# Nature of motion of a pendulum

Consider a pendulum suspended from the ceiling of a lift in free fall , if its displaced from its mean position , what will be its nature of motion?

what i thought was that it would simply stick to the ceiling as the lift's in free fall; but on the contrary what my teacher said was that , as

  "g" = 0 it's  time period  will be "infinity".


Thus it implies it won't move at all.

But is that even possible?

please provide me with some reason or correct me if I'm wrong.

any help appreciated.

cheers.

• Check the forces in the falling lift's referential – Tom-Tom Jan 7 '14 at 13:35
• It may seem counterintuitive, because one rarely experiences a free-fall. But nevertheless: objects in free-fall are weightless! – Nephente Jan 7 '14 at 13:45
• If you really want to give your teacher a hard time, you could argue that due to the gravity gradient the time period will not be infinity and that eventually the axis of minimum momentum of inertia will be aligned vertically. (assuming that this is a non-ideal pendulum, i.e. the rope or connecting rod has mass) – OSE Jan 7 '14 at 15:39

Your teacher is right. It is an experimental fact that being in free-fall is indistinguishable from being in the complete absence of gravitational fields. For instance, consider an astronaut aboard a satellite orbiting the Earth. We've seen videos; the astronauts appear completely weightless. But note that there's still gravity where they are! Satellites are not out in deep space, they're rather close to the Earth, and the effects of gravity are still relatively strong. The reason they seem weightless is because they are in a state of perpetual free-fall. That is to say, all satellites are constantly falling to Earth; the reason they don't reach the surface is because they all have a horizontal/tangential velocity, though this is besides the point.

The point then, is that your lift in free-fall would behave precisely like a lift in deep space. You wouldn't expect a pendulum to be pulled towards the ceiling in deep space, and so you shouldn't expect a pendulum to be pulled towards the ceiling in free-fall!

To look at this from a more classical perspective, note that gravity is a unique force in that it is proportional to the mass of the body on which it acts. Because the acceleration of an object is equal to the force acting on it divided by its mass, we find that the acceleration of an object under gravity is independent of its mass --- this was one of the discoveries of Galileo Galilei, I believe.

What this means is that all objects in our lift --- including the lift itself --- fall towards the Earth with the same acceleration. This means that if the relative motions of the objects in the lift are zero at one instance of time, their relative positions will remain fixed for all future time. So if you displaced the pendulum from the vertical by some angle and let it go, the distance from the bob of the pendulum to the ceiling wouldn't change.

If that distance were to change, the accelerations of the ceiling and the bob of the pendulum would have to differ. If the pendulum were to fly up and hit the ceiling, that would mean that the ceiling was accelerating faster than the pendulum. But as we've established, the lift and the pendulum fall at the same rate --- that's gravity!

$$F_\mathrm{gravity} = mg \,, \qquad F = ma \,, \quad \implies \quad a = g = 9.8 \,\mathrm{ms}^{-2 }$$

• Related: the helium balloon in an accelerating automobile problem. So what would the helium balloon do while the elevator car is accelerating downwards? :-) – Carl Witthoft Jan 7 '14 at 16:00
• This would be right in a homogeneous gravitational field (so $\vec{g}$ independent of position), but in reality there will be a slight difference/gradient which means that the top of the elevator will experience a slightly lower attraction than the bottom. But you would need a very large lift to notice this near the earths surface. – fibonatic Jan 7 '14 at 16:15