Site hosted by Angelfire.com: Build your free website today!
Blog Tools
Edit your Blog
Build a Blog
RSS Feed
View Profile
Open Community
Post to this Blog
« October 2007 »
S M T W T F S
1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31
Entries by Topic
All topics  «
You are not logged in. Log in
Physics for FSU
Tuesday, 16 October 2007
Flywheel pt 1

for the first part,

rotational kinetic energy=.5*moment of inertia*(omega^2)

moment of inertia = .5mr^2 because u r dealing with a solid cylinder.

therefore, KE=.5*(.5mr^2)*(omega^2) 


Posted by physicsforfsu at 10:06 PM EDT
Post Comment | Permalink | Share This Post

View Latest Entries