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What I am looking for is how to determine the bounce vector of a disc hitting the edge of an object. The disc can be at any angle to the edge. I have a program that does the calculation without taking in to account the torque caused by the angle of the disc.

I want the direction that the disc will take after the bounce. If the plane of the disc is perpendicular to the direction of the edge when it hits, the direction of the bounce can be calculated by adding the unit vector from the contact point to the center of the disc to one half the velocity unit vector of the disc. If the disc is not perpendicular to the edge some energy will go to spinning the disc. That will change the direction of the bounce.

Any suggestions?

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  • $\begingroup$ Thank you for responding. I want the direction that the disc will take after the bounce. If the plane of the disc is perpendicular to the direction of the edge when it hits, the direction of the bounce can be calculated by adding the vector from the contact point to the center of the disc to one half the velocity vector of the disc. If the disc is not perpendicular to the edge some energy will go to spinning the disc. That will change the direction of the bounce. $\endgroup$
    – Bob
    Commented Nov 6, 2016 at 16:39
  • $\begingroup$ Ok thanks, I put your comment in the post itself, as comments get cleaned up/ deleted. $\endgroup$
    – user108787
    Commented Nov 6, 2016 at 16:55
  • $\begingroup$ Is the object with an edge something almost unmovable like a wall or a smaller free object like a rectangle of similar mass to the disk? Can we assume the disk or the second object are not initially rotating? Can we assume the second object is stationary or not? $\endgroup$
    – KDP
    Commented Sep 8 at 19:25
  • $\begingroup$ Can we assume an elastic collision and everything is moving on thee same plane so it is a 2D problem? Please let us know how how general you need the answer to be. Be warned, this is quite a complicated and tedious problem and the more you can limit the conditions, the better. $\endgroup$
    – KDP
    Commented Sep 8 at 19:29

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If the disk strikes at an angle, this will not change the direction that the centre of mass (CM) will take when it rebounds. But the disk will also rotate as it rebounds. To be more precise, the collision will change the linear momentum of the CM and also the angular momentum of the disk about the CM.

The code which you have must calculate the impulse = change in linear momentum. Call this vector $P$. The change in angular momentum (also a vector) will be $r \times P$ where $r$ is the position of the point of contact relative to the CM. Change in angular momentum $\Delta L$ is related to change in angular velocity $\Delta \omega$ by $\Delta L = J\Delta \omega$, where $J$ is the moment of inertia of the disk.

The effect of the change in angular momentum could be quite difficult to calculate. The easiest case is when velocity vector of the disk and the impulse vector from the contact are always in the same plane. This is then a 2D problem. However, the moment of inertia has different values depending on which axis is used - it is a tensor. If the disk strikes obliquely, the collision could change the axis about which the disk rotates, as well as changing the rate of rotation.

See :
What is the initial angular momentum of a rigid body given an offset impulsed force?

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  • $\begingroup$ Fortunately, I don’t care about the resulting angular momentum. All that I want to know is whether the z of the bounce is positive or negative, where x is along the edge of the desk, y is across the desk and z is up. Your maintaning the linear momentum agrees with the way my program works. However, when I drop the dog’s squeaky disk on the edge of the desk, there appears to be a significant difference in the bounce between the disc being perpendicular to the edge and the disc rotated about the z axis. $\endgroup$
    – Bob
    Commented Nov 8, 2016 at 4:26
  • $\begingroup$ I am thinking that the energy used in rotating the disc when it is at an angle reduces the velocity of the bounce. Rather than the bounce velocity equaling the incoming velocity, the velocity vector of the impact is smaller than it would be when perpendicular to the edge. That could make a difference in the resulting momentum leading to a difference in the direction of the bounce. What say? $\endgroup$
    – Bob
    Commented Nov 8, 2016 at 4:27
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    $\begingroup$ Your dog's squeaky disk is not an ideal rigid disk, it is soft and flexible - you cannot expect the same results. I can only point you to the equations, which apply to ideal rigid objects. You don't need to follow the equations, you can just vary the parameters and calculation until you get something which looks realistic to you. $\endgroup$
    – sammy gerbil
    Commented Nov 8, 2016 at 16:29
  • $\begingroup$ @sammygerbil is right - the flexibility of the toy disk will greatly affect your observations. $\endgroup$
    – Floris
    Commented Feb 21, 2017 at 0:25

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