Angular momentum in center of mass frame While doing homework I seem to come across a contradiction. The problem is very simple, a ball with radius R is attached to a rope of length L. The pendulum is subject to gravity and oscillates in a plane. I used angular momentum to solve the problem. But when I consider the angular momentum w.r.t. the center of mass G, I get $\vec L_G = \tilde I_G \vec \omega$ and $\frac{d\vec L}{dt}= \vec M_G=0$ since the vectors of the forces have their origin at G. However, it's clear that $\omega$ is not constant. What's wrong with my reasoning?
And in general, when we calculate $\vec L$ with respect to the center of mass, which is rotating, does the formula $\vec L_G = \tilde I_G \vec \omega$ still apply? Is $\omega$ always the total angular velocity? Do I have to calculate the inertia tensor I with respect to a point on the axis of rotation?
 A: If the ball is suspended from its centre of mass (CoM) then it does not rotate as it swings (diagram 1). The forces on it act through the CoM and there is no torque. The ball  maintains the same orientation in space. This can be described as a simple pendulum.
In your case if the ball is rotating as it swings then as you deduce its angular momentum must be changing. This could happen if the string or rod is attached at the rim of the ball (diagram 2). Now it is possible for the ball to rotate around the point of attachment. As this happens the tension in the string or rod does not pass through the CoM of the ball. So there can be a varying torque on the ball which causes a rotation in space. This is a compound pendulum.
Another arrangement is to have the stiff rod attached at two points on the ball, instead of only one (diagram 3). Now the forces on the ball from the rod need not act through the same point, so a torque can be created which rotates the ball in space while it swings.

In another arrangement a heavy ball could rest on the seat of a swing. It is impossible to set the swing oscillating like a pendulum without the ball rolling forward and backward on the seat. A varying torque is required to explain this motion.
If you account for the forces and torques on the ball correctly, the formula $L_G=I\omega$ still applies but $\dot L_G =M_G \ne 0$.
A: Ok, so the ball (and its center of mass CM) rotates around the pivot point $O$ with angular velocity $\omega$ and angular acceleration $\alpha$. The equation of motion follows from $$mL^2\alpha =I_{\text{ball}} \alpha= -mgL\sin\theta $$
The angular momentum of the ball w.r.t point $O$ is $L_{\text{ball}}=I_{\text{ball}} \omega$. Note that the torque is $\dot{L}=-mgL\sin\theta$ from the above relation.
The above discussion was looking from the point $O$. Now let's look from the center of mass.
We have that the point $O$ rotates around CM with the same  angular velocity $\omega$ and angular acceleration $\alpha$. However, $O$ is simply a point with mass zero so $I_{\text{point } O} = 0$. The formula $L = I_{\text{point } O}\cdot\omega$ is true for the point $O$ but simply yields $L_{\text{point } O}=0.$ The torque of point $O$ is similarly zero.
The confusing fact is that the torque and angular momentum of the ball w.r.t. the center of mass is also zero but this is because the angular velocity and acceleration of the ball from CM are zero: the ball does not rotate at all around its center of mass (I'm assuming you are talking about a simple pendulum as in @sammy gerbil's answer.)
A: I assume that you solve the problem calculating the torque from the pivot point, and using that $\tau = I\alpha$, as explained in the mechanodroid's answer.
The problem of calculating the torque from the COM is that it is not an inertial frame. The expression $\tau = I\alpha$ (as the second Newton's Law: $F = ma$) is valid for inertial frames.
In the case of an object without net forces applied, the COM is an inertial frame and the expression for the spin: $L = I\omega$ is valid when calculating from it. Even if the COM is not an inertial frame, but when the external force can be considered as applied on it (as orbiting bodies under gravitational force), $L = I\omega$ is also valid.
