1. Classical Mechanics Lecture 18
Today’s Concepts:
a) Static Equilibrium
b) Potential Energy & Stability

2. Schedule Midterm 3 Wed Dec 11 One unit per lecture!
I will rely on you watching and understanding pre-lecture videos!!!!
Lectures will only contain summary, homework problems, clicker questions, Example exam problems….
Midterm 3 Wed Dec 11

3. Lecture Thoughts

4. Main Points

5. Gravitational Potential Energy

6. Gravitational Potential Energy

7. Gravitational Potential Energy

8. Gravitational Potential Energy

9. Main Points

10. Main Points

11. Stability & Potential Energy
Unstable if CM is “outside of footprint”
footprint
If CM is inside the footprint
CM must be lifted for the object to topple over
Object is stable

12. Statics Problems

13. Statics Problems Into the screen Out of the screen
Into the screen
Out of the screen

14. Static Equilibrium

15. 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 kg
1 m
3 m

16. 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 B) Case C) Same M
300 L T1
Case 1 M
300
L/2 T2
Case 2

17. It’s the same. Why? Balancing Torques q q q q M L T1 d1 Case 1 M T2 d2

18. Checkpoint Clicker Question
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 B) Case C) Same
300
L/2 T2
Case 2 T1 L M
Case 1

19. 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
B) The torque caused by gravity is equal to the torque caused by the tension. Since in Case 1, the lever arm is longer, the force does not need to be as long to equalize the torque caused by gravity.

20. Homework Problem Balance forces: T1 + T2 = Mg T2
T1 = T2 = T
Same distance from CM:
T = Mg/2
So: CM Mg

21. Homework Problem 21

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

23. Clicker Question
What is the moment of inertia of the beam about the rotation axis shown by the blue dot? M A) B) C) d L 23

24. Clicker Question
The center of mass of the beam accelerates downward. Use this fact to figure out how T1 compares to weight of the beam?
ACM T1
A) T1 = Mg
B) T1 > Mg
C) T1 < Mg M a d Mg y x 24

25. Clicker Question
The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam?
ACM T1
A) ACM = da
B) ACM = d /a
C) ACM = a / d M a d Mg y x 25

26. ACM = da Use ACM = da to find a Apply ACM T1 a M Apply d Mg y x
Plug this into the expression for T1

27. 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: M y x d T w 27

28. Applying T
ACM = w2d
Centripetal acceleration d w y x Mg 28

29. We will now work out the general case
Another HW problem:
We will now work out the general case

30. 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? L f
Bill M m y x q f

31. x: Fwall = f y: N = Mg + mg Balance forces: Fwall Balance torques: f N
axis d mg q f y x N

32. This is the General Expression:
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 q f
Lets try it out

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

34. 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 34

35. 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 B) Case C) Same
Case 1
Case 2
A) Because the bottom of the ladder is further away from the wall.
B) The angle is steeper, which means there is more normal force and thus more friction.
C) Both have same mass. 35

36. 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)

37. CheckPoint
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 tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.

39. Homework Draw Free Body Diagram Decompose forces into x,y components
Choose axis for torque calculation
Sum x-component forces
Sum y-component forces
Define torques
Sum of Torques
Relationship between Tx and Ty
Definition of coefficient of friction