1 / 30

Lecture 13

Lecture 13. Goals:. 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 Understand the relationship between force, displacement and work.

svea
Download Presentation

Lecture 13

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 13 Goals: • 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 • Understand the relationship between force, displacement and work Assignment: • HW6 due Wednesday, Feb. 11 • For Thursday: Read all of Chapter 11

  2. Energy -mgDy= ½ m (vy2 - vy02) -mg(yf – yi) = ½ m ( vyf2 -vyi2 ) A relationship between y-displacement and change in the y-speed Rearranging to give initial on the left and final on the right ½ m vyi2+ mgyi = ½ m vyf2 + mgyf We now definemgyas the “gravitational potential energy”

  3. Energy • Notice that if we only consider gravity as the external force then the x and z velocities remain constant • To ½ m vyi2+ mgyi = ½ m vyf2 + mgyf • Add ½ m vxi2 + ½ m vzi2and ½ m vxf2 + ½ m vzf2 ½ m vi2+ mgyi = ½ m vf2 + mgyf • where vi2 = vxi2 +vyi2 + vzi2 ½ m v2terms are defined to be kinetic energies (A scalar quantity of motion)

  4. Energy • If only “conservative” forces are present, the total energy (sum of potential, U, and kinetic energies, K) of a systemis conserved For an object in a gravitational “field” ½ m vyi2+ mgyi = ½ m vyf2 + mgyf U ≡ mgy K ≡ ½ mv2 Emech = K + U Emech = K + U= constant • K and U may change, but Emech = K + Uremains a fixed value. Emech is called “mechanical energy”

  5. m h1 h2 v Example of a conservative system: The simple pendulum. • Suppose we release a mass m from rest a distance h1 above its lowest possible point. • What is the maximum speed of the mass and where does this happen ? • To what height h2 does it rise on the other side ?

  6. Example: The simple pendulum. • 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. y y=h1 y=0

  7. Example: The simple pendulum. • 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 = mgh1at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 h1 y=0 v

  8. Example: The simple pendulum. To what height h2 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 = mgh1 = mgh2 or h1 = h2 y y=h1=h2 y=0

  9. Ub=mgh 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 Car has mass m U=mg2R h ? R ExampleThe Loop-the-Loop … again • To complete the loop the loop, how high do we have to let the release the car? • Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) • Use fact thatE = U + K = constant ! (A) 2R (B) 3R (C) 5/2 R (D) 23/2 R y=0

  10. ExampleThe Loop-the-Loop … again • Use E = K + U = constant • mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R h ? R

  11. . . . . ExampleSkateboard • What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? • Assume we can treat the skateboarder as a “point” • Assume zero of gravitational U is at bottom of the hill m = 25 kg R=10 m 30° R=10 m y=0

  12. . . . . ExampleSkateboard • What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest? • Assume we can treat the skateboarder as “point” • Assume zero of gravitational U is at bottom of the hill m = 25 kg • Use E = K + U = constant Ebefore = Eafter 0 + m g R = ½ mv2 + mgR (1-sin 30°) mgR/2 = ½ mv2 gR = v2 v= (gR)½ v = (10 x 10)½ = 10 m/s R=10 m 30° R=10 m

  13. 1 3 2 h Potential Energy, Energy Transfer and Path • A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless • The ball is dropped • The ball slides down a straight incline • 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

  14. 1 3 2 h Potential Energy, Energy Transfer and Path • A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless • The ball is dropped • The ball slides down a straight incline • 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

  15. . . . . . . ExampleSkateboard • What is the normal force on the skate boarder? N m = 25 kg 60° mg R=10 m 30° R=10 m

  16. . . . . . . ExampleSkateboard • Now what is the normal force on the skate boarder? N m = 25 kg 60° mg R=10 m • S Fr = mar = m v2 / R = N – mg cos 60° N = m v2 /R + mg cos 60° N = 25 100 / 10 + 25 10 (0.87) N = 250 + 220 =470 Newtons 30° R=10 m

  17. vi Elastic vs. Inelastic Collisions • A collision is said to be elastic when energy as well as momentum is conserved before and after the collision. Kbefore = Kafter • Carts colliding with a perfect spring, billiard balls, etc.

  18. A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter • Car crashes, collisions where objects stick together, etc. Elastic vs. Inelastic Collisions

  19. Inelastic collision in 1-D: Example 1 • 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 speed V. • What is the initial energy of the system ? • What is the final energy of the system ? • Is energy conserved? x v V before after

  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 Ebefore-Eafter v No! V x before after

  21. Variable force devices: Hooke’s Law Springs • Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms) • In this spring, the magnitude of the force increases as the spring is further compressed (a displacement). • Hooke’s Law, Fs = - kDs Ds is the amount the spring is stretched or compressed from it resting position. Rest or equilibrium position F Ds

  22. 8 m 9 m Whatis the spring constant “k” ? 50 kg Exercise 2Hooke’s Law (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m

  23. Exercise 2Hooke’s Law 8 m 9 m Whatis the spring constant “k” ? SF = 0 = Fs – mg = k Ds - mg Use k = mg/Ds = 500 N / 1.0 m Fspring 50 kg (A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg

  24. F-s relation for a foot arch: Force (N) Displacement (mm)

  25. m Force vs. Energy for a Hooke’s Law spring • F = - k (x – xequilibrium) • F = ma = m dv/dt = m (dv/dx dx/dt) = m dv/dx v = mv dv/dx • So - k (x – xequilibrium) dx = mv dv • Let u = x – xeq. & du = dx 

  26. m Energy for a Hooke’s Law spring • Associate ½ kx2 with the “potential energy” of the spring • Hooke’s Law springs are conservative so the mechanical energy is constant

  27. Emech K Energy U y Energy diagrams • In general: Ball falling Spring/Mass system Emech K Energy U s

  28. U U Equilibrium • Example • Spring: Fx = 0 => dU / dx = 0 for x=xeq The spring is in equilibrium position • In general: dU / dx = 0 for ANY function establishes equilibrium stable equilibrium unstable equilibrium

  29. Comment on Energy Conservation • 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 • Kinetic energy is not conserved by these non-conservative forces occurring during the collision ! • Momentum along a specific directionis conserved when there are no external forces acting in this direction. • In general, easier to satisfy conservation of momentum than energy conservation.

  30. Lecture 13 Assignment: • HW6 due Wednesday 2/11 • For Monday: Read all of chapter 11

More Related