1. Physics 207: Lecture 14, Pg 1 Lecture 14Goals: Assignment: l l HW6 due Tuesday Oct. 25 th l l For Monday: Read Ch. 11 Chapter 10 Chapter 10   Understand the relationship between motion and energy   Define Kinetic Energy   Define Potential Energy   Define Mechanical Energy   Exploit Conservation of energy principle in problem solving   Understand Hooke’s Law spring potential energies  energy diagrams  Use energy diagrams

2. Physics 207: Lecture 14, Pg 2 Kinetic & Potential energies l Kinetic energy, K = ½ mv 2, is defined to be the large scale collective motion of one or a set of masses l Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy l Mechanical energy, E Mech, is defined to be the sum of U and K l Others forms of energy can be constructed

3. Physics 207: Lecture 14, Pg 3 Recall if a constant force over time then l l y(t) = y i + v yi t + ½ a y t 2 l l v(t) = v yi + a y t Eliminating t gives 2 a y ( y- y i ) = v x 2 - v yi 2 m a y ( y- y i ) = ½ m ( v x 2 - v yi 2 )

4. Physics 207: Lecture 14, Pg 4 Energy (dropping a ball) - mg  y final – y init ) = ½ m ( v y_final 2 –v y_init 2 ) Rearranging to give initial on the left and final on the right ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f We now define mgy ≡ U as the “gravitational potential energy” A relationship between y- displacement and change in the y-speed squared

5. Physics 207: Lecture 14, Pg 5 Energy (throwing a ball) l Notice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f Add ½ m v xi 2 + ½ m v zi 2 and ½ m v xf 2 + ½ m v zf 2 ½ m v i 2 + mgy i = ½ m v f 2 + mgy f where v i 2 ≡ v xi 2 +v yi 2 + v zi 2 ½ m v 2 ≡ K terms are defined to be kinetic energies (A scalar quantity of motion)

6. Physics 207: Lecture 14, Pg 6 When is mechanical energy not conserved l Mechanical energy is not conserved when there is a process which can be shown to transfer energy out of a system and that energy cannot be transferred back.

7. Physics 207: Lecture 14, Pg 7 Inelastic collision in 1-D: Example 1 l A block of mass M is 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 speed V.  What is the initial energy of the system ?  What is the final energy of the system ?  Is energy conserved? v V beforeafter x

8. Physics 207: Lecture 14, Pg 8 Inelastic collision in 1-D: Example 1 What is the momentum of the bullet with speed v ?  What is the initial energy of the system ?  What is the final energy of the system ?  Is momentum conserved (yes)?  Is energy conserved? Examine E before -E after v V beforeafter x No!

9. Physics 207: Lecture 14, Pg 9 Elastic vs. Inelastic Collisions l l A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision. l l How, if no net Force then momentum will be conserved. K before + U  K after + U   E.g. car crashes on ice: Collisions where objects stick together l A collision is said to be perfectly elastic when both energy & momentum are conserved before and after the collision. K before + U = K after + U  Carts colliding with a perfect spring, billiard balls, etc.

10. Physics 207: Lecture 14, Pg 10 Energy l If only “conservative” forces are present, then the mechanical energy of a system is conserved mechanical energy of a system is conserved For an object acted on by gravity K and U may change, K + U remains a fixed value. E mech = K + U = constant E mech is called “mechanical energy” ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f

11. Physics 207: Lecture 14, Pg 11 Example of a conservative system: The simple pendulum. l Suppose we release a mass m from rest a distance h 1 above its lowest possible point.  What is the maximum speed of the mass and where does this happen ?  To what height h 2 does it rise on the other side ? v h1h1 h2h2 m

12. Physics 207: Lecture 14, Pg 12 Example: The simple pendulum. y y= 0 y=h 1  What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum.

