- 18 Apr '08 08:09Suppose you have a closed loop of rope in the shape of a circle of radius 1m. It's floating in space, and rotating in the plane of the circle and about the center of the circle at 1 rad/s. This creates a "centrifugal force" which keeps the loop taut. What is the tension in the rope? Generalize to arbitrary radii and rotation speeds.

Equivalently, we may consider a loop made out of a "springy" material which obey's Hooke's law: suppose we have a loop of material with a rest length of 0m and a spring constant of 1 N/m; what is its equilibrium radius if it rotates as above? Generalize to arbitrary spring constants and rotation speeds.

I think this problem can be generalized to 3 dimensions like so: a beach ball of radius 1m floats in space; its internal air pressure is 1 Pa; what is the "tension" in the ball? The problem is giving a rigorous definition of the tension (or equivalently the spring constant) of 2d surface, which I don't know how to do. - 18 Apr '08 19:20

Further with the beachball analogy, what happens to the stresses on the ball if it starts to rotate?*Originally posted by GregM***Suppose you have a closed loop of rope in the shape of a circle of radius 1m. It's floating in space, and rotating in the plane of the circle and about the center of the circle at 1 rad/s. This creates a "centrifugal force" which keeps the loop taut. What is the tension in the rope? Generalize to arbitrary radii and rotation speeds.**

Equivalently, we may con ...[text shortened]... ion (or equivalently the spring constant) of 2d surface, which I don't know how to do. - 19 Apr '08 00:52

Interesting. This can be brought back down to 2D by imagining a circle of rope held taut by a uniform outward pressure (say 1 N/m), which starts to rotate about one of its diameters. This is trickier because the shape of the rope deforms from a circle. There is then the additional problem of finding the equilibrium shape of the rope. Hmm...*Originally posted by sonhouse***Further with the beachball analogy, what happens to the stresses on the ball if it starts to rotate?** - 22 Apr '08 19:34

there is no such thing as "centrifugal force" ... Think of a stone attached to an end of a rope for example ... if tou spin it and cut the rope the stone with follow the trajectory parallel to the tangent to the circle along which it orbit at the point the rope was cut (i.e. it will not move towards the centre)...*Originally posted by GregM***Suppose you have a closed loop of rope in the shape of a circle of radius 1m. It's floating in space, and rotating in the plane of the circle and about the center of the circle at 1 rad/s. This creates a "centrifugal force" which keeps the loop taut. What is the tension in the rope? Generalize to arbitrary radii and rotation speeds.**

Equivalently, we may con ...[text shortened]... ion (or equivalently the spring constant) of 2d surface, which I don't know how to do. - 22 Apr '08 19:52

That's not quite true, centrifugal force does exist and can be calculated, but it does not exist in all frames of reference. Centripetal force does, however.*Originally posted by 3v1l5w1n***there is no such thing as "centrifugal force" ... Think of a stone attached to an end of a rope for example ... if tou spin it and cut the rope the stone with follow the trajectory parallel to the tangent to the circle along which it orbit at the point the rope was cut (i.e. it will not move towards the centre)...**

The spinning rope is a pretty interesting problem. I have to work out more details, but my initial free body diagram consists of a small section of the rope cut out along two radii, with tension forces perpendicular to the cut surfaces and shear forces parallel to them. Now to balance them! - 22 Apr '08 20:19

http://xkcd.com/123/*Originally posted by 3v1l5w1n***there is no such thing as "centrifugal force" ... Think of a stone attached to an end of a rope for example ... if tou spin it and cut the rope the stone with follow the trajectory parallel to the tangent to the circle along which it orbit at the point the rope was cut (i.e. it will not move towards the centre)...** - 22 Apr '08 20:19

Yes, that's the right approach. I get*Originally posted by PBE6***The spinning rope is a pretty interesting problem. I have to work out more details, but my initial free body diagram consists of a small section of the rope cut out along two radii, with tension forces perpendicular to the cut surfaces and shear forces parallel to them. Now to balance them!**

T = rho.r^2.w^2

Where rho = mass per unit length, r is the radius, and w is the angular velocity.

