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Quiz 1 Re-evaluation Request Due Today May 8

Physics 7B - AB Lecture 6 May 8 Recap on Forces, Impulse, components Angular Momentum Model (Second half of Chap 7) - Rotaional motion - Torque, Angular Momentum. Quiz 1 Re-evaluation Request Due Today May 8. Quiz 2 avergae 8.3 (B - ) Quiz 2 Re-evaluation Request Due May 22.

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Quiz 1 Re-evaluation Request Due Today May 8

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  1. Physics 7B - ABLecture 6May 8Recap on Forces, Impulse, components Angular Momentum Model (Second half of Chap 7) - Rotaional motion - Torque, Angular Momentum

  2. Quiz 1 Re-evaluation Request Due Today May 8 Quiz 2 avergae 8.3 (B-) Quiz 2 Re-evaluation Request Due May 22 Quiz 3 graded, solution up on the web this afternoon, rubrics will follow

  3. Fgnd on block (this is a contact force) FEarth on block Gravitational force (an example of non-contact force, or a.k.a. long range force)

  4. Fgnd on block (this is a contact force) • Net force causes acceleration/change in momentum FEarth on block Gravitational force (an example of non-contact force, or a.k.a. long range force) Newton’s First Law ∑ F on the block = 0 so then ∆v = 0, i.e., ∆p = 0 In other words, the block does not start moving (its velocity, as well as momentum, vector does not change) Newton might say,“ an object at rest will remain at rest”

  5. True Meaning of Newton’s First Law Can an object with zero net force exerted on be moving ?

  6. True Meaning of Newton’s First Law Can an object with zero net force exerted on be moving ? YES! The object is moving with constant velocity that just means its velocity won’t change - magnitude or direction Instantaneous velocity is NON ZERO and acceleration is ZERO. All Newton is saying is that net force causes CHANGES in velocity/momentum.

  7. True Meaning of Newton’s First Law Can an object at rest have zero net force ?

  8. True Meaning of Newton’s First Law Can an object at rest have non zero net force ? YES! It just won’t stay at rest Instantaneous velocity is ZERO and acceleration is NON ZERO. Remember ? net force causes CHANGES in velocity/momentum.

  9. Fgnd on block Assume the slope is frictionless… FEarth on block

  10. Component ofF Earth on block that is parallel to the slope F parallel Earth on block Assume the slope is frictionless… Newton’s First Law ∑ F on the block ≠ 0 then ∆v ≠ 0, i.e., ∆p ≠ 0 The block will move,i.e.,the block will accelerate,i.e., the block’s velocity/momentum will change

  11. Now there is friction Fgnd on block (normal) Fgnd on block (tangential) In real life, there is friction… FEarth on block Newton’s First Law ∑ F on the block = 0 then ∆v = 0, i.e., ∆p = 0 ∑ F on the block ≠ 0 then ∆v ≠ 0, i.e., ∆p ≠ 0

  12. What about Newton’s Third Law ? Fgnd on block (this is a contact force) FEarth on block Gravitational force (an example of non-contact force, or a.k.a. long range force) Fgnd on block =Fblock on gnd FEarth on block =Fblock on Earth

  13. (a) (b) (c) (d) (e) Question I am swinging a pendulum on Jupiter, where the acceleration due to gravity is 25 m/s2 (i.e. 2.5 times bigger than that of Earth). Which represents a possible force diagram of the bob at the end of the pendulum, for when the pendulum is at its lowest point? (Neglect air resistance) path going this way Jupiter

  14. (a) (b) (c) (d) (e) Question I am swinging a pendulum on Jupiter, where the acceleration due to gravity is 25 m/s2 (i.e. 2.5 times bigger than that of Earth). Which represents a possible force diagram of the bob at the end of the pendulum, for when the pendulum is at its lowest point? (Neglect air resistance) path going this way Jupiter

  15. (a) (b) (c) (d) (e) Question I am swinging a pendulum on the moon, where the acceleration due to gravity is 1.6 m/s2 (i.e. one-sixth that of Earth). Which represents a possible force diagram of the bob at the end of the pendulum, for when the pendulum is at its lowest point? (Neglect air resistance) path going this way Moon

  16. (a) (b) (c) (d) (e) Question I am swinging a pendulum on the moon, where the acceleration due to gravity is 1.6 m/s2 (i.e. one-sixth that of Earth). Which represents a possible force diagram of the bob at the end of the pendulum, for when the pendulum is at its lowest point? (Neglect air resistance) path going this way Moon

  17. Water Bottle ExampleWhat causes the bottle to fall on its side??

  18. Rotational Motion Torque  is like Force (though not exactly), except it causes rotational motion of an object. Angular Momentum is like Linear Momentum (though not exactly), except it applies to rotating object.

