# Force due to thermal expansion?

Consider a thin rod which has a disc(or any object) placed at one of its end such that they are just in contact (no attachment).Then if we start heating the rod,what would happen to the disc (or any object) placed at the end....what would the force due to the expansion of the rod(if any)? Please do say if I’m missing something in the question!or missing any bit of concept! Note the object at the end do not expand or consider its specific heat is too high.

• What’s at the other end of the rod? Can you draw a picture? Commented May 9, 2019 at 10:43
• @BobD sure let me show with an image....take any object infact Commented May 9, 2019 at 11:19
• So there's nothing at the other end?
– JMac
Commented May 9, 2019 at 11:55
• @JMac nothing attached but in contact (just placed)...neglect the expansion of object at end. Commented May 9, 2019 at 11:56
• Are the ends of the rod and disk constrained axially, or are they totally unencumbered? This will, of course, make a big difference. Commented May 9, 2019 at 14:08

the force due to the longitudinal expansion (the rod becomes longer) will depend on the rapidity of the heating process and any constraint on the other endpoint of the rod or on the disc: if you keep them pressed together so that the rod expansion is obstructed then you will get enormous pressions in the system and the objects could eventually break or undergo some non trivial modification, for instance buckling of the rod. If the disc is fixed and rod can expand towards the other endpoint, it will do it.

Anyway, for physically reasonable heating processes the forces generated by the longitudinal expansion (if there are no constraints as those mentioned above) will be very small since they are fixed by the acceleration due to the expansion. If you want to study the trajectory then just knowing thiis acceleration will be enough (it will depend on the material and on the heating process) while if you are interested in the force for other reasons you should multuply the disc's mass times the acceleration of the endpoint.

Note that if the heating is very fast and both the disc and the rod are very, very heavy you will get that the dilatation will be somewhat slowed down, similarly to the case in which the system is constrained. Therefore in this limit the acceleration will be less than in the other case, for the same heating speed.

If your disc was a ring that squeezed the rod then the transversal expansion could too generate great strains in the system, as in the case in which the endpoints are held fixed. This would happen when the ring heats up more slowly than the rod or when the ring dilatation is smaller than the rod's (different materials), because the rod becomes thicker and the ring changes its size by a smaller amount.

Back to the main point, you can roughly extimate the acceleration of the disc (with respect to the opposite endpoint of the rod) by taking the longitudinal expansion of the rod (for instance 3.6 millimiters) and dividing it by the square of the time interval in which the rod is heated up (for instance 60 seconds). This is a very rough extimate but will not miss the actual value by a big margin. With these data you have an acceleration of $$10^{-6}$$ m/s^2, so that a disc of 1 kg is pushed with $$10^{-6}$$ Newtons.

To find the acceleration with respect to the lab you just impose that the center of mass does not accelerate, assuming for instance that the rod dilates homogeneously.

• By rapidity do you mean power given to the rod for heating? Commented May 9, 2019 at 12:16
• Ya I understand what you said that it depends on rapidity....is it possible (it depends on our power supplied....but also on the material right?) that it’s expansion is too rapid that the disc as shown in the picture would lose contact with rod and travel further...[provided the disc don’t expand.....or take it such that the disc’s specific heat is so high that the force due to rod’s expansion dominates over the disc’s on total. Commented May 9, 2019 at 12:22
• to your first comment : yeah or any quantity proportional to it, if you prefer. (maybe you could want to define it as the power given to the rod divided the rod's mass or some other rod data in order to make the definition less "rod-dependent"). To your second comment: once the expansion stops or slows down the disc will continue to travel at the speed it reached and it will get detached by the rod in any case. Commented May 9, 2019 at 12:23
• And ....will the disc placed in contact not cause any oppose toward the expansion at that end....also the way u calculated the acceleration is clear but I hope it’s not the total longitudinal expansion, it should be effective expansion at right end..isn’t it? Commented May 9, 2019 at 12:32
• yeah it will oppose, but as I said that will start beeing important only if disc and rod are very, very heavy. At that point is difficult to describe how the expansion is obstructed, because you would need to inspect very carefully the microscopic details of the system. What do you mean with effective expansion at right end? Anyway we are probably saying the same thing.. Commented May 9, 2019 at 12:51