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The Problem

I'm proposing a 2D problem in the x-y plane. Say there is a free-floating disk of radius R and mass Md (no velocity), at which a projectile is shot. The projectile (velocity Vp, mass Mp) has a velocity which is parallel to the x axis. If taking the disk's center for the origin, the projectile contacts the disk at a distance Y above the x axis, and it sticks (no energy lost).

The Question

I'd like to know what the ending velocity and angular velocity of the disk/projectile system is. Using conservation of energy, I end up with two variables, and I can't figure out what else to use to create a system (I'm very rusty on conservation of momentum, but I figure that's where the solution lies?).

Disclaimer

I'm not a student, if you're worried about doing my homework. Back in those days, I wouldn't have blinked at this, but now... software engineering has done me wrong XD

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  • $\begingroup$ Could you post a sketch and the equations you came up with? How did you calculate the second moment of area and the angular momentum? $\endgroup$ – rul30 Dec 29 '17 at 5:46
  • $\begingroup$ I can post a sketch tomorrow. The equation from the conservation of energy yeilds: $\endgroup$ – Corbfon Dec 29 '17 at 6:05
  • $\begingroup$ Ill give you a hint, the mutual center of gravity keeps its trajectory $\endgroup$ – joojaa Dec 29 '17 at 6:18
  • $\begingroup$ (1/2)*MpMv^2 = (1/2)*MdVd^2 + (1/2)*Id*Wd^2 where Wd is a poor substitute for omega $\endgroup$ – Corbfon Dec 29 '17 at 6:48
  • $\begingroup$ @joojaa any more help? "Trajectory" means direction, independent of speed, so are you saying that the shared center of mass's velocity is parallel to the x axis? The only velocity beforehand is parallel to the x axis. I think that is unlikely because any force transmitted to a disk is transmitted along the radius of the disk, which would not be parallel to the x axis. It would be along the at an angle -sin^-1(Y/R) from the center of the disk $\endgroup$ – Corbfon Dec 29 '17 at 8:17
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When it comes to solving any collision problem, it is always important to apply conservation of linear momentum and conservation of angular momentum. This gives us the first two equations.

Conservation of linear momentum

For two rigid bodies 1 and 2, the conservation of linear momentum states the following:

$$m_1 \mathbf{u}_1+ m_2 \mathbf{u}_2 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 $$

$m_i$ is the mass of body $i$, $\mathbf{u}_i$ is the velocity vector of body $i$ just before the collision, and $\mathbf{v}_i$ just after the collision.

Let body 1 be the projectile, and body 2 be the disc. Since the disc is initially stationary, we can simplify this to:

$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$

This is the first equation we’ll need to solve the collision.

Conservation of angular momentum

For angular momentum about an arbitrary point $P$ to be conserved, the following mess of an equation should be (briefly) observed:

$$\left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{u}_1\right)+I_1\omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{u}_2\right)+ I_2\omega_2\mathbf{k} = \left(\mathbf{r}_1-\mathbf{r}_P\right)\times\left(m_1 \mathbf{v}_1\right)+I_1\Omega_1\mathbf{k}+ \left(\mathbf{r}_2-\mathbf{r}_P\right)\times\left(m_2 \mathbf{v}_2\right)+ I_2\Omega_2\mathbf{k} $$

$\mathbf{r}_i$ is the position vector of the centre of mass of body $i$ , $\mathbf{r}_P$ is the position vector of point $P$, $I_i$ is the moment of inertia about the axis of rotation for body $i$, $\omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just before the collision, and $\Omega_i\mathbf{k}$ is the angular velocity vector of body $i$ just after the collision.

By setting $\mathbf{r}_P=\mathbf{r}_2$, noting that body 2 is initially stationary, and noting that body 1 is a particle and thus has no moment of inertia, the equation simplifies to:

$$\mathbf{r}\times\left(m_1 \mathbf{u}_1\right)= \mathbf{r}\times\left(m_1 \mathbf{v}_1\right)+ I_2\Omega_2\mathbf{k}$$

where $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2=-\sqrt{R^2-Y^2}\mathbf{i} + Y\mathbf{j}$.

Note that all the terms are parallel to the (out-of-plane) $z$-axis, i.e. parallel to $\mathbf{k}$. Therefore we can use the dot product to multiply all of the terms by $\mathbf{k}$ without loss of information from the equation:

$$m_1\left(\mathbf{r}\times\mathbf{u}_1\right)\cdot\mathbf{k}= m_1\left(\mathbf{r}\times\mathbf{v}_1\right)\cdot\mathbf{k} + I_2 \Omega_2$$

By noting that: $$\left(\mathbf{a}\times\mathbf{b}\right)\cdot\mathbf{c}= \left(\mathbf{b}\times\mathbf{c}\right)\cdot\mathbf{a} = \left(\mathbf{c}\times\mathbf{a}\right)\cdot\mathbf{b} $$

we can rewrite the equation as:

$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$

where $\mathbf{r^*}=\mathbf{k}\times\mathbf{r}=-Y\mathbf{i}-\sqrt{R^2-Y^2}\mathbf{j}$

In this form, we now have the second of the equations we need to solve the collision problem.

