# Center of mass of angular momentum, torque and similarity between linear and angular quantities

Linear momentum of a system of particles is given by

$$\vec p_{\rm net}=\vec p_1+\vec p_2+\vec p_3+ \ldots + \vec p_n$$

Where $$\vec p_{\rm net}$$ is total linear momentum of the system and expression on RHS is vector sum of the individual particles' linear momenta

We can also write linear momentum of system of particles as $$\vec p_{\rm net}=m~\vec v_c$$ where $$m$$ is total mass of the sytem and $$\vec v_c$$ is velocity of center of mass

My question:

1. Just like we say "the linear momentum of a system of particles is equal to the product of the total mass $$m$$ of the system and velocity of the center of mass", can we say $$\vec L=m (\vec r_c ×\vec v_c)$$ where $$L$$ is total angular momentum of the system of particles, $$m$$ is total mass of the system of particles $$\vec r_c$$ is position vector of center of mass with respect to origin and $$\vec v_c$$ is velocity vector of center of mass?

2. Can we say $$\vec T_{\rm net} = \vec r_c \times \vec F_{\rm net}$$ (Net Torque= Position vector of Center of Mass × Net force vector acting on system of particles) the way we say $$\vec F_{\rm net}=m~\vec a_c$$ for Force?

The equation $$\vec{p} = m \vec{v}$$ tells you that the linear momentum of a system depends on its mass and its velocity, and nothing else.
For rotating objects, this isn't the case. In addition to the mass of the rotating body, the distribution of the mass about the center of rotation is also important. This combined mass and its distribution quantity is represented by the mass moment of inertia. So you get the equation $$\vec L = I~\vec{{\omega}}$$, where $$I$$ is the moment of inertia, and in rotational equations, is analogous to what mass would be in linear equations. That's why it's also sometimes called angular mass.
This is probably easier to demonstrate with Newton's Second Law: $$\vec F = m ~ \vec a$$ tells you that the more massive an object is, the larger the force needed to produce the same acceleration. In the rotational world, this would be $$\vec T = I ~ \vec{{\alpha}}$$, where $$\alpha = \dot \omega$$ is angular acceleration, the time-derivative of angular velocity. This can make intuitive sense if you think about pushing a merry-go-round with some children on it -- it's takes less torque to obtain the same acceleration when the children are close to the center than it is when the children are at the edge. This is because in the former case, the mass is distributed close to the center, giving a smaller moment of inertia. This also probably answers your second question for the rotational analogue to Newton's Second Law. $$\vec T = \vec r \times \vec F$$ is the definition of torque itself.
No your formula for L is wrong, what would you take as vc, v is different for the rotating particles, small near the center of rotation, increasing with distance to the center. when you ad up all these $$m*v=m*r^2*\omega$$ you find a new concept called "Moment of inertia" I $$\vec{L}=I*\vec{\omega}$$ read about Moment of inertia in wikipedia