Physics 221Chapter 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 I11 = I2 2

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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

• I11 = I2 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 pdL/dt = d/dt(r x p)dL/dt = dr/dt x p + r x dp/dtdL/dt = v x p + r x FBut 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

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

= 0 + 4 k + 3 k +0

= 7 k