The third and final equation

We need one more equation to solve this problem. This final equation will arise from the fact that the two bodies stick to one another.

((In your question, you specify that there is no energy loss as a result of the collision. However, it is not actually possible for the total kinetic energy of the system to be conserved if both bodies stick* after the collision. If the collision were perfectly elastic (no energy loss), both bodies must rebound from each other after colliding. For this reason, it is important that conservation of kinetic energy is not applied.

*If the bond that sticks one body to the other is not perfectly rigid (i.e. it is possible to pull apart the body apart as if there was a spring joining them), then it would be theoretically possible for mechanical energy to be conserved. However, this would result in complex oscillating behaviour between the two bodies, much like an undamped mass-spring system.))

If the two bodies are to stick post-collision, it important that velocities of the bodies at the point of contact are the same. This results in the following kinematic condition:

$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{k}\times\mathbf{r}$$

This can be simplified to give the final equation:

$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$

In summary

Three equations have been derived:

$$m_1 \mathbf{u}_1 = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$$

$$m_1\mathbf{r^*}\cdot\mathbf{u}_1= m_1\mathbf{r^*}\cdot\mathbf{v}_1 + I_2 \Omega_2$$

$$\mathbf{v}_1 = \mathbf{v}_2 + \Omega_2\mathbf{r^*}$$

There are three unknowns: $$\mathbf{v}_1 \quad \mathbf{v}_2 \quad \Omega_2$$

These are the velocity of the projectile, the velocity of the centre of the disc, and the angular speed of the disc, respectively, immediately after the collision.

By using vector algebra, it is possible to solve these equation for the three unknowns. Then it may be necessary to perform some substitutions, including:

$$m_1=M_P$$ $$m_2=M_D$$ $$I_2 = \frac{1}{2}M_D R^2$$ $$\mathbf{u}_1=V_P \mathbf{i}$$

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  • $\begingroup$ Wonderful, thank you for your detailed explanation. Could you give a reference for your conservation of angular momentum equation above? I believe that was the piece I have been missing. $\endgroup$ – Corbfon Jan 3 '18 at 20:03
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Suggested approach and a bit of reasoning.

There are multiple ways to view the final situation. You can look at it at the instant of contact and realize that the final velocity of the projectile is Vd + Wd * y. This works, but isn't fun. A better way is to think of the final average velocity of the projectile is just Vd, but it is also stuck to a spinning object off center.

Use the initial masses, inertial moments, and offset y to calculate the final inertial moment (I), neutral axis, and the offset of the projectile from the neutral axis. Call the offset r.

Now there are two energy terms, translational and rotational. Total mass * V^2 + I * W^2 = the original projectile energy.

But we know the ratio of V to W, so we can substitute and get rid of either one.

The collision resulted in some impulse (a force integrated over time). The same impulse caused the rotation rate and the translation rate.

Impulse = M * V
Impulse * r = I * W

so W = M * V * r / I

substitute and solve for V, then W.

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  • $\begingroup$ First, thank you for being the first to step out with an answer! I've got a question about your solution. You said that that L = I * W (where L is angular momentum, and L = M * V * r), which is the conservation of angular momentum. In this case, does that not mean that all of the bullets linear momentum at radius r was translated into the shared mass's new angular momentum? Would there not also be a linear momentum of the shared mass, and if so, does that mean that all of M * V * r did not go toward creating angular momentum, but some created that linear momentum, and thus L < M * V * r? $\endgroup$ – Corbfon Jan 1 '18 at 20:36
  • $\begingroup$ @Corbfon Well, there is some stationary point in space where that is true. The final system has an instantaneous center of rotation. But I went with a direct calculation of CG translation velocity and rotation rate. Parallel axis theorem would show the energy is the same both ways. $\endgroup$ – Phil Sweet Jan 1 '18 at 21:45
  • $\begingroup$ @Corbfon There is only one impulse which governs both the change in CG velocity and the change in rotation rate. But the size of that impulse depends on M, I, and the energy of the projectile. Compare to a thinner disc of larger diameter with 1/2 M but same I. The impulse would be less, but more than 1/2. The disc CG velocity would be greater, but the rotation rate would be less. $\endgroup$ – Phil Sweet Jan 1 '18 at 21:45

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