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Lecture 13

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Lecture 13

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  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