Physics, 06.05.2020 04:59, giulianna41
Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center of mass with an angular velocity of 0.5 rad/s. One of the astronauts then pulls on the rope shortening the distance between the two astronauts to 4 m. What is the change in the rotational kinetic energy (in J) of this system? You may model each astronaut as a point particle.
Answers: 2
![[2\cdot (50\,kg)\cdot (3.5\,m)^{2}] \cdot (0.5\,\frac{rad}{s} ) = [2\cdot (50\,kg)\cdot (2\,m)^{2}] \cdot \omega](/tpl/images/0646/4821/b4e6b.png)
![\Delta K = \frac{1}{2}\cdot \left[2\cdot (50\,kg)\cdot (2\,m)^{2}\cdot \left(1.531\,\frac{rad}{s} \right)^{2}-2\cdot (50\,kg)\cdot (3.5\,m)^{2}\cdot \left(0.5\,\frac{rad}{s} \right)^{2}\right]](/tpl/images/0646/4821/196e7.png)
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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are...
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