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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 .

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problem 2 angular momentum
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
Solution 2 . . . Angular Momentum
  • Guesstimate the formula for the angular momentum?
  • Linear Momentum is mv
  • Angular Momentum is I 
conservation of angular momentum
Conservation of Angular Momentum
  • In the absence of any external torques, the angular momentum is conserved.
  • If = 0 then I11 = I2 2
problem 3 sarah hughes
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
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!
vector cross product
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
Torque as a vector Cross-Product

 = r x F

 = r F sin 

angular momentum
Angular Momentum

L = r x P

L =m v r sin 

dl dt
 = 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

problem 4 cross product
Problem 4 . . . cross product

Given: r = 2 i + 3 j and F = - i + 2 j

Calculate the torque

solution 4 cross product
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

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