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Recently wile solving questions on Thermal Expansion I encountered two types of questions

Type 1: A rod with one half of mass 'm' and the other of mass '2m' is kept on a horizontal smooth surface, it is heated and it expands longitudinally.

In this question centre of mass of system(the whole rod) remains same at the initial position, before and after expansion

Type 2: A uniform rod of mass m is kept on a rough inclined plane and it is heated. The rod expands and the centre of mass moves. Even though the net force on the body is zero. enter image description here I'm not able to analyze that why the centre of mass moves in second case but it doesn't move in the first.

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  • $\begingroup$ Is the system just the rod, or the combined rod and plane? The body seems to suggest it is just the rod. $\endgroup$
    – BowlOfRed
    Jan 23 at 20:03
  • $\begingroup$ @AgniusVasiliauskas That looks like an answer. $\endgroup$ Jan 23 at 20:29
  • $\begingroup$ @Chemomechanics Ok, moved to an answer, by request $\endgroup$ Jan 23 at 22:13

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If a rod is resting on an inclined plane, then there must be sufficient friction for it not to slide entirely. But if the rod changes shape, then the friction on the plane must have been overcome on at least some portions of the interface where the rod is touching.

So you have a partially unknown amount of friction from the ramp applying forces to the rod. I don't think "absence of any external force" would apply.

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  • $\begingroup$ But still the frictional force balances the net effective weight acting along the inclined plane. Therefore, net force is zero , if you consider the whole rod $\endgroup$
    – PinkAura
    Jan 24 at 3:29
  • $\begingroup$ Or i should say, net frictional force balances the weight acting along the inclined plane $\endgroup$
    – PinkAura
    Jan 24 at 5:06
  • $\begingroup$ If the rod has moved, then that's not true. You have no information about how the friction is acting except the result on the rod. You're not given any information about how it acts. Because the rod moves, you know the net force on the rod is not zero at all times. $\endgroup$
    – BowlOfRed
    Jan 24 at 5:25
  • $\begingroup$ Umm, it was a numerical problem, so i guess i solved it, there was this one point (say, P) above l/2 which was fixed . That means above P the frictional force was acting in downward direction and below P frictional force was acting in upward direction. $\endgroup$
    – PinkAura
    Jan 24 at 5:41
  • $\begingroup$ If one point remains fixed , i guess the rod is actually not displacing, it's just some points are displacing from their position. Also, they are moving very slowly so we can assume the acceleration to be 0. $\endgroup$
    – PinkAura
    Jan 24 at 5:44
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Because in the second case stress forces must overcome gravity expanding up the hill, while expansion down the hill is accelerated by gravity, hence com moves down the hill.

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  • $\begingroup$ So will there be asymmetric expansion? $\endgroup$
    – PinkAura
    Jan 24 at 3:27
  • $\begingroup$ Exactly, friction force and gravity cancels each other down the hill, while up the hill it amplifies each other, so this makes expansion more easier down the hill and hence asymmetric. $\endgroup$ Jan 24 at 7:47

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