1. Torque

2. Correlation between Linear Momentum and Angular Momentum Resistance to change in motion: Linear  behavior: Inertia Units M (mass), in kg Angular  Rotational Inerita, Symbol I aka “Moment of Inertia” Has to do with more than just mass, it includes the shape of the object. If x cm is far away, I is bigger. Based on mass and mass distribution.

3. Three Important equations for I : Thin Walled Cylinder I = MR 2 (also valid for an orbiting body) Solid Cylinder I = (1/2)MR 2 (easier because the mass is not so far away) Sphere I = (2/5)MR 2 (marble)

5. Cause of acceleration Linear  Force (Newtons) Angular  Torque ( Nm)

6. Momentum Linear  p = mv ( units kgm/s) Angular  L = I ω (units kgm 2 /s) Bigger mass = harder to stop Faster moving = harder to stop

7. Kinetic Energy Linear  K = (1/2) mv 2 (Units Joules) Angular  K R = (1/2) I ω 2 A rolling ball has translational and rotational Kinetic energy. Remember: m’s become I’s, v’s become ω’s

8. Pivoting Object Take a shape and a piece of paper. Label an x-y axis on the paper. Label “pivot point” on the shape. Place at origin. Label x cm on shape. Draw weight vector. Draw position vector r from pivot to x cm. Move position vector from origin to x cm.

9. Definition: torque Torque is the cause of angular acceleration in the same way that force is the cause of linear acceleration. Symbol τ (pronounced tao) Τ is the cross product between r and F We say τ = r cross F = r x F

10. Relationship between torque and Newton’s second law F = ma rF = rma a is tangential, so a =rα rF = torque, τ = rmrα = mr 2 α = I α ( I=Σm 1 r 1 2 = m 2 r 2 2 …) So Σ τ = I α (α must be in rad/s 2 )

11. Back to picture Draw the angle between the FIRST r vector that starts at the pivot point and the F vector. Then draw the angle between the moved r vector and F. The moved vector is 180-θ. The old r vector was at θ. τ = rF g sin(180-θ) Mathematically, sin(180-θ) = sin θ So τ = rF g sin(θ) in magnitude. Use RHR for direction.

12. On picture, Split weight vector out into components mgcosθ points toward pivot, mgsinθ points perpendicular to make it rotate. mgcosθ gets cancelled out, that’s why we use the sin component, τ = rFsin(θ) That component in this case is spinning it cw, or in the negative direction.

13. Lever arm Draw a line from pivot point across to weight vector so that is crosses weight vector at 90 degrees. That line has a magnitude of rsinθ This is very important. It has a special definition. Its called a lever arm (symbol l ) We can say: τ = l F

14. Cross Product Guided Practice #1 P = 4i + 2j –k Q = -3i + 6j -2k P x Q = Answer: 2i + 11j + 30k

15. Guided Practice #2 Cross Products P = 2.12i + 8.15j – 4.28k N and Q = 2.29i -8.93j – 10.5k m Find P x Q Answer: -124i +12.5j – 37.6k

17. Solution 1) ω O = 0.17 rad/s (v/r) 2) I = (2/5)MR 2 = 0.0144 kgm 2 3) α = 1735 rad/s 4) Δθ =  use quadratic to find time 5) t = 4.25 s

19. Solution Find I  I = ½ MR 2 =.625 kgm 2 Find α  α = 16 rad/s Find Δθ =  Δθ = 800 rad = 127 rev

46. end https://www.youtube.com/watch?v=GLlpi- 0_lB0 https://www.youtube.com/watch?v=GLlpi- 0_lB0