Rolling+&+Angular+Momentum

Hoop or ring = mr² Sphere = (2/5) ** mr ² Rod (rotating around the center) = (1/12)mL ² Rod (rotating around the end) = (1/3)mL ²
 * Rotational Inertia**
 * Disk or cylinder = (1/2) ** mr ²   **



For rolling bodies without slipping: Initial Energy = Final Energy mgh = .5mv² + .5Iw²

For rolling bodies with slipping: Initial Energy = Final Energy mgh = .5mv²



**L = Iw**
or L = p x r = (mv) x r = (mvr)sin(theta) where L is angular momentum, I is rotational inertia, w is angular velocity, p is momentum, r is the moment arm, m is mass, v is linear velocity, and theta is the angle between moment arm and velocity.

Initial L = Final L
 * Conservation of Angular Momentum (in a closed system)**:

Therefore, decreasing the rotational inertia increases angular velocity, because angular momentum stays constant.

Also:


 * Net Torque = (dl)/(dt)**







2002 Mechanics Free Response Number 2 http://www.collegeboard.com/prod_downloads/ap/students/physics/physics_c_m_frq_02.pdf
 * Sample Problems**

2005 Mechanics Free Response Number 3 http://www.collegeboard.com/prod_downloads/ap/students/physics/ap05_frq_physics_c_mech.pdf

2006 Mechanics Free Response Number 3 http://www.collegeboard.com/prod_downloads/ap/students/physics/ap06_frq_physics_c_mech.pdf

http://en.wikipedia.org/wiki/Angular_momentum http://www.lightandmatter.com/html_books/2cl/ch05/ch05.html http://hyperphysics.phy-astr.gsu.edu/Hbase/amom.html
 * References** **and Links**