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I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot \left(\delta\vec \phi \times \vec r_i\right)\\ &= \left(\sum \vec r_i \times\vec F_i \right) \cdot \delta\vec \phi \end{align} $$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

 $$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Hope this helps and tell me if you find some mistakes.

Usually, the motivation behind introducing true is to determine the equilibrium position of a rigid body allowed to rotate about a point. A fundamental theorem of statics says that such a body is at equilibrium iff the resulting torques of all the forces applied on it with respect to the rotation point is zero. Contrary to most derivations, angular momentum is hardly relevant for statics. It is certainly a relevant quantity to consider when a system is invariant by rotation as a consequence of Noether's theorem, but does not shed much insight for statics.

In order to characterise equilibrium, the usual approach for constrained systems is to use the principle of virtual work. Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot \left(\delta\vec \phi \times \vec r_i\right)\\ &= \left(\sum \vec r_i \times\vec F_i \right) \cdot \delta\vec \phi \end{align} $$

The principle of virtual work says that static equilibrium is equivalent to a vanishing virtual work for any possible virtual displacement (obeying the constraints). Therefore, equilibrium is equivalent to: $$ \sum \vec r_i \times\vec F_i = 0 $$ and you naturally introduce torque as the LHS. This proves the aforementioned theorem. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):  $$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Note that the theorem can be extended to any solid body, and you just need to add the prescription that the resulting force is also zero. This guarantees that if torque vanishes about a certain point, it also vanishes about any point.

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot (\delta\vec \phi \times \vec r_i)\\ &= (\sum \vec r_i \times\vec F_i ) \cdot \delta\vec \phi \end{align} $$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

$$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Hope this helps and tell me if you find some mistakes.

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot \left(\delta\vec \phi \times \vec r_i\right)\\ &= \left(\sum \vec r_i \times\vec F_i \right) \cdot \delta\vec \phi \end{align} $$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

$$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Hope this helps and tell me if you find some mistakes.

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \delta W = \sum \vec F_i \cdot (\delta\vec \phi \times \vec r_i) $$

$$ \delta W = (\sum \vec r_i \times\vec F_i ) \cdot \delta\vec \phi $$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

$$ P = (\sum \vec r_i \times\vec F_i ) \cdot \vec \omega $$

Hope this helps and tell me if you find some mistakes.

I personally prefer a derivation using the principle of virtual work where the formula of torque directly comes out. While angular momentum is a natural property to consider for a spherically symmetrical problem, this alternative approach shows its relevance for statics of rigid bodies even when this symmetry is not present.

Take a set of points indexed by $i$ at position $\vec r_i$, on which are applied respectively the forces $\vec F_i$. This gives first formula of vitual work for a general displacement: $$ \delta W = \sum \vec F_i \cdot \delta\vec r_i $$ Furthermore, lets assume the points are rigidly constrained and can only rotate around the origin. Any allowed differential displacement can thus be written as $\delta\vec r_i =\delta\vec \phi \times \vec r_i$ where $\vec \phi$ is the differential angular displacement. Injecting in the work you get: $$ \begin{align} \delta W &= \sum \vec F_i \cdot (\delta\vec \phi \times \vec r_i)\\ &= (\sum \vec r_i \times\vec F_i ) \cdot \delta\vec \phi \end{align} $$

So static equilibrium is equivalent to a vanishing virtual work for any relevant virtual displacement, hence $\sum \vec r_i \times\vec F_i $, the torque naturally pops out. It also explains also the useful power formula for rotation (with angular velocity $\vec \omega$):

$$ P = \left(\sum \vec r_i \times\vec F_i \right) \cdot \vec \omega $$

Hope this helps and tell me if you find some mistakes.