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?
 A: 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
$$\vec{\boldsymbol{\tau}}=\vec{\boldsymbol{r}}\times\vec{\boldsymbol{F}}\;\blacksquare$$
A: 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:
$$ |r_i|^2 = \vec{r_i} \cdot \vec{r_i}$$
And you had written as:
$$ I = \int \vec{r} \cdot \left[ \vec{r_i} \rho(x,y,z) \Delta V_i  \right]$$
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
A: 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.
