Dynamics: conversvation of angular momentum (concept question)

[silentwf]

Homework Statement

This is more of a concept problem rather than solving it. So the questions are as below and the solution provided by the book. (sorry about the large sizes)

http://img190.imageshack.us/img190/6366/201019dynamics.png
http://img130.imageshack.us/img130/5720/201019dynamics2.png
Both of the questions involve one person jumping off an object (whether rotating or not). My question is, in both solutions, the book provides that the person jumping off will have an extra angular velocity other than his/her own, why?
Like in the first picture, the solution uses
$-m_{m}(v-\omegaa)a+m_{p}(\frac{c^{2}}{12}+b^{2})\omega$
Why an extra $\omega$ (velocity) for the man? Why not just v?

And in the second picture where $M_{1}(v+\omega_{3}a)a$, again, why an extra $\omega$ (or velocity)

(for question 2 (with the kids), its just the second part (part b), the extra $\omega_{2}$ i already solved)

Homework Equations

$\Sigma H_{0} = \Sigma H_{1}$

The Attempt at a Solution

(my equations are pretty much equivalent as those provided in the picture, except the part i don't get)

 

[jbsteele]

I guess the angular velocity is used to account for the angular acceleration of the object the person is jumping off, so that the angular acceleration of the person while jumping is taken into account. But why not just use v?

 

[kerimek]

The conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. In the case of these problems, the person jumping off the object is exerting a force on the system which causes an external torque. This results in a change in the angular momentum of the system.

In the first problem, the person jumping off the object has a velocity v and is rotating with an angular velocity ωa. When they jump off, they exert a force on the system which causes an external torque. This external torque causes a change in the angular momentum of the system, resulting in an additional angular velocity ω for the person. This can be seen in the equation -m_m(v-ωa)a, where the negative sign indicates a decrease in angular momentum.

Similarly, in the second problem, the person jumping off the object exerts a force on the system which causes an external torque. This external torque results in a change in the angular momentum of the system, causing an additional angular velocity ω for the person jumping off. This can be seen in the equation M_1(v+ω_3a)a, where the positive sign indicates an increase in angular momentum.

In both cases, the person jumping off the object is exerting a force on the system which causes a change in the angular momentum. This change in angular momentum results in an additional angular velocity for the person jumping off. Therefore, in order to conserve angular momentum, an extra angular velocity must be included in the equations.