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I am attempting to settle a friendly bet. Would a pendulum swing indefinitely in a hypothetical vacuum (i.e. no air resistance) having a hypothetical frictionless bearing (i.e. no energy lost due to friction) assuming the following

  1. The frictionless vacuum is on Earth (9.8 m/s^2).

  2. The pendulum is already in motion and no other external forces other than gravity act on the pendulum.

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Yes. this is a perfect case of a so called perpetuum mobile (see here).

It would respresent a perfect (ideal) non-dissipative system where entropy production $d_iS/dt=0$, in accordance with the 2nd law of thermodynamics. Indeed the first law of thermodynamics (energy conservation) does not say much about this, except that no term for energy loss included.

However, the system must really be non-dissipative, that means no type of friction or dissipative loss of energy in any way, such as friction in the elements of the pendulum etc.

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The real world is full of small effects that only matter when you've eliminated everything else. For example, if the pendulum has a non-zero conductivity its motion through the Earth's magnetic field would cause eddy currents and dissipate energy. This would be a tiny effect, but it would mean the pendulum wouldn't oscillate for ever. I imagine the more creative minds hereabouts could come up with a number of vanishingly small effects that would eventually damp the pendulum.

If you manage to eliminate all these effects then yes, the pendulum will oscillate forever. However you're just asking whether if all sources of energy dissipation are removed will any energy be dissipated, and the answer is obviously no.

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    $\begingroup$ Gravitational wave emission cannot be removed. $\endgroup$ – Michael Brown Jun 24 '13 at 9:32
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    $\begingroup$ Gravitational wave emission can be removed if the pendulum motion has no quadrupole or higher moments. $\endgroup$ – John Rennie Jun 24 '13 at 11:09
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    $\begingroup$ True, but I must confess I can't visualise what kind of "pendulum" that would be. (Not that I'm particularly good at visualising quadrupole moments.) $\endgroup$ – Michael Brown Jun 24 '13 at 11:13
  • $\begingroup$ @JohnRennie: it would have to be a perfect monopole or a dipole. a perfect monopole would be a sphere, so that's out as a pendulum. I have to admit that picturing what a perfect dipole is without a negative charge, but I don't see an easy way to make $\int\rho x^{i}x^{j}x^{k}$ equal zero, but $\int \rho x^{i}$ nonzero. $\endgroup$ – Jerry Schirmer Aug 22 '13 at 1:10
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    $\begingroup$ Actually, I'm asking this as a question: Does a pendulum emit gravitational waves? $\endgroup$ – Emilio Pisanty Nov 18 '13 at 14:39
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As John Rennie points out, there are in practice always small effects that lead to dissipation. Some of these effects may be involved in an essential way such that they cannot be removed. Gravitational waves are such an effect discussed in the comments of John Rennie's answer. There exists another effect that is much larger. First we note that gravity cannot be the only force acting on the pendulum, otherwise the pendulum would be in a free fall and could not work. A pendulum at rest on the Earth's surface will be subject to a normal force from the Earth's surface. As the pendulum swings in the vacuum box, the force exerted by the box on the ground increases and decreases, thereby causing vibrations that move into the Earth. This carries away energy from the pendulum.

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  • $\begingroup$ very true. Also the box cannot be perfectly stiff so that it would flex and dissipate energy to the atmosphere. Even if there was no air, the box woud get hotter through flexure and lose energy through heat. $\endgroup$ – Peter R Jan 22 '16 at 23:43
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It would because the pendulum wouldnt be losing its kinetic energy due to air friction. In a normal scenario the pendulum faces friction from the air around it and continously loses energy to this. In a frictionless envioronment there is no resistive friction hence it goes on
.This motion is referred to as "Simple Harmonic Motion".

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  • $\begingroup$ It is in the vacuum ... $\endgroup$ – user46925 Jan 23 '16 at 1:51

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