# A more interesting Pendulum Paradox

1. Jul 1, 2008

### Count Iblis

Homework problem given by Count Iblis to the PF community:

Consider a box containing a pendulum of total mass M in a spacecraft that is in free falling motion. Suppose that the engines of the spacecraft are turned on and the spacecraft is accelerating at acceleration a, where a is parallel to the orientation of the pendulum, so it can swing with angular frequency omega = sqrt(a/L), where L is the length of the pendulum).

Suppose we don't let the pendulum swing by making sure we don't accelerate the box in the direction perpendicular to the orientation of the pendulum, the acceleration of the spacecraft is at all times parallel to the orientation of the pendulum.

What is the mass of the box?

2. Jul 1, 2008

### RandallB

IMO it cannot be determined without knowing the force F being applied to the box to produce the acceleration “a”.
That force could be measured by supporting the box on a scale in the spacecraft which could weigh it. (Note: scale needs to be based on springs to measure weight; scales based on counterbalanced weights directly measure mass not weight.)

Not sure I see the point of this ‘homework problem’ thread here or in calling it a paradox.
Do you intend to tie it into the Classical Pendulum Paradox thread somehow?

3. Jul 1, 2008

### yuiop

Would a pendulum in a free falling box swing or have an angular frequency in the first place?

4. Jul 1, 2008

### Count Iblis

The box containing the pendulum had a mass of M when it was in weightless conditions in the spacecraft. We are assuming that the pendulum was not moving when the spacecraft started to accelerate.

5. Jul 3, 2008

### Count Iblis

Hint: What is the ground state energy of the pendulum?

6. Jul 3, 2008

### JesseM

In relativity, the inertial mass of a composite system in a box is just the rest mass of all the parts + their kinetic energy + the potential energy between them (for example, an atom has a measurably lower inertial mass than the sum of the rest masses of the particles that make it up, because the potential energy is lower in the bound state). So what seems interesting about this problem is that the kinetic energy of the pendulum is changing, and therefore it might be that the only way to keep the inertial mass of the box constant would be to imagine a pseudo-gravitational "potential energy" based on the pendulum's height (just like in Newtonian physics, a pendulum is exchanging kinetic energy for potential energy as it rises). Is this something like what you were thinking of, Count?

7. Jul 7, 2008

### Count Iblis

JesseM,

that's basically it! We know from quantum mechanics that the pendulum in the ground state has an energy of:

$$E = \frac{1}{2}\hbar\omega$$

And we know that:

$$\omega=\sqrt{\frac{a}{L}}$$

Therefore, the total mass of the box is:

$$M(a) = M(0) + \frac{1}{2}\hbar\sqrt{\frac{a}{L}}$$

I.e. the mass of the box depends on its acceleration. This means that Newton's second law F = m a is nonlinear in a.

8. Jul 7, 2008

### Staff: Mentor

That is only true if you force a constant w. If you start the pendulum swinging freely and then use different accelerations you will find that w changes as your acceleration changes. In that case the mass does not depend on its acceleration and the second law is linear in a.

9. Jul 7, 2008

### RandallB

So you cannot really say what the mass of the box is without knowing the rest mass M(0) of the box as a given.

And the paradox must be that in this approach acceleration indicates a real increase in proper mass much like assuming relativist mass increases are real for E and P. In paradoxal contrast to more popular approaches that only use an unchanging rest mass M0.

I’ll just assume it to be an unresolved paradox, and leave it at that.