PHY221 Ch15: // Axis Theorem and Torque Recall main points: Expression of the Kinetic energy of a rigid body in terms of Kcm and Icm Parallel Axis Theorem Torque and Cross Product Angular acceleration from the torque

PHY221 Ch15: // Axis Theorem and Torque Main Points Kinetic energy of a rigid body in terms of Kcm and Icm In the 3rd slide of Ch 13 we showed that the total kinetic energy of a system is equal to the energy of the Center of Mass (CM) plus the energy RELATIVE to the CM: vCM  CM In our case since we study RIGID bodies, any motion relative to the CM MUST be a rotation around an axis through the CM. And therefore, since : This is a general result that should be used when the work-energy theorem, or the conservation of mech. energy, is used for solving problems involving rigid bodies rotating . Note that any motion can be viewed as the CM motion PLUS a rotation AROUND an axis thru the CM

PHY221 Ch15: // Axis Theorem and Torque Main Points Optional: Direct proof of the preceding result: By definition of the center of mass (see CM chapter) the last term is zero. Note also that we replaced vi/cm by ri/cm since we are dealing with a rigid body. We obtain the fundamental result:

PHY221 Ch15: // Axis Theorem and Torque Main Points Parallel Axis Theorem via KO = Kcm+ 1/2 Icm w2 Using the result on the previous slides, we can arrive at the very useful parallel axis theorem:

PHY221 Ch15: // Axis Theorem and Torque Main Points Torque and Cross product: The torque is defined as: Right Hand Rule: In this course we only look at rotations around a axis of fixed direction, therefore, from studying the right hand rule for a second, we see that the only components of r and F that matter are the x and y components (see picture at right). In the problems you are given, the forces are always in the X-Y plane and the r should always be the vector r=(x,y) and thus  is the angle in the x-y plane. The magnitude of the torque around the z-axis is: x y F  r m O

PHY221 Ch15: // Axis Theorem and Torque Main Points Angular acceleration from the torque, using Newton’s 2nd law:

PHY221 Ch15: // Axis Theorem and Torque Assignment F r F r Take positive values for vectors directed into the page 1. rF 2. –rF 3. rF/2 4. 0 5.NPA CM w/O=? w/CM -w/CM NPA O q CM Parallel Axis theorem requires: Only // axes I/O=I/O’ +Md2 for any O and O’ // Axes and O’=CM // axes or O’=CM