We know the Brachistochrone problem that to find the shape of the curve down which a bead sliding from rest and accelerated by gravity will slip (without friction) from one point to another in the least time.

I wonder if we use a line segment mass on a curve with the same problem. It is clear that the line segment mass will touch two point on the curve. Which curve equation must be for the least time in that case? I do not know how to start.

Note: Line segment mass is released from point A and its finish place is origin (O). the line segment mass is homogen and lenght is $l$ and mass is $m$. no friction is in the system.Point $A(x_1,y_1)$ and Point $B(x_2,y_2)$ are given starting points.

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Thanks for answers

  • 1
    $\begingroup$ There is a technical ambiguity here, is the stick of uniform mass, or is it a mass attached to the middle of a massless stick? You know, the latter is presumably much much simpler. $\endgroup$
    – Ali
    Jul 16, 2013 at 7:37

1 Answer 1


The minimum time for the segment AB will be the minimum time for its center of gravity I (middle of AB) too.
But I is a point in a uniform gravity field without friction.
So the trajectory of I will be a brachistochrone whose equation is known because you know the starting point of I (middle of AB). You then compute the curve that will follow A and B knowing that
Xi = (Xa + Xb)/2 and Yi = (Ya + Yb)/2 and that I follows a known brachistochrone (e.g Xi and Yi verify a known differential equation).

As this is 2 equations for 4 unknowns (Xa,Xb,Ya,Yb), you will need 1 more equation to get a functional relationship between f.ex Ya and Xa.
It is the rigidity : (norm of AB)² = l².

That's how you start and the rest is calculus.

  • 1
    $\begingroup$ This sounds right to me if the mass is concentrated at the middle of the stick, but I assume the OP intended a stick with a uniform distribution of mass. $\endgroup$
    – user4552
    Aug 19, 2013 at 13:42
  • $\begingroup$ It is the movement of the center of gravity that matters. With a uniform distribution it is the middle (assumed here) with another distribution it is another point. All other points are defined by the rigidity condition. So when you know the movement of the center of gravity, you know the movement of all other points of the stick regardless of the mass distribution. $\endgroup$
    – Stan Won
    Feb 17, 2014 at 11:29

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