Recent content by BlameTruth

Update: I figured that the angular acceleration of each of the cylinder would be equal, thus the angular velocity once the cylinder has fallen 8m would be equal in each of the cylinders. So, 2\omega R = v Then I can solve that for omega and plug that into: mgh = \frac{1}{2} mv^2 +...

But it also has rotational kinetic energy which takes away from the translational. So mgh cannot equal just the translational energy.

If you use the equation for rotational kinetic energy, the radius ends up getting canceled out. Which leads me to believe that the rotational kinetic is necessary.

Thank you! So would I be safe in saying that: mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega_1 ^2 Where omega sub 1 is the angular velocity of the falling cylinder? Because then I could just go ahead and say that v=r\omega_2^2 Where omega sub 2 is the angular velocity of the...

Homework Statement Two identical cylinders of mass M and radius R are mounted on frictionless axles as pictured above. Around each cylinder a massless thread is wound. The hanging thread on the left is wound around a third identical cylinder. A box of mass M is attached to the hanging thread on...