>>4632912Here's an exact solution. Assume a rigid rod, length L, mass density rho. Call the velocity of the orange-side end of the rod v_o, and the blue side v_b. Call the orange portal's velocity v_p. The rigidity of the rod requires v_o-v_p=v_b.
There are two forces acting here, the "portal force," which changes the momentum of a bit of mass that passes through (using the point mass rules we already worked out, and taking care of the v*dm/dt part of F=dp/dt) and the rigidity of the rod. The rigidity must act as an ordinary force between the masses on opposite sides of the portal, meaning -m_o*v_o'=m_b*v_b' (prime is a time derivitive). Substituting the rigidity relation from above and replacing the masses with m_o=rho(L-x_b) and m_b=rho*x_b gives: -rho*(L-x_b)(v_b'+v_p')=rho*x_b*v_b'.
Simplifying and replacing x_b ->x(t) and v_p'=-a (constant portal acceleration from zero) gives L*x''(t)+a*x(t)=L*a. Our initial conditions are x(0)=L/2 and x'(0)=0. The solution is then x(t)=L-L/2*Cos(t*sqrt(a/L)). The orange end of the rod passes through the portal at time t=(pi/2)*sqrt(L/a), and thereafter the rod travels at velocity v_f=sqrt(a*L)/2.
