I have a problem that is seriously challenging both my intuition in what might happen and my ability of expressing the maths.


Imagine you have a sphere of radius $R$ with CoM speed $\mathbf{v_0}$ and rotating about its axis at $\mathbf{\omega_0}$. In space. It collides with another identical sphere, at rest at not rotating. What happens to the final linear velocities $\mathbf{v_1}$, $\mathbf{v_2}$, and angular velocities $\mathbf{\omega_1}$ and $\mathbf{\omega_2}$? Both sphere have rough surfaces so there is some exchange of angular velocity.


I've tried normal angular momentum, linear momentum and total (linear + rotational) energy conservation, but I get 4 variables ($\mathbf{v_1}, \mathbf{v_2}, \mathbf{\omega_1}, \mathbf{\omega_2}$) with only 3 equations.

Can I decouple linear and rotational motion since they should not be talking to one another?

What should I take as a point where to calculate angular momenta from?

  • 1
    $\begingroup$ Given that "both spheres have rough surface", you can toss conservation of mechanical energy out the window. Some of the initial mechanical energy will be converted to thermal energy. Also, your equations for conservation of linear and angular momentum are three equations each. Finally, don't forget about the angular momentum due to translation. $\endgroup$ – David Hammen Mar 6 '17 at 0:46

In general the result of a collision is indeterminate; you have to specify more about the dynamics of the collision to get a unique result. This is true even for a 2D collision without rotational motion, where already there aren't enough conservation laws to get a unique result. And it makes sense, because otherwise every scattering cross section in the world would be exactly the same, and particle colliders wouldn't be very interesting!

For example, you can 'decouple linear and rotational motion' by just saying the surfaces of the spheres are frictionless. Making the further restriction that the collision is elastic and that the spheres are incompressible (i.e. there is a 'hard core' repulsion) would give you a unique answer.

Alternatively, you could say that the spheres are 'very rough', so that the extra equation is that the spheres aren't slipping on each other after the collision. You can then add in the incompressibility condition to get a unique answer.

  • $\begingroup$ Mathematically what would the incompressbility condition look like? $\endgroup$ – SuperCiocia Mar 6 '17 at 0:29

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