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Announcements

Announcements. Exam #3 next Friday, April 4 Covers Units 12-18 You must notify me by email by Monday at 5 pm if you have a conflict Online retake window from midnight on Friday until midnight on Sunday. Lecture 18: Rotational Statics part 2. Today’s Concepts: a) Static Equilibrium

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Announcements

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  1. Announcements Exam #3 next Friday, April 4 • Covers Units 12-18 • You must notify me by email by Monday at 5 pm if you have a conflict • Online retake window from midnight on Friday until midnight on Sunday

  2. Lecture 18: Rotational Statics part 2 Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability

  3. ACT: Hanging Mobile A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile? C A) 4 kg B) 5 kgC) 6 kg D) 7 kgE) 8 kg A B Torques must balance: 1 m 2 m 1 kg 1 m 3 m

  4. CheckPoint: Two Signs In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length. A) Case 1 B) Case 2 C) Same T1 T2 300 300 L L/2 M Case 1 Case 2

  5. CheckPointRevote: Two Signs In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length. A) Case 1 B) Case 2 C) Same A) In case 1 the level arm is greater than in case 2 making case 1 have the greater tension. B) In case 2 the lever arm is shorter, so the force of tension must be greater. C) The length does not have anything to do with the magnitude of the tension.

  6. CheckPoint Results: Two Signs In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length. A) Case 1 B) Case 2 C) Same Case 1: Case 2:

  7. Stability & Potential Energy footprint footprint

  8. Calculations T2 T1 CM Mg Balance forces: T1 + T2 = Mg Same distance from CM: T1 = T2 = T So: T = Mg/2

  9. Calculations

  10. These are the quantities we want to find: ACM T1 α M d Mg y x

  11. ACT What is the moment of inertia of the beam about the rotation axis shown by the blue dot? M A) B) C) d L

  12. ACT The center of mass of the beam accelerates downward. Using this fact, how does T1 compare to the weight of the beam? ACM T1 A) T1 = Mg B) T1 > Mg C) T1 < Mg α M d Mg y x

  13. ACT The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam? A) ACM=dα B) ACM = d /α C) ACM= α /d ACM T1 α M d Mg y x

  14. Apply ACM = dα ACM T1 α M d Apply Mg y UseACM = dα to findα Plug this into the expression for T1 x

  15. After the right string is cut, the meter stick swings down to where it is vertical for an instant before it swings back up in the other direction.What is the angular speed when the meter stick is vertical? Conserve energy: T M d ω y x

  16. Centripetal acceleration ACM = ω 2d T Applying d ω Mg y x

  17. Another Problem: We will now work out the general case

  18. General Case of a Person on a Ladder Jordan (mass m) is climbing a ladder (length L, mass M) that leans against a smooth wall (no friction between wall and ladder). A frictional force fbetween the ladder and the floor keeps it from slipping. The angle between the ladder and the ground is θ. How does fdepend on the angle of the ladder and Bill’s distance up the ladder? y x L M Jordan m θ f

  19. Balance forces: x:Fwall=f y: N=Mg + mg y x Fwall Balance torques: L/2 Mg axis d mg θ f N

  20. This is the General Expression: Climbing further up the ladder makes it more likely to slip: M Making the ladder more vertical makes it less likely to slip: d m θ f

  21. If its just a ladder… Moving the bottom of the ladder further from the wall makes it more likely to slip:

  22. CheckPoint: Suspended Beam Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with a black spot. Which of the following configurations best represents the equilibrium condition of this setup? A) B) C)

  23. CheckPoint Results: Suspended Beam Which of the following configurations best represents the equilibrium condition of this setup? A) B) C) If the tension has any horizontal component, the beam will accelerate in the horizontal direction. Objects move in such a way as to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.

  24. T Hy Mg mg Hx Use hinge as axis

  25. T Hy Mg mg Hx y x

  26. m Solve for m

  27. T Hy Mg mg Hx Use hinge as axis Which of these thingsmake T smaller?

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