# How can I find the angular and linear velocity of a 2D body that breaks into two bodies?

Afternoon. This is my first question, so do let me know if I'm doing anything wrong. Looking for help on building a 2D physics game engine with bodies that split in half:

I have a two dimensional body $B_0$, with angular velocity $\omega_0$, linear velocity vector $\vec{v_0}$, mass $m_0$ and moment of inertia $I_0$. If the body splits instantaneously into two bodies $B_1$ and $B_2$, what is the new angular and linear velocity vector of each body?

Each new body has known distance vectors of centre of mass from $B_0$ centre of mass $\vec{r_1} \vec{r_2}$, masses $m_1 m_2$ (which sum to $m_0$), and moment of inertia $I_1 I_2$.

I've had a hard time finding anything online that's related to this, the hard part being the angular momentum that is converted to linear as the pieces fly apart. In essence though could it be seen as an inelastic collision in reverse?

My solution (which I've deduced myself and am hence not completely convinced by):

Linear velocity of a new body is the same, plus $\vec{r} × \omega$ to allow for pieces "spinning off". This gained linear momentum is deducted from the starting angular momentum (using $\vec{L} = \vec{r} \times m\vec{v}$), and the remainder split amongst the two new bodies by the ratios of their moment of intertias. This assumes that $\omega$ is the same sign for both.

Any help appreciated. A formula for the new angular velocities would be ideal.

I'm writing this assuming you're talking about something like a disk that cracks and the pieces move without interacting with each other.

You need to conserve energy:

$m_0v_0^2 = m_1v_1^2 + m_2v_2^2$

$I_0w_0^2 = I_1w_1^2 + I_2w_2^2$

where $v_0$, $v_1$, and $v_2$ are the magnitudes of the velocities $v = \sqrt{v_x^2 + v_y^2}$

You also need to conserve linear and angular momentum:

$m_0\mathbf{v_0} = m_1\mathbf{v_1} + m_2\mathbf{v_2}$

$I_0\mathbf{w_0} = I_1\mathbf{w_1} + I_2\mathbf{w_2}$

or

$m_0v_{0_x} = m_1v_{1_x} + m_1v_{1_x}$

$m_0v_{0_y} = m_1v_{1_y} + m_1v_{1_y}$

$I_0w_{0_z} = I_1w_{1_z} + I_2w_{2_z}$

Now, the hard is figuring out the mechanics of how the system splits. If you start spinning a rock on a string with length $r$, and the string breaks, the rock will be flung in the direction of its tangential velocity $v = rw$ at the instant the string broke. To this, you add the original velicty of the system. You can apply this to your body as well by using the system of centre of masses you provided us with:

$\mathbf{v_1} = \mathbf{r_1}\times\mathbf{w_0} + \mathbf{v_0}$

$\mathbf{v_2} = \mathbf{r_2}\times\mathbf{w_0} + \mathbf{v_0}$

Since the system is in 2D the z dimension just acts to get us a perpendicular product and to determine clockwise and counter-clockwise rotation with negative and positive sings respectively. $\mathbf{r_1}$ and $\mathbf{r_2}$ should always be co-linear and point in opposite directions. You can break these up to get:

$v_{1_x} = r_{1_x}w_0 + v_{0_x}$

$v_{1_y} = -r_{1_y}w_0 + v_{0_y}$

$v_{2_x} = r_{2_x}w_0 + v_{0_x}$

$v_{2_y} = -r_{2_y}w_0 + v_{0_y}$

It's important to note that the x-y components in my development are general and do not account for $\mathbf{r_1}$ and $\mathbf{r_2}$ being co-linear and opposite. That would be accounted for with YOUR initial conditions.

• yes rotating disk that cracks is the exact situation. Also $r_1 r_2$ will indeed be co-linear and opposite. The formulas for velocity look good, but what about $\omega_1$ and $\omega_2$ ? Commented Apr 3, 2015 at 21:54
• I would plug in momentum into rotational energy conservation to solve for one of the angular velocities at a time. Commented Apr 3, 2015 at 22:01
• But surely if the linear velocity changes by $r×\omega$ then some of the rotational momentum has become linear momentum? Commented Apr 3, 2015 at 23:28
• Linear energy/momentum and rotational momentum are usually conserved independently. If it bothers you too much, you could plug it in to a total energy conservation $E_{tot} = E_{rot} + E_{lin}$ Commented Apr 7, 2015 at 23:21

Let's solve this by considering the time-reversed process: two spinning objects collide inelastically. We know that linear and angular momentum are conserved:

$$m_1\vec{v}_1 + m_2\vec{v}_2 = m_0\vec{v}_0$$ $$m_1\vec{r}_1\times\vec{v}_1 + m_2\vec{r}_2\times\vec{v}_2 = m_0\vec{r}_0\times\vec{v}_0$$

The latter can also be written (assuming you have the moments of inertia) as:

$$I_1\vec{\omega}_1 + I_2\vec{\omega}_2 = I_0\vec{\omega}_0$$

The relationship between $\vec{\omega}$ and $\vec{v}$ is $\vec{v}= \vec{r}\times\vec{\omega}$, where $\vec{r}$ is from the axis (or point) of rotation. So you have a system of 6 equations with 6 (coupled) unknowns (each vector has three components). I personally would solve this by writing out the equations in terms of the components of the unknown vectors, and solving using a numerical algorithm (or Mathematica).

• Should that second equation be r1 ... r2 ... r0 instead of r1 all the way through? Commented Apr 3, 2015 at 18:35
• You're right -- corrected. Oh the faults of using copy-paste. Commented Apr 15, 2015 at 0:21

Since the unfractured disk is assumed (it is so assumed, right?) to be of uniform density, it has a center of mass at the geometric center of the disk. Then, since the center of mass of two distinct masses lies on a line connecting the two, the CMs of the two fragments must lie on a diameter of the disk. Not only that, if the radial distance of the CM of $B_1$ is $r_1$, and the radial distance of $B_2$ is $r_2$, then since $m_1$$r_1 = m_2$$r_2$, $r_1$/$r_2$ = $m_2$/$m_1$. However, without knowing the geometries of at least one of the fragments, there is no way to determine the values of $r_1$ and $r_2$. Assuming you do know them, the rest becomes easy. The two fragments can be considered point masses connected by an infinitesimal thread, rotating at $\omega_0$ around the common center of mass. Since the orientations of the two fragments is fixed, each fragment rotates around its own center of mass at $\omega_0$. Each fragment has a tangential velocity proportional to its radial distance from the common CM, so $v_1$ = $\omega_0$$r_0/r_1, and v_2 = \omega_0$$r_0$/$r_2$. Direction, of course, is determined by the angle of the line connecting the two CMs at the moment of release. Furthermore, since there is no interaction between the two fragments during the release, each will maintain a rotation rate $\omega_0$ around its center of mass.