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Following this question: Deriving mass from simple pendulum which is summarized below

Some mass $m$ is release from rest at a horizontal position. $m$ reaches the bottom of its path (so directly under the pivot) at a velocity of $v$ m/s. And the length of the rope $l$. Now with no further information is it possible to derive an equation that can be used to describe the mass of this bob?

And the answer was of-course that the mass is independent of the motion
But I just got curious and set out to try the problem to see if I can tell something about $m$

Fixing my reference frame at the pivot, The Kinetic Energy of the earth-bob system is

$$T=\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}M\dot{y}^2$$

Where $\theta$ is the angle with respect to the vertical, $M$ mass of earth, $y$ displacement of earth in the vertical direction

And potential energy

$$V=-\frac{GMm}{R+y - l\cos\theta}$$

Where $R$ is the radius of earth, So by the conservation of energy

$$\frac{1}{2}ml^2\dot{\theta}^2+\frac{1}{2}M\dot{y}^2=-\frac{GMm}{R+y - l\cos\theta}$$

Solving for $m$ and taking $l\dot{\theta}=v$, at $\theta=90^\circ$

$$m=\frac{M\dot{y}^2}{[\frac{2GM}{R+y-l}-v^2]}$$

Since $$\Delta x\Delta p \geq \frac{\hbar}{2},$$ Applying it for $y$

$$M\dot{y}{y} \geq \frac{\hbar}{2},$$

So can I confidently say that

$$m \geq\frac{\frac{\hbar^2}{4My^2}}{[\frac{2GM}{R+y-l}-v^2]}~?$$

In practice The numerator is almost $0$ and the denominator too is almost $0$ since ($v^2\approx 2gh$), So mass $m$ is indeterminate which implies that it can have any value and it is true

  1. Is my reasoning and derivation right?

  2. Is my application of uncertainty principal right?

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Courage
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1 Answers1

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The introduction of the uncertainty principle is spurious, and in any case is unnecessary as you have already made your point.

In my answer to your previous question I said the motion is independant of the mass of the pendulum bob. However this is only an approximation, and it works only if we take the Earth to be fixed. It's an approximation because objects of different masses do not fall at the same rate. Larger masses actually fall faster than lighter ones for the reason discussed in the answers to Don't heavier objects actually fall faster because they exert their own gravity?.

The approximation assumes that the mass of the Earth is so much greater than the mass of the pendulum bob that the Earth mass can be assumed infinite. In this case the Earth doesn't move in response to the swings of the pendulum. In reality the Earth will move as the pendulum swings, and the pendulum motion will depend on the relative mass of the bob compared to Earth. This means you can calculate the pendulum mass from its motion.

Well, in principle anyway. As you have found, in practice the Earth would move so little that it would be immeasurable.

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John Rennie
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