  19. Rotational motion Definitions…first thing’s first We now focus on rotation rather than translation. • L and p both conserved unless there is an external “push” (i.e. torque or force) • I (Rotational Inertia) and m both measure the resistance • to a “push” - Difference: I can be easily changed, m is constant

  20. Rotational motion Definitions…first thing’s first r : Distance from the origin  R R y component ofR Ry  : Angular displacement Origin  Rx:x component ofR Angular velocity :  = d /dt Angular acceleration:  = d  /dt

  21. Rotational motion Definitions…first thing’s first What about the direction of  ,  ? r : Distance from the origin  R  : Angular displacement Origin  Angular velocity :  = d /dt Angular acceleration:  = d  /dt

  22. Direction of  ,  is given by the RHR Rotation this way Right Hand Rule Rotational motion Definitions…first thing’s first What about the direction of  ,  ? r : Distance from the origin  R  : Angular displacement Origin  Angular velocity :  = d /dt Angular acceleration:  = d  /dt

  23. Rotational motion Definitions…first thing’s first vectors We now focus on rotation rather than translation. • L and p both conserved unless there is an external “push” (i.e. torque or force) • I and m both measure the resistance to a “push” - Difference: I can be easily changed, m is constant

  24. Finding a pivot If the system is rotating, choose the pivot point to be the point the system is actually rotating about. Pivot rotating like this Which way are,pointing ?

  25. Define Torque Force is exerted tangentially on the rim, the rim is at a distance r from the pivot point. Ftangential r Magnitude of Torque = r Ftangential Torque  is like Force (though not exactly), except it causes rotational motion of an object.

  26. Direction of  is given by the RHR Rotation this way Define Torque Force is exerted tangentially on the rim, the rim is at a distance r from the pivot point. Ftangential r Magnitude of Torque  =r Ftangential

  27. Define Angular Momentum L Think of rotational inertia I kind of like mass for now. rotating like this with  Magnitude of Angular Momentum L = I

  28. Define Angular Momentum L Think of rotational inertia I kind of like mass for now. Direction of L is given by the RHR rotating like this with  Magnitude of Angular Momentum L = I Rotation This way

  29. Angular analogue to Impulse is Angular Impulse • Angular Impulse Is related to the net external torque in the following way: Net Angular Impulseext = ∆ L = ∫ext(t)dt If the torque is constant during a time interval ∆t Net Angular Impulseext = ∆ L = ave.ext x ∆ t

  30. Back to Water Bottle ExampleWhat caused the bottle to fall on its side??How to analyze rotational motion

  31. Back to Water Bottle ExampleWhat caused the bottle to fall on its side??How to analyze rotational motion F tangentialEarth on bottle F radialEarth on bottle Fearth on bottle

  32. Back to Water Bottle ExampleWhat caused the bottle to fall on its side??How to analyze rotational motion F tangentialEarth on bottle exerts torque, which causes the bottle to rotate about the pivot point F radialEarth on bottle Radial component has no effect on the rotational motion Fearth on bottle

  33. 20cm 20cm 1 kg M What is the mass M in order for the balance to “balance”,i.e., zero net torque, therefore no change in angular momentum ?

  34. 1kg 20cm 10cm 1 kg Now what happens ?

  35. 1kg 20cm 10cm 1 kg F1kg on rod F1kg on rod Extended Force Diagram Shows the forces at the points where they are actually acting

  36. 20cm 10cm 1 kg 1kg F1kg on rod F1kg on rod 1kg on rod = (0.2m)(1kg)(9.8m/ss) = 1.96mN pointing out of the slide 1kg on rod = (0.1m)(1kg)(9.8m/ss) = 0.98mN pointing into the slide

  37. 20cm 10cm 1 kg 1kg F1kg on rod F1kg on rod Net torque points out of the slide, and so does the ∆L, It will start rotating counter-clockwise!

  38. Next weekMay15 Quiz5 (20min) will cover:Today’s lecture Activities and FNTs from DLM11 and Activities from DLM12Bring Calculator!Closed-book, formulas will be provided.

  39. Be sure to write your name, ID number & DL section!!!!!

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