Will M1 Spin Around M2 When Rope is Cut?

[Davidindenver]

Ok this device is spinning around the center of mass(CM).

CW------CM--------------M1---M2

while at full speed the roap is cut like so
CW------CM-------------- M1---M2

does M1 spin around M2?

It seems simple enough but I can't find the answer.

 

[Domenicaccio]

I think yes, they do spin.

An instant before the break, M1 has a tangential speed v1, and M2 has v2>v1. That should be the same an instant after the break, therefore M1 and M2 will start rotating around their center of mass while going away from the rest of the device.

The rest of the device, unless the point "CM" was fixed in space, will also be shoot away (opposite direction of course) so as the center of mass (the point in space where "CM" was before, but now "CM" is not the center of mass anymore) will remain the same. And the rest of the device will spin too (same verse as M1-M2) so that the angular momentum is conserved.

 

[neu]

The fact that a rope is used makes this very complicated. If you replace th erope with a solid (massless?) bar then yes they would spin around each other, but not necessarily about their centre of mass.

At the instant the bar is cut, M2 will have a higher tangental velocity as:

v_{\perp} = r\frac{d\phi}{d\t} (where \phi is the angle in the plane of motion)

I don't think however that they'll spin about their mutual centre of mass as the spinning motion is intialised only buy the velocity gradient between M1 and M2. This is why I say the bar must be rigid to cause any kind of spinning motion, as if it were rope there would not be constant tension across it as there would in a spin. There would be come sort of ocillatory motion in their spinning.

I hope what I've said makes sense, they're my thoughts, I'm happy to be corrected.

 

[Domenicaccio]

The fact that a rope is used makes this very complicated. If you replace th erope with a solid (massless?) bar then yes they would spin around each other, but not necessarily about their centre of mass.

At the instant the bar is cut, M2 will have a higher tangental velocity as:

v_{\perp} = r\frac{d\phi}{d\t} (where \phi is the angle in the plane of motion)

I don't think however that they'll spin about their mutual centre of mass as the spinning motion is intialised only buy the velocity gradient between M1 and M2. This is why I say the bar must be rigid to cause any kind of spinning motion, as if it were rope there would not be constant tension across it as there would in a spin. There would be come sort of ocillatory motion in their spinning.

I hope what I've said makes sense, they're my thoughts, I'm happy to be corrected.

Uhm...

I think the rope is just as good. The only difference (not considering needlessly complicating issues like rope linear elasticity) is that if the 2 masses were going towards each other, the rope wouldn't prevent it and the distance between M1 and M2 could become less than the original distance. But this is not the case here, M1 and M2 tend to have their distance increase after the break due to tangential speeds, and we can safely assume that the rope in this case guarantees the distance is therefore always the same.

And of course they spin around their new center of mass... what else? That would be true even if the distance wasn't fixed, so even if they were linked with a spring.

 

[neu]

Consider that the two masses are free when released, i.e. there's no rope linking them. Obviously they would travel parallel to each other with M2 having a greater velocity.

Now, back to being connected by the rope. M2 will want to carry on moving at v2 but will be resisted by M1 and be pulled back and diagonally. Hence M2 will slow, and M1 will be accelerated by the rope tension. As the rope is not rigid, there will be times when either M1 or M2 are near free objects as the tension in the rope will be negligable (I think), as the relative velocities of M1 & M2 oscillates. In this scenario a centre of mass would be a crude aproximation.

Now this may or may not be the case depending on the difference in velocity between M1 & M2. You can call this pedantry, but I think it would be a significant effect.

 

[Domenicaccio]

Consider that the two masses are free when released, i.e. there's no rope linking them. Obviously they would travel parallel to each other with M2 having a greater velocity.

Now, back to being connected by the rope. M2 will want to carry on moving at v2 but will be resisted by M1 and be pulled back and diagonally. Hence M2 will slow, and M1 will be accelerated by the rope tension. As the rope is not rigid, there will be times when either M1 or M2 are near free objects as the tension in the rope will be negligable (I think), as the relative velocities of M1 & M2 oscillates. In this scenario a centre of mass would be a crude aproximation.

Now this may or may not be the case depending on the difference in velocity between M1 & M2. You can call this pedantry, but I think it would be a significant effect.

IMHO, you are overcomplicating.

You can also figure out what is the angular momentum of the system M1+M2 relative to its own center of mass (not "CM") one instant before the rope is cut. Such momentum is not zero, and is conserved after the cut.