To parallel @Heopps answer:
Did it really happen?
In spectacular fashion!
In 1965 NASA launched a boilerplate Apollo command module on a Little Joe II rocket to test the Launch Escape System (LES), and got more of a test than they'd bargained for. Due to an erroneous installation of gyros the control vanes on the fins went to full deflection upon launch and caused the rocket to spin up, to the point that centrifugal force broke the motor mounts for the solid motors and the rocket came apart, well below the altitude where the LES was supposed to initiate. The video of that flight is very cool!
Despite the unintended flight profile, the LES got a successful test under a real abort situation.
Note that this example does not truly involve a spacecraft, since this test never reached, and wasn't intended to reach, the Karman line, and used only a boilerplate command module. @Heopps answer steers you to a true spacecraft's unfortunate demise.
EDIT 2018 July 18
To address the spinning femur, a real femur is a lot more complicated than a cylinder.
But first, let me say that for someone who's only dipped a toe (so far!) in the vast ocean of physics, you made a good try! Many would get brain-freeze as soon as they saw an equation, but you didn't. Good for you!
Regarding the cylinder calculation, remember that this object is spinning around its center of mass. If you assume all the mass is concentrated at the ends of the cylinder (which it's not, of course, but assume that for now), then two 125-gram masses are rotating with a radius of gyration of 25 cm. For the gyrating mass and radius of gyration to be 250 gm and 50 cm, the rod would have to be rotating around one end, not the center, and all the mass would have to be at the other end. Instead we have the two 125-gm masses rotating with a radius of gyration of 25 cm, so the equation becomes V = SQRT(4kN*0.25m/0.125kg) = 89.44 m/s, or ~3400 RPM. (This assumes the 4 kN is an ultimate tensile strength) That's actually faster than your calculation indicated! This equation calculates the force applied by only one of the 125-gm masses: the other will apply the same force, just in the opposite direction, so calculating only one of them is sufficient.
If you assume a true thin cylinder with the mass distributed uniformly along its length, the parts of the cylinder closer to the rotational axis are moving at a speed slower than the ends, so they produce less centrifugal force than if they were out at the ends. The net tension produced at the midpoint is half as much (not one third as much!) as the case where the mass is concentrated at the ends, so you could actually rotate faster, by a factor of √2.
Real femurs are somewhere in the middle between the uniform cylinder and the situation above where the mass is concentrated entirely at the ends, but there is a complication.
The complication is that, as you can see from the image, the femur isn't cylindrically symmetrical. It isn't even homogeneous. There's a significant chunk of mass (the head, according to Gray's Anatomy) extending to one side at the upper ("proximal") end. And the shaft is often not straight, though that's not shown here, and the bone isn't homogeneous, so local mass densities aren't constant. When rotating, the asymmetry means that in addition to purely tensile stress (which @TheNate correctly identified stress as the primary metric of interest) there is a bending moment applied to the shaft. At a given cross-section of the shaft (say, for instance, at the center of mass) this bending moment decreases the stress on one side and increases it on the other. Assuming no pre-existing cracks in the bone, the first break would occur where that increased stress occurs and quickly spread as a Griffith crack across the entire shaft. This would happen at a rotation rate slower than those calculated above.