I'm still getting the wrong answer. The key says its 12 rev/min but I'm getting 8.21 rev/minute.
Here's my work:
Conservation of angular momentum (of children and platform) gives:
I\omega_{}1+m\omega_{}1r_{}1 = I\omega_{}2+m\omega_{}2r_{}2
Solve for \omega_{}2,
\omega_{}2=...
Thank you tiny-tim for your help. I've just figured out how to use the Latex symbols, that should make my equations easier to read. The other Latex stuff is a bit more complicated, don't know if I'll need it.
A stepped pulley is essentially a system of connected cocentric disks that have different radii. A stepped pulley (I= 2.0 kg*m^2) is placed on a frictionless axle and then strings are attached to two radii of 20 and 50 cm. The other ends of the strings are attached to a 2.0 and 1.0 kg mass...
Oh wow now I feel stupid. One of those "oh...duh" moments. Anyway thank you very much for your help.
One thing though, is my conversion from 6.0 rev/min to 0.628 rad/sec correct? I tried using the latter in the conservation of angular momentum equation but I ended up getting a slower second...
Four children stand at the edge of a circular horizontal platform that is free to rotate about a vertical axis. Each child has a mass of 35 kg and they are at positions that are a quarter-circle from each other. The platform has a moment of inertia equal to 500 kg*m^2 and a radius of 2.0 m. The...