# Non-rigid body rotational dynamics

I'm attempting to solve the following problem:

Two friends hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about an axis through the center of the rope and perpendicular to the ice. The mass of one friend is $a$ kg and the other has a mass of $b$ kg. The rope of negligible mass is $\ell$ m long and they move at a speed of $v$ m/s.

I'm primarily trying to find the angular momentum of the system of the two skaters. I'm having a hard time thinking about this. All of the tools I have to deal with rotational dynamics are limited to rigid body motion or fixed axis rotation. I've been to the library to try and find other examples and all I can find is something with astronauts where their two masses are the same, which makes the analysis trivial (axis of rotation = center of mass). Here the center of mass is closer to to one of the skaters, but the problem states that the axis of rotation is the center of the rope. Since they both maintain constant tangential velocity, does this mean the axis of rotation has to move?

Any help would be appreciated!

## 1 Answer

For a system of point particles, the definition $$\vec{L}_i=\vec{r}_i\times\vec{p}_i$$ is always true; it's just a definition. I see no reason why that won't work here. The only choice you have to make is where to measure the position vectors $\vec{r}_i$ from. A particularly convenient position from which to measure $\vec{r}_i$ is the rotation axis.

One possibly tricky thing here is that the center of mass accelerates. This means there is always a non-zero net force on the system of the two skaters and rope. I suppose it's exerted by the ice on the skaters.

Also, this situation you're describing sounds to me like it does satisfy the requirements of a rigid body, at least for the specific motion you are describing. If the skaters begin to move toward each other so that the distance between them changes, then it wouldn't be a rigid body.

• I think I'm just in a position where I'm throwing my head against a poorly stated problem. I came to the same conclusion (that some net force has to be acting) but there's no statement about them. The situation is explicitly described as frictionless. A rigid body assumption seems like a queer thing to make for a rope. But the alternative seems like it would be differential equation chaos. Commented Dec 10, 2013 at 6:24
• Ah yes, I missed your 'frictionless' statement in the problem setup. In that case, I'm of the opinion that this problem is misleading. I suppose the author of the problem might be thinking some sort of normal force from the boundary between the ice and ice skate is doing the magic. Realistically, ice skates dig in to the ice surface.
– BMS
Commented Dec 10, 2013 at 6:31
• Yes, all ropes won't act like rigid bodies. But in this case, the system satisfies the definition: relative distances between all components is constant in time. So you can use all the nice equations like $\vec{L}=I\vec{\omega}$.
– BMS
Commented Dec 10, 2013 at 6:37
• Isn't there a problem with that? I mean, the only way that we satisfy the rigid body proposition for the rope is by having some net external force acting on the skaters to keep them rotating about the central axis rather than the center of mass. In which case there would be torques acting on the center of mass, which means a non-constant angular momentum from the perspective of the axis of rotation. Or is it that the torques would act on the axis of rotation and thus not affect the angular momentum? Commented Dec 10, 2013 at 7:00
• If you take the axis of rotation to be the reference point, each skater has a constant $L=rp\sin\theta$, so the net torque on each skater is zero. This can be true even with the horizontal ice force and/or tension force since those forces are parallel or antiparallel to $\vec{r}$. One can make a similar argument for the system of skaters & rope.
– BMS
Commented Dec 10, 2013 at 7:59