# Where is angular momentum supposed to be conserved when dealing with questions on rotation

What I know

Angular momentum is conserved at the point where there is no external torque on the system

When solving questions based on pure rolling (fairly simple concept), if for example, we have a ball that is slipping and not rolling on a rough surface, we are asked to find the velocity when pure rolling starts. Out of the various ways to solve it, one that has always confused me was the conservation of angular momentum method.

Friction acts on the point of contact, so angular moment can be conserved at point of contact.

So the general formula is

$$mv_0r=I\omega +mvr$$

Where, $$\omega=\frac vr$$

But the moment of inertia $$I$$, is taken about the center of mass, and not about the point of contact, to get the right answer.

Now I could have lived with that, perhaps angular momentum always has the moment of inertia taken about the COM. But here is another question:

A uniform rod AB of mass m and length $$2a$$ is falling freely without any rotation under gravity with AB horizontal. Suddenly, the end $$A$$ is fixed when the speed of rod is $$v$$. Find the angular speed of rod with which rod begins to rotate.

$$L_i=L_f$$

$$mva=I\omega$$

$$mva =\frac{m(2a)^2}{3} \omega$$ $$\omega =\frac{3v}{4a}$$

In this case, moment of inertia is calculated about the point of contact A, and not the center of mass.

I just want to know when do we calculate MOI about the COM, and when it’s calculation is about the point of contact. Such questions always confuse me and I want to confirm it once and for all.

$$𝑚𝑣_0𝑟=𝐼𝜔+𝑚𝑣𝑟$$

The RHS is equal the the angular momentum about the point in contact with the ground, where the friction acts. But in order to calculate that, you need the momentum of inertia about that point, which would be done through integration. An easier way to get the angular momentum about any general point, $$A$$, is to calculate the angular momentum about the centre of mass of the system, which is $$I\omega$$, and add to it the angular momentum of the centre of mass about the point $$A$$, which is $$mvr$$. This is what the first example in your question does. In the example of the rod, you have the moment of inertia of the rod known exactly about its end point, so you do not need to calculate it about the centre of mass first and then get the angular momentum about the end point.

• So if I could find the moment of inertia at the point of contact for any object,I can simply write $mv_0r=I\omega$? Jun 14 '20 at 7:08
• Yes, if $I\omega$ is the angular momentum about that point and it is conserved, then you can do that. For example, if you pin the falling rod at its center (instead of the endpoint), we have r = 0 and hence, $\omega=0$, as expected. Jun 14 '20 at 9:22 If you take a "Free Body Diagram" you can see where your equations came from.

The contact force $$F_R$$ act opposite direction to the velocity $$v$$ and also the inertia force $$m\,\dot{v}$$. The inertia torque $$I\,\dot{\omega}$$ act opposite direction to the wheel rotation.

with the external force $$F$$ equal zero you obtain:

$$\sum_M F_y=m\frac{d}{dt}\,v+F_R=0\tag 1$$

$$\sum_M \tau_x=I\,\frac{d}{dt}\,\omega-F_R\,R=0\tag 2$$

and the roll condition

$$v=\omega\,R\tag 3$$

multiply equation (1) with $$R$$ and add to equation (2)

$$m\,R\frac{d}{dt}\,v+I\,\frac{d}{dt}\,\omega=0$$

or

$$m\,R\,\int_{v_0}^v dv+I\,\int d\omega=0$$

$$\Rightarrow$$

$$m\,R\,(v-v_0)+I\,\omega=0\quad,m\,R\,v+I\,\omega=m\,R\,v_0$$

This is your equation, thus the Inertia $$I$$ is the "center of mass inertia"