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

• lorenz0
In summary: So, the angular momentum after collision is ##L_{z_i}\pm\frac{m_iR(R\omega\pm v)}{2|\tau|}\omega_i##.
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

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}##.

#### Attachments

• disk_masses.png
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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
haruspex said:
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?

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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)##.

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

The final angular displacement of a disk hit by two masses depends on several factors, including the initial angular velocity of the disk, the masses of the objects hitting the disk, and the angle at which they hit. It can also be affected by external forces such as friction or air resistance.

## How is the final angular displacement calculated?

To calculate the final angular displacement, you will need to use the principles of conservation of angular momentum. This means that the initial angular momentum of the disk must equal the final angular momentum after the objects hit. You can use the formula Li = Lf, where Li is the initial angular momentum and Lf is the final angular momentum, to solve for the final angular displacement.

## What is the relationship between the masses of the objects and the final angular displacement?

The masses of the objects hitting the disk can greatly affect the final angular displacement. Generally, the more massive the objects, the greater the final angular displacement will be. This is because larger masses have more momentum and can transfer more energy to the disk upon impact.

## Can the final angular displacement be negative?

Yes, the final angular displacement can be negative. This means that the disk will rotate in the opposite direction of its initial rotation. This can happen if the objects hitting the disk have a greater angular momentum in the opposite direction, causing the disk to slow down and eventually rotate in the opposite direction.

## Are there any other factors that can affect the final angular displacement?

Yes, there are other factors that can affect the final angular displacement, such as the angle at which the objects hit the disk. If the objects hit at an angle, it can cause the disk to rotate in a different direction than if they hit directly perpendicular to the disk. Additionally, external forces such as friction or air resistance can also affect the final angular displacement.

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