A question involving Torque I have a few questions which have all seemingly originated from the following thought:
Question 1
It is known that torque is given by τ=r × F. If we use the formulae F=mdv/dt and v=ω × r, we arrive at:
τ=r × m(dω/dt × r + ω × dr/dt)=-m(r × r × dω/dt)+m(v × v)=$0$
My first thought was: Why is this happening?
I concluded that the torque is due to the external force that we apply and not the net force that is a acting on a body. More specifically in the case of a door, there are two forces acting on the door: the force that we apply on the door and the force that prevents the door from breaking off its hinges when we exert the force; and the torque we compute is the cross product between the force that we exert and r. (I have answered this question of mine, but the questions which it leads to, I haven't been able to answer).
Question 2
If my earlier deductions are true shouldn't angular momentum be always conserved (irrespective of whether the torques is applied or not)? More specifically, if we plug in l=r × p in dl/dt we arrive at r × F. However, here F is the net force (according to the definition of angular momentum in my textbook) and not the external force that we exert on the body. Thus, dl/dt=$0$. Therefore, why isn't angular momentum written to be conserved in all cases in every textbook that I have read?
Question 3
This final question is a bit disconnected. Whenever a body revolves around a point, the force on the body has to be pointing towards the center of revolution. So how can one explain that in order to make a door move, we receive the greatest success when we apply the force perpendicular to the door? (Explanation must not use methods that have the above shortcomings.)
 A: Question 1-2
Your math is all wrong: the cross product is neither associative nor commutative.
$$  \overrightarrow{A}  \times  \overrightarrow{B}=- \overrightarrow{B}  \times  \overrightarrow{A}$$
$$ \overrightarrow{A}  \times  \big( \overrightarrow{B}  \times  \overrightarrow{C} \big)=  \big(\overrightarrow{A}  .\overrightarrow{C}\big)\overrightarrow{B} -\big(\overrightarrow{A}  .\overrightarrow{B}\big)\overrightarrow{C}$$
Question 3:
If the force points toward the axis of rotation it has no torque and as a result there is no change in rotation...
A: There appears to be a mistake in your proof that $\tau = 0$:
You appear to assume that
$$ \vec r \times m(d\vec\omega/dt \times \vec r) = - (\vec r \times \vec r) \times md\vec \omega/dt$$
and then say that this is equal to 0 as $\vec r \times \vec r$ is 0. However, this is incorrect. The triple vector product rule states that
$$ \vec A \times (\vec B \times \vec C) = \vec B(\vec C\cdot \vec A) - \vec C(\vec A\cdot \vec B)$$
and
$$ (\vec A \times \vec B) \times \vec C = \vec B(\vec C\cdot \vec A) - \vec A(\vec B\cdot \vec C) $$
which are not equivalent.
Using the triple vector product rule, we get
$$ \vec r \times m(d\vec \omega/dt \times \vec r) = - \vec r \times m(\vec r \times d\vec \omega/dt) = -\vec r(\vec r \cdot d\vec \omega/dt) + d\vec \omega/dt(\vec r\cdot \vec r)$$
$d\vec \omega/dt$ is perpendicular to $\vec r$, so $\vec r \cdot d\vec \omega/dt = 0$, and $\vec r \cdot \vec r = r^2$. This gives us
$$ \vec \tau = mr^2 d\vec \omega/dt = Id\vec \omega/dt = d\vec l/dt$$
Where I is the moment of inertia of a point mass m at distance r from the axis of rotation. So this shows that torque is the rate of change of angular momentum, as we would expect.
As for question 3:

Whenever a body revolves around a point, the force on the body has to be pointing towards the center of revolution

The force that constrains the body to circular motion points towards the centre, not the force causing the rotation. In your door example, the inwards radial force is the tension within the door material between the door handle and the hinges.
The reason we get the most success with a force perpendicular to the door is shown in the original equation
$$ \vec \tau = \vec r \times \vec F$$
Or in terms of magnitudes
$$\tau = rF sin(\theta) $$
where $\theta$ is the angle between our applied force and (in this example) the surface of the door. This shows that we generate the greatest torque when $\theta = 90°$, I.E the force is perpendicular to the door (and also shows why we push as far away from the hinges as we can).
A: First some notation to make things simpler. I use $\boldsymbol{\dot{\omega}} = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{\omega}$, and $\boldsymbol{\dot{v}} = \tfrac{{\rm d}}{{\rm d}t} \boldsymbol{v}$.
Here are some points to consider:

*

*In $\boldsymbol{\tau} = \boldsymbol{r} \times \boldsymbol{F}$ and $\boldsymbol{F} = m \boldsymbol{\dot{v}}$, the vector $\boldsymbol{F}$ is not the same thing. In the first expression, $\boldsymbol{F}$ designates the equipollent torque from a specific force. In the second expression, $\boldsymbol{F}$ is the net force applied to a body. Those are only the same if there is only one force applied on a body, but this is not stated in the question. If there are other forces to consider they need to be included in in the net force.
Overall you have to be careful when using variables such as $\boldsymbol{F}$ and $\boldsymbol{r}$ as you need to be specific of what they mean (when they measure).


