Hole Drilled into Sphere: Motion?

[FallenApple]

Say I have a solid sphere of mass on a horizontal surface. If suddenly there was a spherical hole(shape of a sphere) off center, will the sphere suddenly move? And there is no friction between the ground and the sphere. I'm assuming that it will. But my reasoning is this, the sphere suddenly having lost mass will have it's CM change locations. This will make it unstable or stable depending on the location. If a hole appeared sideways, then the CM will be off centered relative to the vertical axis going through the point of contact, causing the new object to rotate forever.

If the hole is on axis going through the contact point, then it will oscillate if the hole is above the original CM. The new object will not oscillate if the hole is below the CM, it will just fall and rotate in one direction forever since its unstable.

Is this the correct logic?

 

[BvU]

There is nothing relating the new CM to the old one. No reason it should want to go to the same location as the old one, no force that would do such a thing.
[edit] recanting. See Oro below.

 

[FallenApple]

There is nothing relating the new CM to the old one. No reason it should want to go to the same location as the old one, no force that would do such a thing.

What I am saying is that if I have a Sphere with a hollow below the center, then the CM is higher for this object compared to that of a regular solid sphere.

This sphere with low location for hollow would be like putting a bowling pin that was set upside down. It would want to tilt and fall over. Since this should be an unstable equilibrium.

 

[Nugatory]

Drilling a hole through a ball is one way of rearranging the mass inside the ball to to move the center of gravity around.

If the ball is initially in equilibrium on a frictionless surface and the center of mass moves around, there are three possibilities:
1) If the new center of mass ends up directly above where the old one was, the ball is still in equilibrium so it won't move, but the equilibrium is now unstable - if we disturb the ball even slightly, we'll end up in case #3 below.
2) If the center of mass ends up directly below where it had been, the ball is now in a stable equilibrium. It won't move, and if it is disturbed it will tend to return to its original position.
3) If the center of mass is displaced sideways so that it is neither directly above or below where it had been, gravitational forces will produce a torque around the center of the ball and it will rotate until the ball ends up in #2 above.

 

[A.T.]

And there is no friction between the ground and the sphere.

If there is no friction, the sphere won't roll away, even if you move the COM sideways. It will then just oscillate in place.

The new object will not oscillate if the hole is below the CM, it will just fall and rotate in one direction forever since its unstable.

Assuming no loses at all (nor friction, no deformation etc.), if you nudge it from the unstable equilibrium, it will keep spinning in place in the same direction, but at varying speed that periodically goes down to the initial nudge speed.

 

[CWatters]

...the CM will be off centered relative to the vertical axis going through the point of contact, causing the new object to rotate forever.

It might rotate but not forever. It rotates towards a stable position. It might overshoot and oscillate a bit but eventually will comet to a stop if there is any friction. See case 3) Nugatory describes.

 

[CWatters]

Do we need to consider how the mass in the hole is removed? You might have to apply conservation of momentum to the problem. eg if the mass in the hole(s) is/are ejected horizontally and tangentially using a spring or explosives (for example), and there is no friction, then it might be possible for the object to rotate indefinitely.

 

[A.T.]

eventually will comet to a stop if there is any friction.

The OP states no friction between the ground and the sphere. But friction with air or deformation losses (internal friction) will obviously stop it as well. It's not clear if the OP asks about a completely lossless case.

 

[Dadface]

It's something you can easily confirm with a quick experiment using for example a table tennis ball with a blob of plasticine stuck on it. Its not a spherical hole but the same principle applies, namely adding the plasticine alters the centre of mass and the equilibrium.

 

[hackhard]

Is this the correct logic?

no it is not
motion takes place only in 1st case

 

[FallenApple]

Drilling a hole through a ball is one way of rearranging the mass inside the ball to to move the center of gravity around.

If the ball is initially in equilibrium on a frictionless surface and the center of mass moves around, there are three possibilities:
1) If the new center of mass ends up directly above where the old one was, the ball is still in equilibrium so it won't move, but the equilibrium is now unstable - if we disturb the ball even slightly, we'll end up in case #3 below.
2) If the center of mass ends up directly below where it had been, the ball is now in a stable equilibrium. It won't move, and if it is disturbed it will tend to return to its original position.
3) If the center of mass is displaced sideways so that it is neither directly above or below where it had been, gravitational forces will produce a torque around the center of the ball and it will rotate until the ball ends up in #2 above.

Interesting. So what is the mechanism for 3? I know that if I release a pendulum from the horizontal position, it will rotate over 180 degrees, reach the same height, and then rotate back again.

But for the modified sphere, the center of mass will rotate down to the bottom, where there it would still have kinetic energy, so should raise up to the same height.

 

[A.T.]

Interesting. So what is the mechanism for 3?

See post #8.

 

[FallenApple]

See post #8.

Oh ok. I see. I should have been more clear about the surface being frictionless. But its interesting to note what happens when there's friction as well.

 

[A.T.]

Oh ok. I see. I should have been more clear about the surface being frictionless.

You were clear about that. The question was if there are any other losses. If not, then it will move forever as described in post #5.