What is the final angular displacement of a disk hit by two masses?

[lorenz0]

Homework Statement

A disk of mass $m=10kg$ e raggio $R=0.5 m$ is initially at rest on a horizontal plane without friction. Two point masses $m_1=5kg$ and $m_2=1kg$ move with speed $v_1=2m/s$ and $v_2=4m/s$ the first from left to right and the second from right to left. At a certain istant $m_1$ and $m_2$ hit the disk in a completely inelastic collision.
Find: 1) Velocity of the center of mass after the collision; 2) Angular velocity of the disk after the collision; 3) If the disk had a rod passing through its center of mass and perpendicular to the the disk find its angular velocity after the collision and the impulse given by the rod in the collision; 4) If now the rod where to apply a braking torque $\tau=-0.2 N\cdot m$ to the disk, after how many turns would the disk (with the two masses) stop?

Relevant Equations

$\vec{L}=\vec{r}\times\vec{p}=I\vec{\omega}, \omega_f^2=\omega_i^2+2\alpha\Delta\theta$

1) By conservation of linear momentum: $m_1 v_1-m_2v_2=(m+m_1+m_2)v_{cm}\Rightarrow v_{cm}=\frac{m_1}{m+m_1+m_2}v_1-\frac{m_2}{m+m_1+m_2}v_2=\frac{3}{8}\frac{m}{s}$;

2) By conservation of angular momentum: $-Rm_1v_1-Rm_2v_2=I_{total}\omega=(I_{disk}+m_1R^2+m_2R^2)\omega$ so
$-(m_1v_1+m_2v_2)R=(\frac{1}{2}mR^2+m_1R^2+m_2R^2)\omega\Rightarrow\omega=-\frac{m_1v_1+m_2v_2}{\frac{m}{2}+m_1+m_2}=-\frac{14}{11}\frac{rad}{s}$;

3) To find the impulse given by the rod we can use the impulse theorem so $I=(m+m_1+m_2)v_{cm}=16\cdot\frac{3}{8}\frac{kg\cdot m}{s}=6 \frac{kg\cdot m}{s}$. Now, to find the angular velocity in this case I would do the same analysis as in (2) and say the angular velocity is thus the same but I am not sure so I would appreciate a comment about how to think about this case.

4) $\omega_f^2=\omega_i^2+2\alpha\Delta\theta$ which in our case gives $0=\omega_i^2-2\frac{|\tau|}{I_{tot}}\Delta\theta\Rightarrow\Delta\theta=\frac{I_{tot}\omega_i^2}{2|\tau|}\omega_i^2=\frac{(\frac{1}{2})mR^2+m_1R^2+m_2R^2)\omega_i^2}{2|\tau|}\omega_i^2=\frac{(5+5+1)(\frac{1}{2})^2}{2\frac{1}{5}}(\frac{14}{11})^2 rad=\frac{245}{22} rad$ so the number of turns is $n_{turns}=\frac{\frac{245}{22}}{2\pi}=\frac{245}{44\pi}$.

 

[haruspex]

In part 2, you found the initial angular momentum of the disc+masses about a fixed point on the line through the disc's centre and parallel to the initial motions. E.g. about the point where the disc's centre started.
You then equated this to an angular momentum of the combined system after the collisions, about that point. But does that give you the rotation rate of the disc about its centre? What are the contributions to that angular momentum?

 

[lorenz0]

In part 2, you found the initial angular momentum of the disc+masses about a fixed point on the line through the disc's centre and parallel to the initial motions. E.g. about the point where the disc's centre started.
You then equated this to an angular momentum of the combined system after the collisions, about that point. But does that give you the rotation rate of the disc about its centre? What are the contributions to that angular momentum?

Thanks for your interest in my question; I have been thinking about the problem and I think I have found the correct way to do part (2):

the center of mass of the system does not coincide with the center of mass of the disk and it is located at height $0.625$ from the lowest point of the disk i.e. ath height $h=0.125 m$ above the center of the disk. So, we have that $L_{z_i}=-(R-h)m_1v_1-(R+h)m_2v_2$ and $L_{z_f}=(I_{disk_{cm}}+mh^2+m_1(R-h)^2+m_2(R+h)^2)\omega - (R+h)(m+m_1+m_2)v_{cm}$ and from the conservation of angular momentum $L_{z_i}=L_{z_f}$ we can find $\omega=\frac{(R+h)(m+m_1+m_2)v_{cm}-(R-h)m_1v_1-(R+h)m_2v_2}{m(\frac{1}{2}R^2+h^2)+m_1(R-h)^2+m_2(R+h)^2}$
and the answer I got in part (2) above is the answer to question 3, where the $disk+m_1+m_2$ system rotates around an axis through its center. Is this correct?

 

[haruspex]

Your equation looks right, but in my experience it is generally unhelpful to find the CoM of such a system. Just work with an axis at the original mass centre position of the disc. The moments of the two point masses after collision are $m_iR(R\omega\pm v)$.