*Newton's second law specifically describes the motion of the center of mass where $\boldsymbol{F}_{\rm net} = m \boldsymbol{\dot{v}}_{\rm com}$, regardless of where the forces are applied.


*So my stating $\boldsymbol{v}_{\rm com} = \boldsymbol{\omega} \times \boldsymbol{r}_{\rm com}$ you are describing the kinematics of a pivoted rigid body where $\boldsymbol{r}_{\rm com}$ defines the location of the center of mass of the body relative to the pivot.


*But by stating $\boldsymbol{\tau}_{\rm pivot} = \boldsymbol{r}_{\rm com} \times \boldsymbol{F}_{\rm com}$ you are defining the torque about the pivot of a force through the center of mass. But what is important for a pivoted body is the torque of the pivot force about the center of mass, which would be $\boldsymbol{\tau}_{\rm com} = -\boldsymbol{r}_{\rm com} \times \boldsymbol{F}_{\rm pivot}$.
The general theme of carelessness in throwing expressions together in dynamics is rather dangerous. I recommend drawing a free body diagram force and including all relevant forces/torques in the calculation.


*The calculation of acceleration needs specific labels, and is thus $\boldsymbol{\dot{v}}_{\rm com} = \tfrac{\rm d}{{\rm d}t}( \boldsymbol{\omega} \times \boldsymbol{r}_{\rm com}) = \boldsymbol{\dot{\omega}} \times \boldsymbol{r}_{\rm com} + \boldsymbol{\omega} \times \boldsymbol{v}_{\rm com} $


*Finding the dynamics of a pivoted body requires two equations. Newton's 2nd law, and Euler's law of rotation. In the specific scenario of a pivoted body with a single applied force about the center of mass $\boldsymbol{F}_{\rm com}$, you need to consider the pivot reaction force also $\boldsymbol{F}_{\rm pivot}$
$$ \begin{aligned} \text{linear} &\\
  \boldsymbol{F}_{\rm net} & =m \boldsymbol{\dot{v}}_{\rm com}  \\ \hline
 \boldsymbol{F}_{\rm pivot} + \boldsymbol{F}_{\rm com} & = m \left( \boldsymbol{\dot{\omega}} \times \boldsymbol{r}_{\rm com} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{r}_{\rm com}) \right) \\
\end{aligned} $$
$$ \begin{aligned} \text{rotational} & \\
   \boldsymbol{\tau}_{\rm net} & = \mathrm{I}_{\rm com} \boldsymbol{\dot{\omega}} + \boldsymbol{\omega} \times ( \mathrm{I}_{\rm com} \boldsymbol{\omega}) \\ \hline -\boldsymbol{r}_{\rm com} \times \boldsymbol{F}_{\rm pivot} & = \mathrm{I}_{\rm com} \boldsymbol{\dot{\omega}} + \boldsymbol{\omega} \times ( \mathrm{I}_{\rm com} \boldsymbol{\omega})
\end{aligned} $$
The above two equations (6 component equations) need to be solved for $\boldsymbol{F}_{\rm pivot}$ and $\boldsymbol{\dot{\omega}}$ (6 unknown components) in order to understand how the system behaves.


*The net torque about the center of mass is not zero, contrary to the question asked.
$$ \begin{aligned}\boldsymbol{\tau}_{{\rm net}} & =-\boldsymbol{r}_{{\rm com}}\times\boldsymbol{F}_{{\rm pivot}}\\
 & =-\boldsymbol{r}_{{\rm com}}\times m\left(\boldsymbol{\dot{\omega}}\times\boldsymbol{r}_{{\rm com}}+\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times\boldsymbol{r}_{{\rm com}}\right)\right) 
\end{aligned} $$
regardless on how the vector triple products are applied $a \times (b \times c) = b \times (a \times c) - c \times (a \times b) = b  (a \cdot c) - c  (a \cdot b) $. Only if the pivot is on the center of mass with $\boldsymbol{r}_{{\rm com}}=0$ then the net torque is zero.
