# Pendulum forces in orbit

On Earth a pendulum is governed by a set of equations, relating the angle of the swing, the mass etc.

However with a pendulum in orbit there is no fixed point (assume 2 masses attached via a string). If a force is acting on one of the masses (pulling it away from the other mass - analogue to gravity) how does the other mass respond. Intuitively the system would swing like a pendulum, however the net result of this force must in fact be the translation of the masses and string to another location in space.

I've boiled down to a simple problem:

In the above the square mass experience a force along the dotted line away from the round mass. With the offset of theta, what is the force that acts to rotate the square mass so the force aligns with the string?

• If you are in orbit with this device at zero-g (free fall) then technically its not a 'pendulum' anymore. For a pendulum its assumed you have gravity as a restoring force. No gravity, no restoring force, no pendulum. Commented Feb 11, 2015 at 0:19
• So what dynamics govern this problem then? Commented Feb 11, 2015 at 0:30
• It's a '2-body rigid dynamics' problem, but highly non-linear since the two masses will only create tension in the string when they are somehow forced in opposite directions from one another. When that's not happening they are essentially separate bodies acting on their own with slack in the string. Predicting motion really depend on where forces are applied. And whether the string is inelastic or elastic. If elastic the string can provide restoring force if tension occurs. Commented Feb 11, 2015 at 0:36
• So (from the image in the question) if the square is forced leftward, but no force is applied to the sphere there would be no tension? Commented Feb 11, 2015 at 0:40
• There would be tension. If you accelerate the block to the left, the ball would follow under tension of the string. Then you have essentially 'created' gravity. The dynamics would be a harmonic oscillator (pendulum) with natural frequency of SQRT(L/a) where L is the string length and a is the acceleration. This all assumes the string is inelastic. Commented Feb 11, 2015 at 0:44

If you are in orbit with this device at zero-g (free fall) then technically it's not a 'pendulum' anymore. For a pendulum, it is assumed you have gravity as a restoring force. No gravity, no restoring force, no pendulum.

It's rather a '2-body rigid dynamics' problem, but highly non-linear since the two masses will only create tension in the string when they are somehow forced in opposite directions from one another. When that's not happening they are essentially separate bodies acting on their own with slack in the string. Predicting motion really depends on where forces are applied. And whether the string is inelastic or elastic. If elastic the string can provide restoring force if tension occurs.

But if there is a leftward acting force on the block, there would be tension in the string since the ball has inertia. If you accelerate the block to the left, the ball would follow under tension of the string. Then you have essentially 'created' gravity. The dynamics would be a harmonic oscillator (pendulum) with natural frequency of $$(L/a)^{1/2}$$ where $$L$$ is the string length and $$a$$ is the acceleration. This all assumes the string is inelastic.

Note the force acting on the block would need to be perpendicular to the surface and act through the block's center of mass to prevent the block from rotating.

• Thanks again! Once we got down to the tension being applied due to the mass I started understanding the problem a little more clearly. Commented Feb 11, 2015 at 1:21