I Rotating sphere which separates into hemispheres

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Hi all,

The scenario I'm considering is a solid sphere (of uniform density) rotating with constant angular velocity when it abruptly splits into two hemispheres along a cut which contains the rotation axis. The hemispheres will begin to separate; if, for example, we consider the rotation to be counterclockwise, then the upper hemisphere will seem to slide to the right (against the rotation) until they separate.

My question is, which of these (if either) is the correct differential equation to describe the motion? Force considerations (in an angularly-accelerating, co-rotating frame) show that the only forces parallel to the direction of sliding are the fictitious centrifugal, Coriolis, and Euler forces. Computing these carefully, we have
[tex]\frac{d^2\ell}{dt^2}=\ell(t)\left(\frac{\frac{2}{5}R^2\omega_0}{\ell(t)^2+\frac{83}{320}R^2}-\frac{\frac{3}{4}R(\frac{83}{320}R^2+\frac{3}{4}\ell(t)^2)}{\ell(t)(\ell(t)^2+\frac{83}{320}R^2)\sqrt{\ell(t)^2-(\frac{3}{4}R)^2}}\frac{d\ell}{dt}\right)^2.[/tex]
Here [itex]\ell(t)[/itex] is the separation of the CM of the hemispheres.

If I assume kinetic energy is conserved (which it appears it must be, by symmetry and the fact that the normal here is an internal force), then we instead obtain the first-order differential equation
[tex]\frac{dr}{dt}=\frac{\sqrt\frac{2}{5}Rr\omega_0}{\sqrt{\frac{83}{320}R^2+r^2}},[/tex]
where [itex]r[/itex] is the distance between the centers (not CM) of the hemispheres.

The latter equation is decidedly simpler, but does not seem to produce the right results, even in limiting cases. The former is qualitatively accurate, which is reassuring, but can only be solved numerically (I think - I'd be very glad if anyone can prove me wrong!).

If the energy conservation approach is wrong, could someone please explain to me why? The normal force can't transfer energy from one hemisphere to the other, because the system has order 2 rotational symmetry. There is nowhere else that the energy can go, so it just transfers between the two types of kinetic energy. (I'm assuming also that the total kinetic energy is the sum of the rotational kinetic energy about the CM and the translational kinetic energy of the CM, which I've seen elsewhere is valid.)

Any help anyone can give me in unconundruming this conundrum will be greatly appreciated!

Best,
QM
 

kuruman

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If I understand the problem correctly, angular momentum about the axis of rotation must also be conserved. The CM of the hemispheres will move in opposite directions such that ##m v r_{cm}=I\omega##, where ##r_{cm}## is the distance of a hemisphere's CM to the axis.
 
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If I understand the problem correctly, angular momentum about the axis of rotation must also be conserved. The CM of the hemispheres will move in opposite directions such that ##m v r_{cm}=I\omega##, where ##r_{cm}## is the distance of a hemisphere's CM to the axis.
Hi Kuruman!

Thank you for your response. Yes, I forgot to mention, but angular momentum conservation is implicitly used to derive both of the differential equations above. For the first, I do not assume conservation of energy - merely that there are centrifugal, Coriolis, and Euler forces acting on the upper-right hemisphere (WLOG) and obtain an equation of motion from Newton's second law. For the second, I use energy and angular momentum conservation together, which allows me to eliminate angular velocity from the equation.

I've sketched both my approaches in the attached LaTeX .pdf. Please let me know if you see any way forward, a mistake I've made, etc.

Best,
QM
 

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kuruman

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Perhaps I am not seeing something so please explain what it is. The picture below shows the sphere at the moment of the split looking down on the sphere. The rotation axis is perpendicular to the screen. At that moment the of the split the CM of the left hemisphere is moving down and the CM of the right hemisphere is moving up. Furthermore, the force that one hemisphere exerts on the other is zero (that's what a split means). So according to Newton's 1st law once the force between hemispheres is turned off, each center of mass at distance ##r_{cm}## from the axis of rotation will move in a straight line with the velocity it had just before the split. Did I misinterpret something?

