1. 11/1 Rotation, Torque  Begin Rotational Dynamics Read Ch 9 sec 1-3  Suggested problems from Ch 7 22, 35, 36, 37, 53, 55  HW: 11/1 HW Handout “Extended FBD and Torque” due Wed 11/6 (Will post soon)  Exam 3: Thursday, 11/14

2. Rotation vectors: right hand rule clockwise, counterclockwise into page = clockwise Thumb gives direction, only two choices, always parallel to the axis. out of page = counterclockwise

3.  Equilibrium:  Net Force = 0 as usual  Net Torque = 0 also  Torque is a twisting force  only two possible directions CW and CCW  Torque CW = Torque CCW for equilibrium  Torque = Force X Lever Arm (more later) Torque and the 2nd Law

4. Predict the tension in the strings: 0.5m 1.0 m M bar = 1.0 kg 1.0 kg T = 10 N in both Equilibrium sec. 9.2

5. Now predict the tension in the strings: 0.8m 1.0 m M bar = 1.0 kg 1.0 kg T2T2 T1T1

6. Free Body Diagram 0.8m 1.0 m F net = ma = 0 gives T 1 + T 2 = 20 N but its not enough info. T 2,bar T 1,bar 0.5m T s,bar 10 N W E,bar 10 N Weight of the bar at the its center of mass.

7. Free Body Diagram Apply  net = I ,  = 0 0.8m 1.0 m First pick an axis, any axis!!! T 2,bar T 1,bar 0.5m T s,bar 10 N W E.bar 10 N  net = 10(0.2) + 10(0.5) - T 1 (1) = 0 T 1 = 7N T 2 +7 = 10 + 10, T 2 = 13N (F net = 0) Axis pos out of page

8. Free Body Diagram Apply  net = I ,  = 0 0.8m 1.0 m First pick an axis, any axis!!! T 2,bar T 1,bar 0.5m T s,bar 10 N W E.bar 10 N  net = -10(0.5) - 10(0.8) + T 2 (1) = 0 T 2 = 13 N T 1 + 13 = 10 + 10, T 1 = 7N (F net = 0) Axis pos out of page

9. CM in Lab Where is the center of mass of the bar? O D F1F1 F2F2 F3F3 F 4 = m bar g but where is it applied?  net = I   = 0  net = 0 equilibrium

10. Sample problem: The mass of the meter stick is 1 kg. Find the mass of the other block. 2 kgm = ? 1 kg meter stick