Finding the ratio $\dfrac{\Delta E_X}{\Delta E_Y}$

Question

These two cylinders are put on top. Once we heat them with an amount of $\Delta T$, the gravitational potential energy of $X$ and $Y$ are increasing by $\Delta E_X$ and $\Delta E_Y$ . Find the ratio $\dfrac{\Delta E_X}{\Delta E_Y}$

I tried to find their potential energy

$$U_X = mgh \tag{1}$$

and

$$U_Y = mg(2h) = 2mgh \tag {2}$$

However, there should be something I'm missing.

Regards!

In physics, different assumptions are made while solving different problems. So if you have been asked to assume linear expansion, this question can be solved easily. — Shreyas Pimpalgaonkar, Jun 21 '18 at 12:24

score 1 · Accepted Answer · edited Jun 21 '18 at 15:22

1

Yes, so see. Sorry for my presentation skills.

If you heat the boxes with temperature $T$, they expand - and the expansion of each of them is proportional to their lengths and the change in temperature.

So as the lengths are same, they both will expand by the same amount.

Let's say each of them expands by length $l$.

So the COM of lower one goes higher by $\dfrac l2$, and the COM of upper one goes higher by $\dfrac l2+l = \dfrac{3l}{2}$

Since the density of the rods is still same, we can calculate the Potential difference directly by mgh (h of COM)

delta $E($upper rod$) = m\cdot g\cdot\left(\dfrac{3l}{2}\right)$

delta $E($lower rod$) = m\cdot g\cdot\dfrac l2$

Ratio $= \dfrac13$

edited Jun 21 '18 at 15:22

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answered Jun 21 '18 at 12:08

Shreyas Pimpalgaonkar

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Your final answer is truly wrong. – Hamilton Jun 21 '18 at 12:19
Could you tell what the answer is supposed to be? – Shreyas Pimpalgaonkar Jun 21 '18 at 12:21
That's correct now. Could you use latex? – Hamilton Jun 21 '18 at 12:42
Ah my bad, i had written Y/X instead of X/Y. Okay I'll use LaTeX. Did you understand the concept btw? – Shreyas Pimpalgaonkar Jun 21 '18 at 12:43
1

Absoulety, that was awesome! However, I believe that it will seem more clear once you use latex. – Hamilton Jun 21 '18 at 12:46

Finding the ratio $\dfrac{\Delta E_X}{\Delta E_Y}$

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