Physics 221 Chapter 11

1. Physics 221Chapter 11

2. Problem 2 . . . Angular Momentum • Guesstimate the formula for the angular momentum? • A. mv • B. m • C. I  • D. 1/2 I 

3. Solution 2 . . . Angular Momentum • Guesstimate the formula for the angular momentum? • Linear Momentum is mv • Angular Momentum is I 

4. Conservation of Angular Momentum • In the absence of any external torques, the angular momentum is conserved. • If = 0 then I11 = I2 2

5. All about Sarah Hughes . . . Click me!

6. Problem 3 . . . Sarah Hughes • A. When her arms stretch out her moment of inertia decreases and her angular velocity increases • B. When her arms stretch out her moment of inertia increases and her angular velocity decreases • C. When her arms stretch out her moment of inertia decreases and her angular velocity decreases • D. When her arms stretch out her moment of inertia increases and her angular velocity increases

7. Solution 3 . . . Sarah Hughes • B. When her arms stretch out her moment of inertia increases and her angular velocity decreases • I11 = I2 2 • So when I increases,  decreases!

8. Vector Cross-Product • A X B is a vector whose: • magnitude = |A| |B| sin  • direction = perpendicular to both A and B given by the right-hand rule. • Right-hand rule: Curl the fingers of the right hand going from A to B. The thumb will point in the direction of A X B

9. Torque as a vector Cross-Product  = r x F  = r F sin 

10. Angular Momentum L = r x P L =m v r sin 

11.  = dL/dt Proof L = r x p dL/dt = d/dt(r x p) dL/dt = dr/dt x p + r x dp/dt dL/dt = v x p + r x F But v x p = 0 because p = mv and so v and p are parallel and sin 00 = 0 dL/dt = r x F  =dL/dt

12. Problem 4 . . . cross product Given: r = 2 i + 3 j and F = - i + 2 j Calculate the torque

13. Solution 4 . . . cross product  = r x F  = (2 i + 3 j) x (- i + 2 j)  = - 2 i x i + 4 i x j - 3 j x i + 6 j x j  = 0 + 4 k + 3 k +0  = 7 k