1
$\begingroup$

I'm a programmer and I'm trying to create a physically accurate game. I'm not an expert on physics so if I'm missing the correct terms please excuse me.

What I'm trying to do is to simulate an arrow hitting and deflecting off a wall

So imagine an arrow and an impulse is applied to it on any point. I need to find out how much of that impulse will cause the arrow to rotate and how much of it to translate seperately.

I know the rotational kinetic energy is (1/2) I w^2 and I for a rod is (1/12) M L^2 around the center but I'm not sure about what to do if the axis is around a point between the center and the end which is the case in my game.

Also I dont know what do the rotational and translational kinetic energies add up to. All I have is a velocity change that I can equate.

Thanks in advance

$\endgroup$
  • $\begingroup$ This is a very much more difficult calculation than you imagine. Many factors have an effect on what happens : the hardness of the wall, the flexibility of the arrow, the angle at which the arrow strikes, etc. A better solution would be to make a video recording then try to imitate the effect in your program. $\endgroup$ – sammy gerbil Oct 15 '16 at 20:31
  • $\begingroup$ I understand the deflection is very complicated and I'm implementing a less accurate version. Instead of calculating the resistance on the tip, I'm just rotating the arrow and then apply the right force/torque combination. What I need is just the last part. $\endgroup$ – Sangratura Oct 15 '16 at 20:35
  • $\begingroup$ I dont want help with the deflection, please read well. I want to know how a force on a leverage affects the object in terms of rotational and translational components. $\endgroup$ – Sangratura Oct 15 '16 at 20:36
0
$\begingroup$

If the impulse force $P$ (force x time) is applied at the centre of mass (CM), this will cause only a change in the linear momentum of the CM without affecting any rotation. You can solve this part of the problem using a vector diagram, with the 3 sides of a triangle representing the initial and final momenta and the impulse (=change in momentum).

If the impulse is applied at perpendicular distance $x$ from the CM it is equivalent to applying the same impulse $P$ at the CM, along with a torque impulse of $xP$ which causes a change in angular momentum about the CM. If the impulse is not in the same plane as the initial rotation, a vector diagram is required for this calculation also.

$\endgroup$
  • $\begingroup$ In the second case, shouldn't P + xP be equal to the P in the first case ? So that the total energy is distributed between rotation and linear movement? $\endgroup$ – Sangratura Oct 15 '16 at 21:06
  • $\begingroup$ Impulse is not the same as energy. See Off-center impulse equations and related questions linked by ja72. $\endgroup$ – sammy gerbil Oct 15 '16 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.