Calculating omega as a function of time for a flywheel

[dannyR]

hiya all, I've done a experiment which was hanging a mass from a light string wrapped around the axis of a flywheel. The mass was released and the flywheel began to rotate.

during calculations I've found out it would be great to have ω as a function of time and I've been stuck about how to get this.

could I do a force diagram using F=ma?, but then I am unsure of the mass "m".
would I use the mass which is falling and add the moment of inertia of the flywheel or is this very wrong? :(.

or could I use energy stored such as

mgh=1/2Iω²+1/2mr²ω²+K

mgh, loss in potential energy of the falling mass

kinetic energy in the flywheel

kinetic energy in the falling mass

where K would be the frictional force i think it would be proportional to ωr

could i replace h the height the mass has fallen by using the F=ma bit i talked about above then replace h=1/2at² then solve for t or ω??


ive thought about this a lot and always been stopped by not knowing how to calculate something or use F=ma with moment of inertia stuff could someone please point me in the right direction?

Thanks a lot Danny

 

[jbsteele]

:)To answer your question, yes, you can use the equation F=ma to solve for ω as a function of time. You will need to use the equation of motion for the flywheel, which is given by: Iω(t) + mr2ω'(t) = mgh - K, where I is the moment of inertia, m is the mass of the hanging mass, r is the radius of the flywheel, g is the gravitational acceleration, h is the height of the hanging mass, and K is the friction force. You can then solve this equation for ω'(t) and integrate it to get ω(t).