The angular momentum of the two masses is computed independent of the skater - you were given the total angular momentum of the skater (including arms and hands which are normally considered part of the person) and ONLY have to compute the moment of inertia / angular momentum of the masses. A point mass at the end of a string has $$I=mr^2$$ as you know. The arms of the skater were already accounted for, and the mass of the weights is not distributed along the arms, it is all at the end.
Angular momentum is $I\omega$. You should now be able to compute it from $I_{total}=I_{skater}+I_{masses}$, and $\omega$ is given. It will, of course, not change when the skater pulls in her arms - conservation of angular momentum, and there is no external torque on the skater-plus-masses system.
The moment of inertia of the masses does change when the skater pulls in her arms - you can compute it for the masses, but not for the arms (which are also coming closer). That is a problem with the question - you must assume a massless arm if you want to compute the moment of inertia when the arms are pulled in.
And you need the moment of inertia for the last part, since you can write the angular kinetic energy as
$$KE = \frac12 I \omega^2$$
So it is not enough to know $L$, you actually need to be able to compute the new angular velocity. And for that you must make a simplifying assumption (massless arms).
On that assumption, you can compute the increased kinetic energy from the above (because you know the new angular velocity from the new moment of inertia).
Before:
$$\omega = 10 rad/s\\
I_{skater}=50 kg m^2\\
I_{weights} = 10 kg m^2\\
KE = \frac12 I_{total} \omega^2$$
and you should be able to figure the rest from here...
You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass.
So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C is (rather messy) $$ \boxed{ \begin{aligned}
\sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\
\sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right)
\end{aligned} } $$
The sum of the forces part equates to mass times acceleration of the center of mass. If the COM is not used, these extra terms appear to account for the change.
To help you the laws of motion can be summarized as follows:
- Linear momentum is defined as mass times the velocity of the center of mass $$\vec{p} = m \vec{v}_{C}$$
- Angular momentum at the center of mass is defined as rotational inertia at the center of mass times angular velocity
$$\vec{L}_C = I_C \vec{\omega}$$
- The net forces acting on a body equal the time derivative of linear momentum
$$ \sum \vec{F} = \frac{{\rm d}}{{\rm d}t} \vec{p} = m \vec{a}_C$$
- The net torques acting on a rigid body about the center of mass equal the time derivative of angular momentum at the center of mass $$\sum \vec{\tau}_C = \frac{{\rm d}}{{\rm d}t} \vec{L}_C = I_C \vec{\alpha} + \vec{\omega} \times I_C \vec{\omega}$$
- To transfer these quantities to a different location A with $\vec{r}=\vec{r}_C -\vec{r}_A$ use the following rules
$$\begin{aligned}
\vec{v}_A & = \vec{v}_C + \vec{r} \times \vec{\omega} \\
\vec{a}_A & = \vec{a}_C + \vec{r} \times \vec{\alpha} + \vec{\omega} \times \left( \vec{r} \times \vec{\omega} \right) \\
\vec{L}_A & = \vec{L}_C + \vec{r} \times \vec{p} \\
\sum \vec{\tau}_A & = \sum \vec{\tau}_C + \vec{r} \times \sum \vec{F} \\
\end{aligned}$$
whereas forces, linear momenta, angular velocity and angular acceleration are shared with the entire rigid body and thus do not change from point to point.
To the above you can add the vector form of the parallel axis theorem with
$$ I_A = I_C - m [\vec{r}\times] [\vec{r}\times]$$
where $[\vec{r}\times]$ is the 3×3 skew symmetric matrix for the cross product operator $\begin{pmatrix}x\\y\\z\end{pmatrix}\times = \begin{vmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{vmatrix}$
This comes out of the momentum transformation from C to A, but it is not the complete picture. To see what happens you have to look at the following 6×6 spatial inertia matrix:
$$\begin{aligned}\vec{p} & =m\vec{v}_{C}=m\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\vec{L}_{A} & =I_{C}\vec{\omega}+m\,\vec{r}\times\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\begin{pmatrix}\vec{p}\\
\vec{L}_{A}
\end{pmatrix} & =\begin{vmatrix}m & -m\,\vec{r}\times\\
m\,\vec{r}\times & I_{C}-m\,\vec{r}\times\,\vec{r}\times
\end{vmatrix}\begin{pmatrix}\vec{v}_{A}\\
\vec{\omega}
\end{pmatrix}
\end{aligned}$$
Best 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.