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Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignment: l HW6 due Wednesday, Feb. 11 l For Thursday: Read all of Chapter 11 Chapter 10 Chapter 10 

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Presentation on theme: "Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignment: l HW6 due Wednesday, Feb. 11 l For Thursday: Read all of Chapter 11 Chapter 10 Chapter 10 "— Presentation transcript:

1 Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignment: l HW6 due Wednesday, Feb. 11 l For Thursday: Read all of Chapter 11 Chapter 10 Chapter 10  Understand the relationship between motion and energy  Define Potential Energy in a Hooke’s Law spring  Develop and exploit conservation of energy principle in problem solving Chapter 11 Chapter 11  Understand the relationship between force, displacement and work

2 Physics 207: Lecture 13, Pg 2 Energy - mg  y= ½ m (v y 2 - v y0 2 ) - mg  y f – y i ) = ½ m ( v yf 2 -v yi 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 as the “gravitational potential energy” A relationship between y-displacement and change in the y-speed

3 Physics 207: Lecture 13, Pg 3 Energy 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 terms are defined to be kinetic energies (A scalar quantity of motion)

4 Physics 207: Lecture 13, Pg 4 Energy l If only “conservative” forces are present, the total energy () of a system is conserved (sum of potential, U, and kinetic energies, K) of a system is conserved For an object in a gravitational “field” E mech = K + U l K and U may change, but E mech = K + U remains a fixed value. E mech = K + U = constant E mech is called “mechanical energy” K ≡ ½ mv 2 U ≡ mgy ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f

5 Physics 207: Lecture 13, Pg 5 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

6 Physics 207: Lecture 13, Pg 6 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.

7 Physics 207: Lecture 13, Pg 7 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

8 Physics 207: Lecture 13, Pg 8 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

9 Physics 207: Lecture 13, Pg 9 Example The Loop-the-Loop … again l To complete the loop the loop, how high do we have to let the release the car? l Condition for completing the loop the loop: Circular motion at the top of the loop (a c = v 2 / R) l Use fact that E = U + K = constant ! (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

10 Physics 207: Lecture 13, Pg 10 Example The Loop-the-Loop … again l Use E = K + U = constant l mgh + 0 = mg 2R + ½ mv 2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R R h ?

11 Physics 207: Lecture 13, Pg 11 l What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? l Assume we can treat the skateboarder as a “point” l Assume zero of gravitational U is at bottom of the hill R=10 m.. m = 25 kg Example Skateboard.. R=10 m 30 ° y=0

12 Physics 207: Lecture 13, Pg 12 l What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? l Assume we can treat the skateboarder as “point” l Assume zero of gravitational U is at bottom of the hill R=10 m.. m = 25 kg Example Skateboard.. R=10 m 30 ° l Use E = K + U = constant E before = E after 0 + m g R = ½ mv 2 + mgR (1-sin 30 ° ) mgR/2 = ½ mv 2 gR = v 2  v= (gR) ½ v = (10 x 10) ½ = 10 m/s

13 Physics 207: Lecture 13, Pg 13 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

14 Physics 207: Lecture 13, Pg 14 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

15 Physics 207: Lecture 13, Pg 15 l What is the normal force on the skate boarder? R=10 m.. m = 25 kg Example Skateboard.. R=10 m 30 °.. N mg 60 °

16 Physics 207: Lecture 13, Pg 16 l Now what is the normal force on the skate boarder? R=10 m.. m = 25 kg Example Skateboard.. R=10 m 30 °  F r = ma r = m v 2 / R = N – mg cos 60 ° N = m v 2 /R + mg cos 60 ° N = / (0.87) N = =470 Newtons.. N mg 60 °

17 Physics 207: Lecture 13, Pg 17 Elastic vs. Inelastic Collisions l A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. K before = K after  Carts colliding with a perfect spring, billiard balls, etc. vvivvi

18 Physics 207: Lecture 13, Pg 18 Elastic vs. Inelastic Collisions l A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. K before  K after  Car crashes, collisions where objects stick together, etc.

19 Physics 207: Lecture 13, Pg 19 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

20 Physics 207: Lecture 13, Pg 20 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!

21 Physics 207: Lecture 13, Pg 21 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 COM  mv = mv 1f + mv 2f  v 1 = v 1f + v 2f l COE  ½ mv 1 2 = ½ mv 1f 2 + ½ mv 2f 2  v 1 2 = v 1f 2 + v 2f 2 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 Soln 1: v 1f = 0 and v 2f = v 1 Soln 2: v 2f = 0 and v 1f = v 1

22 Physics 207: Lecture 13, Pg 22 Variable force devices: Hooke’s Law Springs l Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) l In this spring, 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

23 Physics 207: Lecture 13, Pg 23 Exercise 2 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

24 Physics 207: Lecture 13, Pg 24 Exercise 2 Hooke’s Law 50 kg 9 m 8 m What is the spring constant “k” ?  F = 0 = F s – mg = k  s - mg Use k = mg/  s = 500 N / 1.0 m (A) 50 N/m(B) 100 N/m(C) 400 N/m (D) 500 N/m mg F spring

25 Physics 207: Lecture 13, Pg 25 F-s relation for a foot arch: Force (N) Displacement (mm)

26 Physics 207: Lecture 13, Pg 26 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

27 Physics 207: Lecture 13, Pg 27 Energy for a Hooke’s Law spring l Associate ½ kx 2 with the “potential energy” of the spring m l Hooke’s Law springs are conservative so the mechanical energy is constant

28 Physics 207: Lecture 13, Pg 28 Energy diagrams l In general: Energy K y U E mech Energy K s U E mech Spring/Mass system Ball falling

29 Physics 207: Lecture 13, Pg 29 Energy diagrams Spring/Mass/Gravity system Energy K K y UgUg E mech U spring U Total m Force y -mg spring & gravity spring alone

30 Physics 207: Lecture 13, Pg 30 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

31 Physics 207: Lecture 13, Pg 31 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.

32 Physics 207: Lecture 13, Pg 32 Lecture 13 Assignment: l HW6 due Wednesday 2/11 l For Monday: Read all of chapter 11


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