I'm working on a sandbox game that allows the player to build a spacecraft and attach thrusters in arbitrary locations on the hull of the spacecraft. The game is set completely in 2D. (I'm a programmer, not a physicist, so I apologize if my terminology is all wrong).

I have need to actually calculate my own physics. Say the player makes a simple spacecraft as follows:


where "A" are the thrusters pointing down, "O" is the cockpit and just so happens to be the center of mass as well, and "=======" is just some metal bar joining everything together. We can completely ignore the existence of any gravity.

Both thrusters are identical and, when turned on, produce an upward force of F at the point of the thrusters. The torque about the CoM for the left one completely cancels out the torque of the right one - hence the net torque is zero. The resultant force at the CoM (I believe) is 2*F up. Therefore, the ship accelerates up - there is no sideways acceleration and no torque to cause the craft to rotate.

What if only the right-side thruster is fired while the left one is left off? There would be a net anti-clockwise torque on the craft but I believe there is also some resultant force on the CoM as well? Torque I know how to calculate, but what about this resultant force vector?

I'm trying to relate this to the experience of flicking the tip of a pen resting on a table top. There would be torque because I observe the pen spinning, but there would also be some resultant force because the pen is also sliding down the table.



Newton's second law $\vec F_\text{net}=d\vec p/dt$, or the-more-convenient-for-simulations $d\vec v=(\vec F_\text{net}/m)\,dt$, still applies. It doesn't matter where the force is applied on the body; the instantaneous acceleration of the center of mass will be the same as if the forces were applied at the center of mass.

As for your pen example, try doing the experiment. Flick a pen very hard perpendicular to its axis. You'll find the center of mass (initially) goes in a straight line parallel to the direction of your flicking force. You might also notice that the pen doesn't seem to go as far compared to flicking it at the center of mass; this is primarily due to the smaller time of contact when flicking it on the edge; the pen rotates away from your finger.

  • $\begingroup$ Thanks for the quick response! I guess in my simulations, I noticed the craft picked up some sideways motion and thought I was missing something in my calculations of the net force at CoM. Turns out it was due to F being applied over time - allowing time for the craft to rotate hence also rotating F to give a small horizontal component. $\endgroup$ – Tangalisk Sep 1 '14 at 9:33
  • $\begingroup$ Yes, that sounds correct. $\endgroup$ – BMS Sep 1 '14 at 17:31

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