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I'm currently studying 2D rigid body dynamics. To describe the translational motion of a 2D rigid body, the linear momentum form of Newton's second law is used:

$$\vec{F}_{\rm ext}=m_G\vec{a}_{G/O}$$

where $m_G$ is the mass of the rigid body and $\vec{a}_{G/O}$ is the acceleration of the CM of the rigid body relative to the origin O of the reference frame.

However, in describing the rotational motion of the rigid body relative to its CM, the angular momentum form of Newton's second law is always used:

$$\vec{\tau}=I_G\vec{\dot{\omega}}$$

The question I have is, what's the advantage of using the angular momentum form of Newton's second law over using the linear momentum form in deriving the equation of motion for the rotational motion of the rigid body? Couldn't you derive an equivalent equation using the linear momentum form? Why do all textbooks tend to use the angular momentum form?

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When working with a rotating object, angular quantities (instead of linear) are used because every point in the object has a different linear velocity and acceleration (as vectors)

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Couldn't you derive an equivalent equation using the linear momentum form?

Yes you can.

what's the advantage of using the angular momentum form of Newton's second law over using the linear momentum form in deriving the equation of motion for the rotational motion of the rigid body?

If the rigid body is composed of N particles, there is an external force acting on every particle. On top of this, there is force of constraints that makes body rigid. This is the force, which makes all the particles to keep their distances from each other. You do not know this force, you only know its effect.

The angular formulation is convenient, because the constraints are satisfied automatically and you do not need to deal with them. It also reduces the amount of equations you need to deal with from 3N (3 equations for a particle) to just 4 (3 equations for motion of CM plus 1 equation for a 2D rotation).

In linear form case, you have 3N equations $$\vec{F}_{i;\text{ext}}+\vec{F}_{i;\text{constr}}=m_i\vec{a}_i,$$ where $\vec{F}_{i;\text{ext}}$ is external force acting on ith particle, $\vec{F}_{i;\text{constr}}$ is internal force of constraint, $m_i$ is mass of ith particle and $\vec{a}_i$ its acceleration.

To get rid of the forces of constraint you can first make a cross product with $\vec{r}_i,$ which is position vector from axis of rotation to the ith particle $$\vec{r}_i\times\vec{F}_{i;\text{ext}}+\vec{r}_i\times\vec{F}_{i;\text{constr}}=m_i\vec{r}_i\times\vec{a}_i,$$ Now, let us add all particles up $$\sum_i\vec{r}_i\times\vec{F}_{i;\text{ext}}+\sum_i\vec{r}_i\times\vec{F}_{i;\text{constr}}=\sum_im_i\vec{r}_i\times\vec{a}_i.$$ The first sum is simply the torque acting on the body $\vec{\tau}$. The second term is torque produced by constraints themselves. This is zero, as constraints are supposed to keep the distance of the points of the body fixed relative to themselves and they are not supposed to accelerate the rotation of the body (see below). The right hand side can be simplified in the case of rigid body using the angular acceleration $\vec{\epsilon}$, which is the same for all particles and perpendicular to $\vec{r}_i$ (the rotation is 2D) $$\sum_im_i\vec{r}_i\times\vec{a}_i=\sum_im_i\vec{r}_i\times\left(\vec{r}_i\times\vec{\epsilon}\right)=\vec{\epsilon}\sum_im_ir_i^2=I\vec{\epsilon}.$$ Thus $$\vec{\tau}=I\vec{\epsilon}.$$

If you would want to use the linear form, you would need to go through this whole derivation again and again. It is better to just remember the result and go from there.

Note:

The better reason why $$\vec{r}_i\times\vec{F}_{i;\text{constr}}$$ is zero, is to realize that $\vec{F}_{i;\text{constr}}$ are simply internal forces of the system and internal forces produce no torque on the system due to 3rd Newton law.

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