\right)_{obj} = v_P(t) + \omega \times \textbf{q}_{world}$$

etc...but I feel I'm already off track here, because it seems like I should have a rotation matrix in there to represent q relative to P...

Other question, isn't $$\omega \times L$$ equal to 0 (since the vectors are parallel)...?

5. Oct 19, 2009

### D H

Staff Emeritus
There is a big difference between the frame in which a vector is represented versus the frame from which the vector's time derivative is observed. The displacement vector from point A to point B can be expressed in many different frames. Those different representations may look vastly different, but in the end they all represent the same vector. The time derivatives of this displacement vector as observed from two different reference frames are substantively different vectors.

That said, one can use transformations between representations to derive the relation

$$\left(\frac{d\mathbf q}{dt}\right)_{\text{inertial}} = \left(\frac{d\mathbf q}{dt}\right)_{\text{rotating}} + \boldsymbol \omega \times \mathbf q$$

Start with the transformation of a vector q expressed in some reference frame "I" to a reference frame "R", with the two frames sharing a common origin:

$$\mathbf q_R = \boldsymbol T_{I \to R} \mathbf q_I$$

Here $\mathbf q_I$ and $\mathbf q_R$ are the representations of the vector q in the I and R reference frames, and $\boldsymbol T_{I \to R}$ is the transformation matrix from frame I to frame R.

Differentiating with respect to time,

$$\frac{d\mathbf q_R}{dt} = \boldsymbol T_{I \to R} \frac{d\mathbf q_I}{dt} + \frac{d\boldsymbol T_{I \to R}}{dt} \mathbf q_I = \boldsymbol T_{I \to R} \frac{d\mathbf q_I}{dt} + \frac{d\boldsymbol T_{I \to R}}{dt} {\boldsymbol T_{I \to R}}^T \mathbf q_R$$

Without derivation (the derivation is a multi-page mathout) , the time derivative of the transformation matrix is given by

$$\frac{d\boldsymbol T_{I \to R}}{dt} = \boldsymbol T_{I \to R}\boldsymbol S(\boldsymbol \omega_{I \to R:I}) = -\boldsymbol S(\boldsymbol \omega_{I \to R:R}) \boldsymbol T_{I \to R}$$

where $\boldsymbol S(\mathbf a)$ is the skew symmetric matrix generated by the vector a. $\boldsymbol \omega_{I \to R:I}$ and $\boldsymbol \omega_{I \to R:R}$ are the representations in frames I and R respectively of the angular velocity of frame R with respect to frame I. Note that the standard representation of a body's angular velocity is the angular velocity of the body frame with respect to inertial, expressed in body frame coordinates, or $\boldsymbol \omega_{I \to R:R}$.

Applying the above to the time derivative of q,

$$\frac{d\mathbf q_R}{dt} = \boldsymbol T_{I \to R} \frac{d\mathbf q_I}{dt} - \boldsymbol S(\boldsymbol \omega_{I \to R:R}) \boldsymbol T_{I \to R} {\boldsymbol T_{I \to R}}^T \mathbf q_R = \boldsymbol T_{I \to R} \frac{d\mathbf q_I}{dt} - \boldsymbol S(\boldsymbol \omega_{I \to R:R}) \mathbf q_R$$

Solving for the time derivative of $\mathbf q_I$,

$$\boldsymbol T_{I \to R} \frac{d\mathbf q_I}{dt} = \frac{d\mathbf q_R}{dt} + \boldsymbol S(\boldsymbol \omega_{I \to R:R}) \mathbf q_R = \frac{d\mathbf q_R}{dt} +\boldsymbol \omega_{I \to R:R} \times \mathbf q_R$$

6. Oct 19, 2009

### Franco_v

The first equation only applies to 2D planar motion. The second equation applies to more general 3D motion.

Note that in the 2D case the angular momentum (L) is acting in the same direction as the angular velocity (w), so wxL = 0.

7. Oct 20, 2009

### D H

Staff Emeritus
No. Did you read the earlier responses? But you did remind me of something that went unanswered:

The moment of inertia is treated as a scalar quantity in freshman physics problems for the simple reason that freshmen do not have the mathematical background to deal with the general problem of rotation.

The inertia tensor is a second order tensor, a 3x3 symmetric matrix. Angular momentum is related to angular velocity by $\mathbf L = \boldsymbol I \boldsymbol \omega$. These will be parallel in the special case of rotation about an eigenaxis of the inertia tensor. The general case, in which the instantaneous rotation axis is not an eigenaxis, angular velocity and angular momentum are not parallel.

8. Oct 20, 2009

### Kricket

Thanks, DH, for writing that all out. So what you're saying is, for example, if I had a teapot that I threw into the air, and I use the earth as my fixed world reference frame, (and we ignore translation for the time being) then I would pick "I" as the earth-frame, "R" as the teapot-frame, and $$\omega$$ would be the rotation of the teapot frame relative to the earth - i.e. the angular velocity vector expressed in the earth's basis...?

And then for the torque example, L must be expressed in the teapot's basis, so I get the cross product of two vectors expressed in different bases..?

Or perhaps not, given your explanation of why $$\omega$$ and L aren't likely to be parallel. But in the special case of a sphere, they always would be, right?

9. Oct 20, 2009

### Kricket

Ah, no, sorry, no funny mixing of references in the cross product. :) Reference frames are a real PITA.

Intuitively, what does this quantity represent? If I just have some random object in space with no angular acceleration, the formula seems to suggest that torque will be non-zero...

10. Oct 20, 2009

### D H

Staff Emeritus
Almost. ω is the angular velocity vector of the teapot with relative to the earth, but it is expressed in the teapot's basis. Why this basis? The inertia tensor is in the teapot's basis. The inertia tensor is constant in a frame fixed with respect to the teapot. It is not constant when expressed in inertial coordinates. Because the inertia tensor is in the teapot's basis, the angular velocity must also be in the teapot's basis for the product Iω to be valid.

Correct.