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Lecture 23

Lecture 23. Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room CH Sections 604 605 606 609 610 611 (TA: Moriah T.,  Abdallah C., Tamara S.) VV Sections 602 603 607 608 612 (TA: Eric P., Dan C., Zhe D.) Format: Closed book, three 8 x11” sheets, hand written

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Lecture 23

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  1. Lecture 23 • Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room • CH Sections 604 605 606 609 610 611 (TA: Moriah T.,  Abdallah C., Tamara S.) • VV Sections 602 603 607 608 612 (TA: Eric P., Dan C., Zhe D.) • Format: Closed book, three 8 x11” sheets, hand written • Electronics: Any calculator is okay but no web/cell access • Quiet room: Test anxiety, special accommodations, etc. • Chapters Covered • Chapter 9: Linear momentum and collision (not 9.9) • Chapter 10: Rotation about fixed axis and rolling • Chapter 11: Angular Momentum (not 11.5) • Chapter 12: Static equilibrium and elasticity

  2. Basic Concepts and Quantities • Momentum, Angular Momentum • Linear Momentum, Impulse • Angular Momentum (magnitude and direction) • Torque • Collisions: Elastic & Inelastic • Center of Mass • Rotational Motion (1-axis) • Angular displacement (Δq) / Velocity(w), Acceleration (a). • Moments of Inertia • Rotational Kinetic Energy • Conservation Laws: Energy, Momentum, Angular Momentum • Static Equilibrium • Elasticity: Young’s, Shear, Bulk Modulus

  3. Chapter 9 : Momentum and Momentum Conservation Ix = Ix

  4. Chapter 9

  5. Chapter 9 • Center of Mass

  6. Chapter 10 v = w R s R    At a point a distance R away from the axis of rotation, the tangential motion: • s (arc) =  R • vT (tangential) = R • aT =  R

  7. Chapter 10

  8. Chapter 10

  9. Chapter 11

  10. Chapter 12

  11. Approach to Statics: • In general, we can use the two equations to solve any statics problems. • When choosing axes about which to calculate torque, choose one that makes the problem easy.... • However if there is acceleration, then restrict rotation axis to center of mass (as well as for translation).

  12. Changes in length: Young’s modulus • Young’s modulus: measures the resistance of a solid to a change in its length. F elasticity in length A L0 L

  13. Changes in volume: Bulk Modulus V0 F Vi + V • Bulk modulus: measures the resistance of solids or liquids to changes in their volume. Volume elasticity

  14. Changes in shape: Shear Modulus Applying a force perpendicular to a surface

  15. ComparisonKinematics Angular Linear

  16. Comparison: Dynamics Angular Linear m I = Si mi ri2 F = a m t = r x F = a I L = r x t = I w p = mv W = F •x W =  D K = WNET K = WNET

  17. Momentum and collisions • Remember vector components • A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline. • How far along the incline do the joined blocks slide?

  18. Momentum and collisions • Remember vector components • A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a q=30° frictionless incline. • Momentum parallel to incline is conserved • Normal force (by ground on cart) is  to the incline mvi cos q + m 0 = 2m vf  vf = vi cos q / 2 = 4.4 m/s Now use work-energy 2mgh + 0 = ½ 2m v2f d = h / sin q = v2f / (2g sin q) mvi cos q mvi

  19. Example problem: Going in circles B ri wi • A 2.0 kg disk tied to a long string undergoes circular motion on a frictionless horizontal table top. The string passes through a hole and then hangs vertically. The disk starts out at 5.0 rev/sec 0.50 m away from hole. If you pull slowly down on the string so that the final radius is 0.25 m, what is the final angular velocity? • No external torque so angular momentum is conserved. • Ibwb = Iawa • I = mR2 • (0.50)2wb = (0.25)2wa • 4 wb = wa = 20 rev/sec

  20. Spinning ball on incline • A solid disk of mass m and radius R is spinning with angular velocity w. It is positioned so that it can either move directly up or down an incline of angle q (but it is not rolling motion). The coefficient of kinetic friction is m. At what angle will the disk’s position on the incline not change? w q

  21. Spinning ball on incline N f w q mg • A solid disk of mass m and radius R is spinning with angular velocity w. It is positioned so that it can either move directly up or down an incline of angle q (but it is not rolling motion). The coefficient of kinetic friction is m. • At what angle will the disk’s position on the incline not change?

  22. Spinning ball on incline N f w q mg • A solid disk of mass m and radius r is spinning with angular velocity w. It is positioned so that it can either move directly up or down an incline of angle q (but it is not rolling motion). The coefficient of kinetic friction is m. While spinning the disk’s position will not change. • How long will it be before it starts to roll? • This will occur only when w=0.

  23. Example Problem • A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown. • How fast is the bullet going when it leaves? • What is the tension in the rod just after the bullet exits? Method: Impulse I = area under curve

  24. Example Problem • A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown. • How fast is the bullet going when it leaves?

  25. Example Problem • A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown. • What is the tension in the rod just after the bullet exits?

  26. Statics: Example A sign of mass M is hung 1 m from the end of a 4 m long beam (mass m) as shown in the diagram. The beam is hinged at the wall. What is the tension in the wire in terms of m, M, g and any other given quantity? wire q = 30o 1 m SIGN

  27. Statics: Example T Fy 30° X Fx 2 m mg Mg 3 m Process: Make a FBD and note known / unknown forces. Chose axis of rotation at support because Fx & Fy are not known • F = 0 0 = Fx – T cos 30° 0 = Fy + T sin 30° - mg - Mg z-dir: Stz = 0 = -mg 2r – Mg 3r + T sin 30° 4r (r = 1m) The torque equation get us where we need to go, T T = (2m + 3M) g / 2

  28. Center of Mass Example: Astronauts & Rope • Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope. (1) Does he/she get back to the ship ? (2) Does he/she meet the other astronaut ? M = 1.5m m

  29. Example: Astronauts & Rope m M = 1.5m • There is no external force so if the larger astronaut pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it.

