Your Comments. That s the plan


 Baldwin Barrett
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1 Your Comments I love physics as much as the next gal, but I was wondering. Why don't we get class off the day after an evening exam? What if the ladder has friction with the wall? Things were complicated enough before adding rotation urgh!!!!!!!!! Please just cancel the 3rd exam and final and let's call it a semester!!! What exactly is a footprint? The prelecture made it seem like it was an intuitive concept, but what's the real definition? I really found this a difficult topic, except for the center of mass and stability. Please walk through an example or two so I can get the hang of this. As usual, I'm doing this prelecture after the exam, and my brain is totally fried. They help though. I may not be doing very well in this class, but I sure am learning a lot. This would be a good lecture to review all of the topics of rotations. I think a lot of people (myself included) have been focusing on the topics on the midterm and have neglected some of the new material That s the plan I have a weather related question. Sandy was a hurricane, but has transitioned into an extratropical cyclone, and an effect of this transition is that the storm has increased in size and (slightly) decreased in wind speed while central pressure has simultaneously decreased. Does conservation of momentum best explain why the storm has increased in size while winds have decreased? (rotation slows down about the central axis, so radius of rotation must increase) Possibly may also have to do with energy Mechanics Lecture 18, Slide 1
2 Since you asked.. I have heard the term center of gravity used, and it seems to be applicable here but it wasn't mentioned in the prelecture. Can you go over how the term center of gravity would be applied in the center of mass/footprint problems? Center of mass is the average position of stuff weighted by mass: X CM m x m x m m Center of gravity is the average position of stuff weighed by its weight: X CG w x w 1 2 w x... w... If weight can be written like w 1 =m 1 g then they are the same. If the strength of gravity varies across the object they can be different. We don t consider Center of Gravity in Physics 211 Mechanics Lecture 10, Slide 2
3 I have absolutely no idea how to do the "Three Masses" problem for the homework in section 16 that's due Friday. Could you please go over this in lecture, or at least give a hint on how to start it. T f m a 1 s fr a I s s s R s s T 1 T T 1 m a s T T R I 2 1 d d d a R s d 1 T T 2 1 m a d 2 f T 2 a m g 1 7 m m m 2 5 h h d s T 2 m h g m g T m a h 2 h
4 Physics 211 Lecture 18 Today s Concepts: a) Static Equilibrium b) Potential Energy & Stability Mechanics Lecture 18, Slide 4
5 Clicker Question 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? A) 4 kg B) 5 kg C) 6 kg D) 7 kg E) 8 kg 1 m 2 m 1 m 3 m 1 kg Mechanics Lecture 18, Slide 5
6 Clicker Question In which of the static cases shown below is the tension in the supporting wire bigger? In both cases M is the same, and the blue strut is massless. A) Case 1 B) Case 2 C) Same T T M L M L/2 Case 1 Case 2 Mechanics Lecture 18, Slide 6
7 It s the same. Why? Case 1 Case 2 T 1 q d 1 T 2 d 2 q M L M L/2 Balancing Torques L L MgL T L sin q 0 1 Mg T sin T 1 Mg sin q T 2 Mg sin q Mechanics Lecture 18, Slide 7
8 CheckPoint 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 T 1 T M L Case 1 L/2 Case 2 Mechanics Lecture 18, Slide 8
9 CheckPoint 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) the length is longer. B) Both strings have to provide an equal torque to hold the beam. In case 2, the perpendicular distance from the pivot is less which means a greater force is needed to hold the beam up. C) the center of mass for the rod is the same in both cases. therefore the force due to gravity is the same. to be static, the tension must equal of gravity, therefore tension force is the same. Mechanics Lecture 18, Slide 9
10 Homework Problem T 1 T 2 Same distance from CM: T 1 T 2 T CM Balance forces: T 1 T 2 Mg So: T Mg/2 Mg Mechanics Lecture 18, Slide 10
11 Homework Problem Mechanics Lecture 18, Slide 11
12 These are the quantities we want to find: T 1 A CM M d Mg y x Mechanics Lecture 18, Slide 12
13 Clicker Question What is the moment of inertia of the beam about the rotation axis shown by the blue dot? A) B) I I 1 12 M d M L 2 2 d L M C) 1 I M L M d Mechanics Lecture 18, Slide 13
14 Clicker Question The center of mass of the beam accelerates downward. Use this fact to figure out how T 1 compares to weight of the beam? A) T 1 Mg T 1 A CM M B) T 1 > Mg d C) T 1 < Mg Mg y x Mechanics Lecture 18, Slide 14
15 Clicker Question The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam? A) A CM d T 1 A CM M B) A CM d / d C) A CM / d Mg y x Mechanics Lecture 18, Slide 15
16 Apply F ext M A C M A CM d Mg T MA 1 CM T Mg MA 1 CM T 1 A CM M Apply I ext d M gd I I A CM d Mg y A CM g Md I 2 Use A CM d to find x Plug this into the expression for T 1 Mechanics Lecture 18, Slide 16
17 After the right string is cut, the meterstick 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: 1 M gd 2 I 2 M gd I 2 T d M y x CM demos Mechanics Lecture 18, Slide 17
18 Applying F ext M A C M T T Mg M d 2 Centripetal acceleration T Mg M d 2 d A CM 2 d y Mg x Mechanics Lecture 18, Slide 18
19 Another HW problem: We will now work out the general case Mechanics Lecture 18, Slide 19
20 General Case of a Person on a Ladder Bill (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 f between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is f. How does f depend on the angle of the ladder and Bill s distance up the ladder? Bill m L M f y x f q Mechanics Lecture 18, Slide 20
21 Balance forces: x: F wall f y: N Mg mg Balance torques: L mgd cos q Mg cos 2 q F Lsin q 0 wall L/2 F wall F wall mg d L Fwall Mg cot 2 q d M g f m g cot q L 2 f axis f N d q mg Mg y x Mechanics Lecture 18, Slide 21
22 This is the General Expression: d M g f m g cot q L 2 Climbing further up the ladder makes it more likely to slip: Making the ladder more vertical makes it less likely to slip: M d m Lets try it out f q Mechanics Lecture 18, Slide 22
23 If its just a ladder d M g f m g cot q L 2 f Mg 2 cot q Moving the bottom of the ladder further from the wall makes it more likely to slip: Mechanics Lecture 18, Slide 23
24 CheckPoint In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest? A) Case 1 B) Case 2 C) Same Case 1 Case 2 Mechanics Lecture 18, Slide 24
25 CheckPoint In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest? A) Case 1 B) Case 2 C) Same A) The ladder is most likely to slip when theta is increased so by this logic, when the angle is greater, the force of friction must also be greater. Case 1 Case 2 B) In Case 2, the ladder is more vertical and pushes against the ground more, causing there to be a greater normal force. Friction is a constant times this normal force. Mechanics Lecture 18, Slide 25
26 CheckPoint 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 an black spot. Which of the following configurations best represents the equilibrium condition of this setup? A) B) C) Mechanics Lecture 18, Slide 26
27 CheckPoint Which of the following configurations best represents the equilibrium condition of this setup? A) B) C) B) Center of mass of the beam is directly beneath the axis of rotation of the rope. C) The center of mass is lowest in C, minimizing gravitational potential energy. Mechanics Lecture 18, Slide 27
28 Stability & Potential Energy I don't understand what is meant by 'inside the footprint'. footprint footprint Mechanics Lecture 18, Slide 28
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