# Conservation of angular momentum in broken pendulum

I'm having troubles in understanding the correct way to use the conservation of angular momentum. Consider the pendulum in the picture with a neglegible mass rope and a slender rode of mass M and lenght L. Suppose that the rope is cut off when it is passing through the vertical. What will the motion of the rod the be?

I'm totally ok with the fact that the center of mass will behave as a point and therefore will follow a parabola, but how do I interpret correctly the rotational motion that the rod follows?

From Koenig theorem

$L=L'+L_{cm}=I_{cm} \omega +L_{cm}$

Where L' is the angular momentum measured in c.o.f. frame and therefore is rotational. But in this case is $\omega\neq 0$ ? If so I don't see why, since the rod, before the cutting is not rotating around its c.o.f., so why should it do so after ?

One thing that I was thinking about it that the velocity of the rod is different in the points near the rope and in the points further from the rope. But how is this difference of velocities translated in to an angular velocity?

• i think there is some relation between rate of change w.r.t. to time in fixed frame and in rotating frame of reference-that may help in clarifying the physical picture-as the frame attached to rod seems to be rotating. Mar 5, 2016 at 10:55

Well, the rod is rotating around its center of mass before the rope is cut. The angle between the rod and the vertical axis that passes through the rod's center of mass is changing with respect to time, which means, the rod is rotating. However, since the rod's center of mass rotates around point P with the same angular velocity as the rod's angular velocity around the center of mass of the rod, the rod is always oriented along the rope; which I guess is why you think it is not rotating before the rope is cut.

1. The rod is rotating about its centre of mass. Observe the orientation of either end of the rod and see how it changes with time. At the extreme position, the rod was tilted. But at lowest point, the rod is completely vertical. This is because the rod is rotating. It just seems like there is no rotation, because at any instant th angular velocity of the rod's centre about the point of suspension of the thread is equal to the angular velocity of the rod about its own centre. The centre of the rod about the suspension point of the thread and the tip of the rod about the rod's centre have the same "rate of angle change" at any instant of time, giving the illusion that it is not rotating.

2. Now, we need to account for the angular velocity acquired by the rod. Clearly, the velocity and angular velocity of the rod are both zero at the extreme position. So how did the rod gain angular momentum along the way(about it centre of mass)? About the rod's centre of mass, the torque of gravity is zero. The torque of the tension of the string appears to be zero, but it is not. Let us assume that the rod is attached to the thread with help of a hook. The hook cannot be attached to the rod at exactly one point. There will be a small area of contact. Thus, there are multiple contact points at which different forces act, the sum of all these forces being equal to the tension in the string. So the torque of the pulling tension is non-zero, even when taken about the rod's COM , as the line of action of all these forces will not pass through the centre. This is the reason for the change in angular velocity of the rod. It is wrong to conserve the angular momentum before the string is cut. You may, however do it after the string has been cut. Hope this helps.

The speed of rod sections will be proportional to the distance from the center. Therefore, when string is cut, the points of the rod that were closer to the center will move forward slower than the more distant ones. This is what I think constituates the nature of angular momentum (I have heard that it is not fundamental and can be reduced to simple, translational momentum and attraction of the bodies that rotate around each other).

Imagine a rod of the full length, with no string. The elements of the rod at the center will not experience the translational motion at all. You'll see immediately that the rod keeps rotating.

Given the linear distribution of speeds, we can easily make sure that the center of the rod, which is easy to find, will fly at the average speed. This confirms that the ends will rotate into the opposite directions. This will conserve the momenutm -- the rod will keep rotating at somewhat hither angular speed. This is what expect.

That is, you compute the speeds of the rod ends, take them with respect to the rod center speed. These will be like v and -v. Then, look a notebook for the angular momentum of the rod given its I and v. You will solve it for W.