SplitSphere.png
 

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A.T.

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...the upper hemisphere will seem to slide to the right (against the rotation) until they separate...
Are you saying that they will remain in sliding contact after the split for a while, and not separate immediately?
 
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Are you saying that they will remain in sliding contact after the split for a while, and not separate immediately?
Yes, that is correct.
Perhaps I am not seeing something so please explain what it is. The picture below shows the sphere at the moment of the split looking down on the sphere. The rotation axis is perpendicular to the screen. At that moment the of the split the CM of the left hemisphere is moving down and the CM of the right hemisphere is moving up. Furthermore, the force that one hemisphere exerts on the other is zero (that's what a split means). So according to Newton's 1st law once the force between hemispheres is turned off, each center of mass at distance ##r_{cm}## from the axis of rotation will move in a straight line with the velocity it had just before the split. Did I misinterpret something?

View attachment 240210
I could be wrong (hence why I'm asking here), but the description you give seems to be valid only for the instant that the split occurs. The hemispheres should continue rotating as well as sliding immediately after they separate. Consider an orange thrown into the air which suddenly gets cut in two the way I described. You can imagine that the slices will continue to rotate (albeit more slowly) as they separate.

I think my point is best made in the following approach. A solid hemisphere of radius [itex]R[/itex] has its center [itex]\frac{3}{8}R[/itex] above the center of its circular face, so the two hemispheres must exert a force [itex]\frac{3}{16}mv^2/R[/itex] to maintain its centripetal acceleration (the CM is moving in a circle of radius [itex]\frac{3}{8}R[/itex]). At the moment of separation this force is set to zero. Applying Newton's 2nd law to each hemisphere dictates the angular velocity immediately after separation is unchanged; gradually, however, the normal force between the hemispheres slows down the angular velocity. If the hemispheres could pass through each other, I can see how they would travel in a straight line, but obviously they can't. I can definitely make this argument crisp, but it is tedious; suffice it to say the hemispheres must slide after separation.

I think what's happening in the energy-conserving ODE is that the sphere continuing to rotate as it did before separation conserves both [itex]J[/itex] and [itex]K[/itex], so there is no reason for the hemispheres to separate. I thought that maybe considering the hemispheres to rotate instantaneously about their CMs precisely at [itex]t=0[/itex] would help, but I'm not sure about that assumption.

Best,
QM
 
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Hold on a second - I think we might both be right. The energy-conservation ODE doesn't apply at [itex]r=0[/itex] because that's the very small window during which the spheres separate. Then, moments later, the rotation of each hemisphere causes them to collide inelastically, losing some kinetic energy but not angular momentum. From this point on, kinetic energy and angular momentum are conserved, leading to an ODE like the second one I found, but with a different total kinetic energy and [itex]r>0[/itex]. Does this sound reasonable?
 

kuruman

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I guess my original thought was too simplistic. I would agree that energy is not conserved but angular momentum is. I need to think some more about this.
 

A.T.

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Are you saying that they will remain in sliding contact after the split for a while, and not separate immediately?
Yes, that is correct.
Consider the rotating frame with constant angular velocity where the sphere is initially at rest:

- Initially you have only the Centrifugal forces, pulling the halves apart. So they should separate immediately.

- As soon they start moving apart, there will be Coriolis forces, which deflect their paths, but don't create any torques around their CMs. So their orientations should remain constant, and their cut surfaces should remain parallel, while they move apart. The only way to collide again would be after the Coriolis forces bend their paths by more than 90°, provided the halves are too large to pass each other.
 
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I believe this is possible; as an extreme case, consider hemispheres that are a shell of negligible mass and a heavy point mass near the center of the circular face. The CM is then very close to the point mass and the hemisphere will not have traveled far before colliding with the other.

Could my original scenario be realised by rotating both hemispheres about their CM at the instant [itex]t=0[/itex] and watching the time evolution from there? We then don't have any linear velocity of the CM, so total [itex]J,K[/itex] are different but their conservation at all times is still valid.
 

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