Second part follows later. Although - rest length of zero? I know that makes the calculation easier, but it doesn't sound very realistic. - 22 Apr '08 20:24

I considered an n-sided polygon with masses for corners and massless rigid rods for edges, and then calculated the tension in the rods in the limit that n goes to infinity. But I think there should be a simpler way.*Originally posted by PBE6***That's not quite true, centrifugal force does exist and can be calculated, but it does not exist in all frames of reference. Centripetal force does, however.**

The spinning rope is a pretty interesting problem. I have to work out more details, but my initial free body diagram consists of a small section of the rope cut out along two radii, with tension forces perpendicular to the cut surfaces and shear forces parallel to them. Now to balance them! - 22 Apr '08 20:38 / 1 edit
*Originally posted by mtthw***Yes, that's the right approach. I get**

T = rho.r^2.w^2

Where rho = mass per unit length, r is the radius, and w is the angular velocity.

Second part follows later. Although - rest length of zero? I know that makes the calculation easier, but it doesn't sound very realistic.*GregM checks his reasoning.*-- Are you sure you're not missing a factor of 2.pi?

And if you like you can generalize the second problem to arbitrary rest lengths. - 22 Apr '08 20:47 / 1 edit

I was just about to say the same thing. It seems like the centripetal acceleration should be w^2.r (directed inwards), so the centripetal force would be m.w^2.r. Since m = rho.L, and L = 2.pi.r, shouldn't the centripetal force be: 2.pi.rho.w^2.r^2?*Originally posted by GregM**GregM checks his reasoning.*-- Are you sure you're not missing a factor of 2.pi?

And if you like you can generalize the second problem to arbitrary rest lengths.

Someone may have to step me through this part, too. For my infinitesimal slice, the tension in the rope is directed tangentially, while the shear forces are directed radially. I would expect that the shear forces would keep the slice of rope in place, not the tension. Physical intuition gives me the opposite answer though. - 22 Apr '08 20:50

I don't*Originally posted by GregM**GregM checks his reasoning.*-- Are you sure you're not missing a factor of 2.pi?*think*so, but it's always possible. There are factors of 2pi floating around though - have I defined the terms in the same way as you? The mass of my rope will be 2pi.r.rho in my terms. - 22 Apr '08 21:20

But remember that the tension isn't directed inwards.*Originally posted by PBE6***I was just about to say the same thing. It seems like the centripetal acceleration should be w^2.r (directed inwards), so the centripetal force would be m.w^2.r. Since m = rho.L, and L = 2.pi.r, shouldn't the centripetal force be: 2.pi.rho.w^2.r^2?**

Someone may have to step me through this part, too. For my infinitesimal slice, the tension in the rope is d ...[text shortened]... of rope in place, not the tension. Physical intuition gives me the opposite answer though.

Here's my argument. Consider a short arc ABCO, where O is the centre, and ABC is the arc. Angle AOB = BOC = a.

Each end of the arc has a force T applied, and the component of that force in the direction BO is T sin a.

The acceleration of the arc is mw^2r. And the length of the arc is 2ar, so m = 2ar.rho.

So 2T.sin(a) = 2a.r.rho.w^2.r

As a -> 0, sin(a) ~ a, giving the answer I gave above.

You can ignore the shear forces - by symmetry they must be zero. - 23 Apr '08 13:08

I tried going through your solution and it looks right to me.*Originally posted by mtthw***But remember that the tension isn't directed inwards.**

Here's my argument. Consider a short arc ABCO, where O is the centre, and ABC is the arc. Angle AOB = BOC = a.

Each end of the arc has a force T applied, and the component of that force in the direction BO is T sin a.

The acceleration of the arc is mw^2r. And the length of the arc is 2ar, so m ...[text shortened]... g the answer I gave above.

You can ignore the shear forces - by symmetry they must be zero. - 23 Apr '08 17:08

So for the second part, the tension force will stretch the springy material until the tension and the retractive forces balance:*Originally posted by PBE6***I tried going through your solution and it looks right to me.**

T = rho.w^2.r^2

Fr = k.x = k.(2.pi.r)

rho.w^2.r^2 = k.(2.pi.r)

r = 2.pi.k / rho.w^2 - 24 Apr '08 09:25

That was (roughly) my first thought (also noting the solution r = 0).*Originally posted by PBE6***So for the second part, the tension force will stretch the springy material until the tension and the retractive forces balance:**

T = rho.w^2.r^2

Fr = k.x = k.(2.pi.r)

rho.w^2.r^2 = k.(2.pi.r)

r = 2.pi.k / rho.w^2

*But*...if it's elastic it doesn't make sense to solve in terms of the mass/unit length any more, as this will vary as it stretches.