# Derive Torque $\tau = \mathbf{r} \times \mathbf{F}$ from $\tau= I\alpha$

I have started learning about rotational mechanics about a month back. I am trying to derive everything from the fundamentals.<br>


First off, I derive the relation for the moment of inertia using the Kinetic energy argument. ($$I= \int_0^m r^2 dm$$)
From here by analogy to linear dynamics, I state Angular momentum $$\mathbf{L}= I \boldsymbol{\omega}$$ Again, as an analogy to Newton's 3rd Law, I say Moment of Force (Torque) = rate of change of angular momentum= $$I \boldsymbol{\alpha}$$

Now, I am having trouble proving that $$I \boldsymbol{\alpha} = \mathbf{r} \times \mathbf{F}$$ ($$I, \boldsymbol{\alpha}$$ and $$\mathbf{r}$$ are w.r.t the same axis). I am able to prove it easily for a point mass system, where the force is applied on the mass directly, but unable to prove it for a general Rigid body. $$I$$=moment of inertia

$$\boldsymbol{\alpha}$$= Angular acceleration vector

$$\boldsymbol{\omega}$$=Angular velocity vector

$$\mathbf{r}$$ = Position vector

$$\mathbf{F}$$ = Force vector

EDIT
I tried something new but it seems contradictory...

torque= I $$\alpha$$ = $$\alpha \int_0^m r^2 dm$$ =$$\alpha ( \int_0^m r \times r dm)$$

now using integration by parts,
torque=$$\alpha (r \int_0^m r dm - \int_0^m ((dr/dm) \int_0^m r dm)dm$$

By using the relation ,
F= $$\alpha \int_0^m r dm$$
torque=$$\alpha (r (F / \alpha ) - (F/ \alpha ) (\int_0^r dr)$$ =$$\alpha (r (F / \alpha ) - r (F / \alpha )) =0$$

this is certainly not true as torque may not always be 0. What am I doing wrong here? Also can someone answer my original question with a similar approach?

It isn't always true that $$\vec{\boldsymbol{\tau}}=I\vec{\boldsymbol{\alpha}}$$. The general form of torque is $$\vec{\boldsymbol{\tau}}=\dfrac{d}{dt}\vec{\boldsymbol{L}}$$ From definition of angular momentum $$\vec{\boldsymbol{L}}=\vec{\boldsymbol{r}}\times \vec{\boldsymbol{p}}$$ gives \begin{align*} \vec{\boldsymbol{\tau}}&=\dfrac{d}{dt}(\vec{\boldsymbol{r}}\times \vec{\boldsymbol{p}})\\ \vec{\boldsymbol{\tau}}&=\vec{\boldsymbol{r}}\times \dfrac{d\vec{\boldsymbol{p}}}{dt}\\ \end{align*} and from Newton's second law of motion $$\vec{\boldsymbol{F}}=\dfrac{d}{dt}\vec{\boldsymbol{p}}$$ then $$\vec{\boldsymbol{\tau}}=\vec{\boldsymbol{r}}\times\vec{\boldsymbol{F}}\;\blacksquare$$

• well this method raises a similar question of why L=r $\times$p (i started with the definition L=I $\omega$), Still thanks for your effort :) Nov 10, 2020 at 13:15
• Examine any intro to math physics textbook and you'll find the answer readily. Anyway, the angular momentum is defined as the cross product $L = r \times p$ because $L$ will always be perpendicular to the plane containing $r$ and $p$. Thus, the equation $L = I\omega$ is more subtle than you probably know: the inertia $I$ is actually a rank 2 tensor, also called a matrix, so only in very simple cases does $L$ point in the same direction as $\omega$ (i.e. when I is a diagonal matrix). en.wikipedia.org/wiki/Moment_of_inertia#Inertia_tensor Nov 10, 2020 at 13:58
• Nice explanation here farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html Nov 10, 2020 at 14:00
• @saitama because $L=I\omega$ isn't the definition of angular momentum.
– Ken
Nov 11, 2020 at 6:34

## Context

If we were being more explicit, we define torque about some origin $$\mathrm{O}$$ around this origin we can rewrite an inertia element of some mass element some where as:

$$\Delta I_{i} = |\vec{r_i}|^2 \Delta m_i$$

Where,

$$\Delta I_i \text{ is the inertia of the mass element}$$

$$\Delta{\vec{r_i}} \text{ is the magnitude of position vector from the defined origin O}$$

$$\Delta{ m_i} \text{ is the mass element we are considering}$$

Though we cant directly apply the definition of inertia to the whole system, we can indeed apply the definition of it for a point mass on to a small mass element. We can think of shrinking the mass element such that the whole mass is concentrated in a small enough region that we could approximate it.

Better yet, a more intelligent way would be to write this expression using density. Assume we have an object with varying density $$\rho$$ which depends on $$(x,y,z)$$ then we can say the mass of that element in some n-volume around region of inspection is given by $$\rho \Delta V_i$$ (*). This turns our equation to this form:

$$\Delta I_i = |r_i|^2 \rho(x,y,z) \Delta V_i$$

Now, given a rigid body, we can think of as partitioning it into small n-volume elements of $$\Delta V_i$$ and finding the inertia of each and summing over that. So, what we can do is sum over all the n-volume elements using an n-dimensional integral (**).

$$I = \int |r_i|^2 \rho(x,y,z) \Delta V_i$$

## The problem in your post:

In the second method, you had a step where you had done the following:

$$|r_i|^2 = \vec{r_i} \cdot \vec{r_i}$$

$$I = \int \vec{r} \cdot \left[ \vec{r_i} \rho(x,y,z) \Delta V_i \right]$$

And integrated the above quantity by parts, however, when you do this you are integrating a vector field ($$\int \vec{r} \Delta V_i$$)over a volume. This operation does not make any sense by the mathematics that I know of / can find by googling. The closest thing I found was this quora post.

References:

For a good explanation of how to derive this in detail go see Kleppner and Kolenkow around page-245

Note: n-volume is the sort of generalization of volume

CLUE:

Radial force on a point:m.$$\omega^2$$r

Tangential force on it:mr$$\alpha$$ (Torque due to it :m.$$r^2.\alpha$$)

Sum over all particles for torque.

$$\tau^{net}=\sum[...]=I.\alpha$$

Think about whether or not radial force provides any torque or not.

• this method (tang force= mr $\alpha$ ), looks obvious for a point mass m in circular motion of radius r, where force is applied on the point mass tangentially, but how is this relation valid for a continuous mass( rigid body)? (i'm assuming that by r, you mean to refer to position vector of point of application of force) Nov 10, 2020 at 4:14
• @saitama If you have understood that part, For a rigid body it is easy, We just have to sum/integrate it for every possible point on the rigid body. Nov 10, 2020 at 4:21
• ok, so I have managed to understand why F=mr $\alpha$ for a rigid body, But in third line you say summing over all particles......gives torque. How does summing tangential force give torque? even their dimensions are also differenent Nov 10, 2020 at 5:10
• @saitama I've made an edit(I am summing torque not force), does it seem clear now? Tell me if it doesn't. Nov 10, 2020 at 5:14
• ok now i have got it. Thanks. Also you may look into the new approach i tried (the edit in the question), and tell me what is wrong there, although I have got the answer to my main question by your method. Nov 10, 2020 at 5:37