What is the tension in the string?

Question

Two pucks m1=m2=1kg connected by a string L=1m slide on frictionless ice. Initial velocites v1=2v2=2m/s in direction perpendicular to the string.

O --------2m/s-------->
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1m
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O---1m/s--->

Answer 1

They do, of course, rotate about their common center of mass as the 2nd answer points out, but the velocities were wrong. The transformation is to the common CM moving to the right at 1.5 m/s. The relative velocities are
v1' = 0.5 m/s
v2' = -0.5 m/s
fc1 = fc2 = m * v^2 / r = 1 * 0.5^2 / 0.5 = 0.5 N.

Answer 2

I have to disagree with Obiwan's solution: He is assuming that you can pretend that the slower object is at rest, and the faster is circling around it. This is not true, as both are moving.

Here's how to do it right:

Do a Galilean frame transformation, to the center-of-mass frame. In that frame
v1' = + 1 m/s
v2' = - 1 m/s

It is then clear that both m1 and m2 are circling about the point in the middle, with radius = L/2 = 0.5 m. Each one of them requires centripetal force:
mv^2/radius = (1 kg)(1 m/s)^2/(0.5 m)
= 2 N

So the string tension must be 2 N.
(It doesn't have to be doubled.)

That's the answer in the center-of-mass frame. But force is an invariant under Galilean transformation (the pre-Einsteinian version of relativity), so the answer is the same in the original fame. Hence, T = 2 N.

Answer 3

The mass moving at the higher velocity will rotate about the mass moving at the lower velocity. We can consider the slower one to be at rest, and the faster one moving about it in a circle. Then,the relative velocity will be 2 m/s - 1 m/s = 1 m/s. It will be at a distance, or radius, of 1 m, and its centripetal accelertion, by definition, is v^2/r. So the centripetal force, or the tension on the string, is mv^2/r.
Substituting, m=1, v=1, r=1, we get the force is 1 kg m/sec^2, and that is by definition, 1 newton.
So the tension in the string is 1 newton.