13. Physics 207: Lecture 14, Pg 13 Example: The simple pendulum. v h1h1 y y=h 1 y=0  What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing

14. Physics 207: Lecture 14, Pg 14 Example: The simple pendulum. y y=h 1 =h 2 y=0 To what height h 2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point. E = mgh 1 = mgh 2 or h 1 = h 2

15. Physics 207: Lecture 14, Pg 15 Potential Energy, Energy Transfer and Path l A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless 1. The ball is dropped 2. The ball slides down a straight incline 3. The ball slides down a curved incline After traveling a vertical distance h, how do the three speeds compare? (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell h 13 2

16. Physics 207: Lecture 14, Pg 16 Example The Loop-the-Loop … again l l To complete the loop the loop, how high do we have to let the release the car? l l Condition for completing the loop the loop: Circular motion at the top of the loop (a c = v 2 / R) l l Exploit the fact that E = U + K = constant ! (frictionless) (A) 2R (B) 3R(C) 5/2 R(D) 2 3/2 R h ? R Car has mass m Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case. Also recall the minimum speed at the top is U b =mgh U=mg2R y=0 U=0

17. Physics 207: Lecture 14, Pg 17 Example The Loop-the-Loop … again l l Use E = K + U = constant l l mgh + 0 = mg 2R + ½ mv 2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R R h ?

18. Physics 207: Lecture 14, Pg 18 Example – Fully Elastic Collision l Suppose I have 2 identical bumper cars. l One is motionless and the other is approaching it with velocity v 1. If they collide elastically, what is the final velocity of each car ? Identical means m 1 = m 2 = m Initially v Green = v 1 and v Red = 0 l l COM  mv 1 + 0 = mv 1f + mv 2f  v 1 = v 1f + v 2f l l COE  ½ mv 1 2 = ½ mv 1f 2 + ½ mv 2f 2  v 1 2 = v 1f 2 + v 2f 2 l l v 1 2 = (v 1f + v 2f ) 2 = v 1f 2 +2v 1f v 2f + v 2f 2  2 v 1f v 2f = 0 l l Soln 1: v 1f = 0 and v 2f = v 1 Soln 2: v 2f = 0 and v 1f = v 1

19. Physics 207: Lecture 14, Pg 19 Variable force devices: Hooke’s Law Springs l Springs are everywhere, l The magnitude of the force increases as the spring is further compressed (a displacement). l Hooke’s Law, F s = - k  s  s is the amount the spring is stretched or compressed from it resting position. F ss Rest or equilibrium position

20. Physics 207: Lecture 14, Pg 20 Hooke’s Law Spring l For a spring we know that F x = -k s. F(x) s2s2 s s1s1 -k s relaxed position F = - k s 1 F = - k s 2

21. Physics 207: Lecture 14, Pg 21 Exercise Hooke’s Law 50 kg 9 m 8 m What is the spring constant “k” ? (A) 50 N/m(B) 100 N/m(C) 400 N/m (D) 500 N/m

22. Physics 207: Lecture 14, Pg 22 F vs.  x relation for a foot arch: Force (N) Displacement (mm)

23. Physics 207: Lecture 14, Pg 23 Force vs. Energy for a Hooke’s Law spring l F = - k (x – x equilibrium ) l F = ma = m dv/dt = m (dv/dx dx/dt) = m dv/dx v = mv dv/dx l So - k (x – x equilibrium ) dx = mv dv l Let u = x – x eq. & du = dx  m

24. Physics 207: Lecture 14, Pg 24 Energy for a Hooke’s Law spring l Associate ½ ku 2 with the “potential energy” of the spring m l l Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

25. Physics 207: Lecture 14, Pg 25 Energy diagrams l In general: Energy K y U E mech Energy K u = x - x eq U E mech Spring/Mass system Ball falling 0 0

26. Physics 207: Lecture 14, Pg 26 Equilibrium l Example  Spring: F x = 0 => dU / dx = 0 for x=x eq The spring is in equilibrium position l In general: dU / dx = 0  for ANY function establishes equilibrium stable equilibrium unstable equilibrium U U

27. Physics 207: Lecture 14, Pg 27 Comment on Energy Conservation l We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.  Mechanical energy is lost:  Heat (friction)  Bending of metal and deformation l Kinetic energy is not conserved by these non-conservative forces occurring during the collision ! l Momentum along a specific direction is conserved when there are no external forces acting in this direction.  In general, easier to satisfy conservation of momentum than energy conservation.

28. Physics 207: Lecture 14, Pg 28 Comment on Energy Conservation l We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.  Mechanical energy is lost:  Heat (friction)  Deformation (bending of metal) l Mechanical energy is not conserved when non-conservative forces are present ! l Momentum along a specific direction is conserved when there are no external forces acting in this direction.  Conservation of momentum is a more general result than mechanical energy conservation.