Inconsistency of PE to KE conversion in moving reference frames

Question

Here's a nice trick question to keep you amused over the weekend.

A trolley of unit mass with light frictionless wheels is released to roll down a ramp onto a smooth level surface. The PE lost equals the KE gained, so if the trolley drops 5m vertically, its speed on the level surface will be circa 10 m/s (rounding up fractionally), a gain of 50J.

Now consider the same experiment viewed by a moving observer. Say one passing in a car at 10m/s. To that observer the trolley's speed increased from 10m/s to 20m/s as it rolled down the ramp, so its KE rose from 50J to 200J, an increase of 150J, even though the PE lost was still 50J as before.

How do you account for the discrepancy, especially if the ramp is fixed to the ground?

Answer 1

Let's do this properly and then take the approximation that you are taking from the beginning. So, let's assume that the object that is rolling down is of mass $m$ and the ramp is of mass $M$. Let's work in a generic inertial frame where both are moving horizontally with a velocity of $u$. The object rolls down the ramp and then the object and the ramp have the final velocities of $v$ and $V$ respectively. The standard application of momentum conservation and energy conservation would give us the following equations:

$$mgh+\frac{1}{2}(m+M)u^2=\frac{1}{2}mv^2+\frac{1}{2}MV^2$$$$(m+M)u=mv+MV$$One can easily solve them to give $$v=u+\sqrt{\frac{2gh}{1+\frac{m}{M}}}$$$$V=u-\frac{m}{M}\sqrt{\frac{2gh}{1+\frac{m}{M}}}$$

Now, if you only wish to speak of the velocities, we can take the limit $\frac{m}{M}\to 0$ simply by looking at the expressions and obtain that $$v=u+\sqrt{2gh}+\mathcal{O}\Big(\frac{m}{M}\Big)$$$$V=u+\mathcal{O}\Big(\frac{m}{M}\Big)$$

These results match with what one would naively expect anyway--that if the ramp is infinitely more massive than the object then the speed of the ramp wouldn't change at all. However, here is the trick. If you want to speak of the kinetic energy of the ramp, you need to multiply the square of its speed with its mass. So, we need to do this carefully for the change in the speed of the ramp is suppressed by the mass of the ramp but if we are calculating a quantity in which we are also multiplying the mass of the ramp in the numerator, we better do it properly. So, let's properly evaluate $\frac{1}{2}MV^2$ (and $\frac{1}{2}mv^2$ while we are at it).

So, from the exact expressions of $v$ and $V$, we can write $$\frac{1}{2}MV^2=\frac{M}{2}\bigg(u-\frac{m}{M}\sqrt{\frac{2gh}{1+\frac{m}{M}}}\bigg)^2=\frac{1}{2}Mu^2-mu\sqrt{\frac{2gh}{1+\frac{m}{M}}}+mgh\Bigg(\frac{\frac{m}{M}}{1+\frac{m}{M}}\Bigg)$$ $$\frac{1}{2}mv^2=\frac{m}{2}\bigg(u+\sqrt{\frac{2gh}{1+\frac{m}{M}}}\bigg)^2=\frac{1}{2}mu^2+mu\sqrt{\frac{2gh}{1+\frac{m}{M}}}+mgh\Bigg(\frac{1}{1+\frac{m}{M}}\Bigg)$$

Now taking the limit $\frac{m}{M}\to 0$ we obtain$$\frac{1}{2}MV^2=\frac{1}{2}Mu^2-mu\sqrt{2gh}+\mathcal{O}\Big(\frac{m}{M}\Big)$$ $$\frac{1}{2}mv^2=\frac{1}{2}mu^2+mu\sqrt{2gh}+mgh+\mathcal{O}\Big(\frac{m}{M}\Big)$$

So, in the limit where $\frac{m}{M}\to 0$, to the leading order, the change in the kinetic energy of the ramp and the object are $$\Delta \text{KE}_{\text{ramp}}=-mu\sqrt{2gh}$$ $$\Delta \text{KE}_{\text{object}}=mu\sqrt{2gh}+mgh$$

So, the point is that the "additional" change in kinetic energy that the object seems to acquire in the case where $u\neq 0$ (in comparison to the case where $u=0$) is exactly the same as the loss in the kinetic energy of the ramp which would have been zero if $u$ had been zero. This "additional" change arises from a cross-term which survives at the zeroth-order even when we take the limit of $\frac{m}{M}\to 0$. So, that is why, the kinetic energy of the ramp does change in this process even if the ramp is infinitely heavy.

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In fact, we should have done this procedure all along even in a frame where $u=0$ for we do not really know beforehand that the change in the kinetic energy of the ramp must vanish in the leading order.

Answer 2

So in a frame moving at $u$, the initial kinetic energy of the mass is:

$$ T_i = \frac 1 2 mu^2 $$

and after sliding to velocity

$$v = \sqrt{2gh}$$

the final kinetic energy is:

$$ T_f = \frac 1 2 m(u+v)^2 = T_i + \frac 1 2 mv^2 + muv $$

So there is a frame dependent discrepancy in the change in the kinetic energy:

$$ \Delta_T = muv$$

The missing energy is provided by the normal force in the horizontal direction:

$$ F = mg\sin{\theta} $$

which acts over a $u$-dependent distance of:

$$ d = ut $$

where

$$ t = \frac{h/\sin{\theta}}{v/2}$$

The work done by this force is:

$$ W = Fd = mg\sin{\theta} \frac{h/\sin{\theta}}{v/2} = muv = \Delta_T $$.

Answer 3

The answer is that in response to the movement of the trolley the earth recoils fractionally. If the earth's recoil speed measured in the earth's original rest frame is dv, its change in KE is proportional to dv² which is vanishingly small, and can be neglected when considering the conversion of PE to KE. Measured in any other frame moving at speed V in the direction of motion of the recoil, the loss in the earth's KE is proportional to dv² - 2Vdv, which does not vanish in the way that dv² does.