Lecture 12

1. Lecture 12 Goals: • Chapter 8: Solve 2D motion problems with friction • Chapter 9: Momentum & Impulse • Solve problems with 1D and 2D Collisions • Solve problems having an impulse (Force vs. time) Assignment: • HW5 due Tuesday 10/19 • For Monday: Read through Ch. 10.4

2. T T y vT q x mg mg The swing….a test axis of rotation at top of swing vT = 0 Fr = m 02 / r = 0 = T – mg cosq T = mg cosq T < mg at bottom of swing vT is max Fr = m ac = m vT2 / r = T - mg T = mg + m vT2 / r T > mg

3. Example, Circular Motion Forces with Friction (recall mar = m |vT |2 / rFf≤ms N ) • How fast can the race car go? (How fast can it round a corner with this radius of curvature?) mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2 r

4. Example • Only one force is in the horizontal direction: static friction x-dir: Fr = mar = -m |vT |2 / r = Fs= -ms N(at maximum) y-dir: ma = 0 = N – mg N = mg vT = (ms m g r / m )1/2 vT = (ms g r )1/2 = (0.5 x 10 x 80)1/2 vT = 20 m/s y N x Fs mg mcar= 1600 kg mS = 0.5 for tire/road r = 80 m g = 10 m/s2

5. Zero Gravity Ride A rider in a “0 gravity ride” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

6. Banked Curves In the previous car scenario, we drew the following free body diagram for a race car going around a curve on a flat track. n Ff mg What differs on a banked curve?

7. N mar mg Banked Curves Free Body Diagram for a banked curve. Use rotated x-y coordinates Resolve into components parallel and perpendicular to bank y x Ff q ( Note: For very small banking angles, one can approximate that Ff is parallel to mar. This is equivalent to the small angle approximation sin q = tan q, but very effective at pushing the car toward the center of the curve!!)

8. Navigating a hill Knight concept exercise: A car is rolling over the top of a hill at speed v. At this instant, • n > w. • n = w. • n < w. • We can’t tell about n without knowing v. At what speed does the car lose contact? This occurs when the normal force goes to zero or, equivalently, when all the weight is used to achieve circular motion. Fc = mg = m v2 /r  v = (gr)1/2½ (just like an object in orbit) Note this approach can also be used to estimate the maximum walking speed.

9. Locomotion of a biped: Top speed

10. How fast can a biped walk? What about weight? A heavier person of equal height and proportions can walk faster than a lighter person A lighter person of equal height and proportions can walk faster than a heavier person To first order, size doesn’t matter

11. How fast can a biped walk? What can we say about the walker’s acceleration if there is UCM (a smooth walker) ? Acceleration is radial ! So where does it, ar, come from? (i.e., what external forces are on the walker?) 1. Weight of walker, downwards 2. Friction with the ground, sideways

12. How fast can a biped walk? Given a model then what does the physics say? Choose a position with the simplest constraints. If his radial acceleration is greater than g then he is on the wrong planet! Fr = m ar = m v2 / r < mg Otherwise you will lose contact! ar = v2 / r  vmax = (gr)½ vmax ~ 3 m/s ! (So it pays to be tall and live on Jupiter) Olympic record pace over 20 km 4.2 m/s (Lateral motion about hips gives 2.3 m/s more)

13. Impulse and Momentum: A new perspectiveConservation Laws : Are there any relationships between mass and velocity that remain fixed in value?

14. Momentum Conservation • Momentum conservation (recasts Newton’s 2nd Law when net external F = 0) is an important principle (usually when forces act over a short time) • It is a vector expression so must consider Px, Py and Pz • if Fx (external) = 0 then Px is constant • if Fy (external) = 0 then Py is constant • if Fz (external) = 0 then Pz is constant

15. Inelastic collision in 1-D: Example • A block of massMis initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a final speed V. • In terms of m, M,andV : What is the momentum of the bullet with speed v ? x v V before after

16. v V x before after Inelastic collision in 1-D: Example What is the momentum of the bullet with speed v ? • Key question: Is x-momentum conserved ? P After P Before

17. Exercise Momentum is a Vector (!) quantity • Yes • No • Yes & No • Too little information given • A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction • In regards to the block landing in the cart is momentum conserved?

18. Exercise Momentum is a Vector (!) quantity • x-direction: No net force so Px is conserved. • y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) Py is not conserved (system is block and cart only) Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart What is the final velocity of the cart? 2 kg 5.0 m 30° 10 kg 7.5 m

19. j i Exercise Momentum is a Vector (!) quantity 1) ai = g sin 30° = 5 m/s2 2) d = 5 m / sin 30° = ½ aiDt2 10 m = 2.5 m/s2Dt2 2s = Dt v = aiDt= 10 m/s vx= v cos 30° = 8.7 m/s • x-direction: No net force so Px is conserved • y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it lands in the cart. Initial Final Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s V’x = 1.4 m/s N 5.0 m mg 30° 30° 7.5 m y x

20. Lecture 12 • Assignment: • HW5 due Tuesday 10/19 • Read through 10.4