  30. Example: Astronauts & Rope m M = 1.5m (2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!) In all cases the center of mass will remain fixed because they are initially at rest and there is no external force. To find the position where they meet all we need do is find the Center of Mass

  31. Forces and rigid body rotation • To change the angular velocity of a rotating object, a force must be applied • How effective an applied force is at changing the rotation depends on several factors • The magnitude of the force • Where, relative to the axis of rotation the force is applied • The direction of the force A B C Which applied force will cause the wheel to spin the fastest?

  32. Leverage • The same concept applies to leverage • the lever undergoes rigid body rotation about a pivot point: B C A Which applied force provides the greatest lift ?

  33. More on torques • You need to change the tire on you car. You use a tire wrench which allows you to apply a pair of forces. • (A) What is the torque produced by a tire wrench of length L, given an applied couple of magnitude F, acting on a lug nut (point F) as shown in the figure? • (B) Assume the lug nut is stuck What is the torque acting on the wheel, if the lug nut is a distance r from the center? Image courtesy John Wiley & Sons, Inc.

  34. Wheel wrench • 1. tF = (L/2) F + (L/2) F = LF • 2. tF = r F sinf + r F sin ( p-q )= = LF • 3. tF = L F + 0 F • Notice the torque is the same everywhere. f f

  35. For Thursday • Chapter 13 (Newton’s Law of Gravitation)

  36. Momentum & Impulse • 0 • ½ v • 2 v • 4 v • A rubber ball collides head on (i.e., velocities are opposite) with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed immediately after the collision?

  37. Momentum & Impulse • A rubber ball collides head on with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed after the collision? (a) 0 (b) ½ v (c) 2 v (d) 4 v

  38. Momentum, Work and Energy • A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is 0.80 m. (See the diagram.) • To what maximum height above the surface will the ball/block assembly rise after the collision? (g=9.8 m/s2) A. 2.2 cm B. 4.4 cm C. 11. cm D. 22 cm E. 44 cm F. 55 cm

  39. Momentum, Work and Energy • A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is .80 m. (See the diagram.) • To what maximum height above the surface will the ball/block assembly rise after the collision? A. 2.2 cm B. 4.4 cm C. 11. cm D. 22 cm E. 44 cm F. 55 cm

  40. Momentum, Work and Energy ( Now with friction) • A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface. • If mstatic = 0.54, how far can the spring be compressed and the block remain stationary (i.e., maximum static friction)? • S F = 0 = k u - f = k u - m N u = m mg/k = 0.54 (0.40x10 N) / 270 N/m= 0.0080 m

  41. Momentum, Work and Energy ( Now with friction) • A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface. The spring is compressed 0.20 m • If mkinetic = 0.50 and the block is 9.8 m away from the unstretched spring, how high with the clay/block pair rise? • Emech (at collision) = Uspring + Wfriction = ½ k u2 - m mg d • 1/2 m v2 = 135(0.04)-0.50(0.40x10.)10.=(540-20) J=520 J • v2 = 1040/0.40 m2/s2 • Now the collision (cons. of momentum) and the swing.

  42. Momentum and Impulse • Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)? (a) 121 (b) 286 (c) 490 (d) 623 (e) 767

  43. Momentum and Impulse • Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)? (a) 121 (b) 286 (c) 490 (d) 623 (e) 767 Dp = sqrt(2x9.8x12.3)x50

  44. Momentum & Impulse • Suppose that in the previous problem, the positively charged particle is a proton and the negatively charged particle, an electron. (The mass of a proton is approximately 1,840 times the mass of an electron.) Suppose that they are released from rest simultaneously. If, after a certain time, the change in momentum of the proton is Dp, what is the magnitude of the change in momentum of the electron? (a) Dp / 1840 (b) Dp (c) 1840 Dp

  45. Momentum & Impulse • Suppose that in the previous problem, the positively charged particle is a proton and the negatively charged particle, an electron. (The mass of a proton is approximately 1,840 times the mass of an electron.) Suppose that they are released from rest simultaneously. If, after a certain time, the change in momentum of the proton is Dp, what is the magnitude of the change in momentum of the electron? (a) Dp / 1840 (b) Dp (c) 1840 Dp

  46. Conservation of Momentum • A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision? (a) 10.8 m/s (b) 11.2 m/s (c) 12.4 m/s (d) 12.8 m/s

  47. Conservation of Momentum • A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision? (a) 10.8 m/s (b) 11.2 m/s (c) 12.4 m/s (d) 12.8 m/s

  48. Conservation of Momentum • Sean is carrying 24 bottles of beer when he slips in a large frictionless puddle. He slides forwards with a speed of 2.50 m/s towards a very steep cliff. The only way for Sean to stop before he reaches the edge of the cliff is to throw the bottles forward at 20.0 m/s (relative to the ground). If the mass of each bottle is 500 g, and Sean's mass is 72 kg, what is the minimum number of bottles that he needs to throw? (a) 18 (b) 20 (c) 21 (d) 24 (e) more than 24

  49. Momentum and Impulse • A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs? (a) The bag provides the necessary force to stop the person. (b) The bag reduces the impulse to the person. (c) The bag reduces the change in momentum. (d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person. (e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.

  50. Momentum and Impulse • A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs? (a) The bag provides the necessary force to stop the person. (b) The bag reduces the impulse to the person. (c) The bag reduces the change in momentum. (d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person. (e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.

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