Rotational Motion - Two spheres orbiting around a common barycenter

[Rosengrip]

Homework Statement

Two spherical bodies with equal mass m1=m2=1000 kg are orbiting around a common barycenter in a weightless environment because of the gravitational attraction.

A: Show how the frequency of orbiting is related to the distance between two bodies.

B: We connect two bodies with a massless rod. There is a spring connected to the center of this rod (axis around which the bodies rotate) with a weight on one end. Mass of weight is 1 kg. What is the spring coefficient, if the weight doesn't move during rotation? Assume the frequency of rotation is one revolution per day.

C: Assume the weight attached to a spring is oscillating during rotations with small amplitudes. What is the spring coefficient, if the weight oscillates 12 times during 1 revolution?

Sketch for better understanding:
[URL]http://www.shrani.si/f/26/TO/4cz4TVm7/skica1.png[/URL]

Homework Equations

F = (Gm₁m₂)/r²
F = ks
Equations for oscillating motion

The Attempt at a Solution

A: Force of gravity is a centripetal force.

for body1: m₁w²r = (Gm₁m₂)/R² (r = distance for body1 to barycenter, R = distance between to bodies, w = angular velocity of body1)

Out of this equation I get the frequency of rotation for body1:

f = SQRT[ (Gm₂) / (4R²rPi²) ]
It's the same for the other body.

B: Force of spring is a centripetal force

mw²s = ks (s = distance of weight from barycenter and the length of spring, m = mass of weight)

k = mw²

Here I didn't include both attractive forces from two bodies, which complicate things a bit. What I got here is way too simple so I'm pretty sure I'm missing something.

C: don't really have an idea how to do this, any tips would be cool :)

I'm pretty sure my soulution of A is right, not sure about B though. So I need someone to review this. Thanks.

 

[gneill]

B: We connect two bodies with a massless rod. There is a spring connected to the center of this rod (axis around which the bodies rotate) with a weight on one end. Mass of weight is 1 kg. What is the spring coefficient, if the weight doesn't move during rotation? Assume the frequency of rotation is one revolution per day.

Interesting. There's no mention of the resting length of the spring, nor of the equilibrium position of the weight from the center of rotation. I suppose one could come up with a formula for the spring coefficient which depends upon the distance of the equilibrium point from the barycenter and the resting length of the spring.

If the resting length of the spring is taken to be zero, then there will be no centrifugal force acting on the weight if it starts from there. So the spring coefficient could be any value! It's indeterminate.

If the spring constant is infinite (the spring is a rigid rod), then the weight is guaranteed not to move no matter what!

If the resting length of the spring is zero but the weight is nudged away from the center, then presumably it could settle down at a distance r from the barycenter that would depend upon the value of k. Your choice of either r or k will determine things.

I'm not sure how one is "expected" to interpret this. Is there an obvious choice?

Oh, by the way, I don't think it's valid to ignore the gravitational forces acting on the weight if its equilibrium position not a small fraction of distance between the spheres; The magnitudes of the acceleration terms become similar when r is more than about 15% of R.

 

[Rosengrip]

The question is confusing in my native language too, I've sent an email to my mentor about this and I'll update it as soon as I get the answer.

Thanks for the reply though.

 

[gneill]

The question is confusing in my native language too, I've sent an email to my mentor about this and I'll update it as soon as I get the answer.

Thanks for the reply though.

Thinking about it some more, I think that any position for the weight other than the center of mass of the two spheres is going to cause the assembly to have a new center of mass, and it won't be halfway between the spheres! This will make things a tad more complicated.

If one interprets the weight having zero motion as meaning not even orbiting about the center of mass, then the only possible choice would be for it to remain at the current center of mass.

 

[rwisz]

I would like to first commend you for taking a thorough and analytical approach to solving these problems. Your equations and equations for oscillating motion are correct in general, but there are a few things that could be clarified or expanded upon to provide a more complete solution.

For part A, it is important to note that the frequency of orbiting is directly proportional to the angular velocity, and thus the distance between the two bodies. As you correctly stated, the frequency can be calculated using the equation f = SQRT[(Gm2)/(4R2rPi2)]. However, it may also be helpful to mention that the angular velocity, w, can be calculated using the equation w = v/r, where v is the linear velocity of the bodies. This relationship between frequency and distance is known as Kepler's third law.

For part B, your approach is generally correct, but as you mentioned, you are missing the attractive forces between the two bodies. The spring coefficient, k, can be calculated using the equation k = mw2 - (Gm1m2)/R2, where w is the angular velocity and R is the distance between the two bodies. This takes into account the additional centripetal force from the gravitational attraction between the two bodies.

For part C, the spring coefficient, k, can be calculated using the equation k = (mω2)/[r(1+cos(2πn))], where m is the mass of the weight, ω is the angular frequency of the oscillations, r is the distance between the weight and the barycenter, and n is the number of oscillations during one revolution. This takes into account the additional centripetal force from the gravitational attraction between the two bodies, as well as the periodic nature of the oscillations during one revolution.

In conclusion, your approach and equations are generally correct, but it may be helpful to clarify and expand upon certain aspects to provide a more complete solution. Keep up the good work in your studies of rotational motion!