The question comes from writing an oversimplified version of a physics in a 2D world as a computer simulation. I know that I can make something that looks not too horrible, but it'd be great to use this as a learning opportunity.

Let's say we have a 2D object (a polygon) that is falling with a specific acceleration. Here, to make it easier I define acceleration as +1 unit every 1 tick of the clock. It is trivial to figure out where each corner of the polygon is, $v(t)=v(t-1)+a$ and $p(t)=p(t-1)+v(t)$.

But once one of the points is anchored the object should start drawing a circle, with the anchored point being a centre. The question that I have is how to figure out the distance traveled in each tick of the clock.

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Intuitively I see that if the centre of the object (we assume that its mass is uniform, and thus it is the centre of mass) is on the same horizontal line as the anchor, then at that instance the centre of the mass should fall freely, and the closer the centre of the mass comes to the vertical line with the anchor point, the slower its fall is. This screams to me that $\cos(\text{centre angle})$ is going to be my friend. With that I could cobble together a way to "rotate fall" from the stationary position.

My bigger question is what do I do with the speed of free fall once one point becomes anchored. How do I translate the motion into a circular one? Am I approaching it wrongly?

  • $\begingroup$ Welcome to this community! Are you assuming that the polygon is rigid, ie its internal angles are unchangeable? Or that only its sides are rigid but their relative angles may change? $\endgroup$ – pglpm Jun 6 '20 at 9:37
  • $\begingroup$ @pglpm The polygon can be rotated as the whole, but the internal angles and lengths of segments cannot be changed. At least in this simplified approach for now. $\endgroup$ – v010dya Jun 6 '20 at 10:17
  • $\begingroup$ Then you can use the equations for a rigid body, and your body (its mass centre) is fully equivalent to a pendulum under gravity: en.wikipedia.org/wiki/Pendulum Or? $\endgroup$ – pglpm Jun 6 '20 at 10:19

In the instant when the object hits the anchor, all of its linear momentum should be considered as turned into angular momentum. You can then solve the rotational analog of Newton's second law using the angular momentum as your initial condition and considering the torque as caused by gravity.


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