Physics 221 Chapter 11

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

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

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

All about Sarah Hughes... Click me!

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

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

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

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

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

 = 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 0 0 = 0 dL/dt = r x F  = dL/dt

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

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  = k + 3 k +0  = 7 k