1. Collisions and Rotations
Schedule
Midterm 3 Friday April 24
Collisions and Rotations
Units 10-18
Final Exam
Units 1-19
Section 001
Section 002

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

3. Midterm Exam 3
Take home exam portion on all units to help prepare you for the Final exam and mitigate any possible poor performance on Final exam…
Section 002 can submit take-home and corrected midterm exam 3 up until the time of the Final exam for section 001

4. Midterm 3 Sample Exams
Midterm 3 sample exams may be found on Canvas under the Files menu exams/sample_exams/midterm3

5. Lecture Thoughts

6. Main Points

7. Gravitational Potential Energy

8. Gravitational Potential Energy

9. Gravitational Potential Energy

10. Gravitational Potential Energy

11. Main Points

12. Main Points

13. 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

14. Statics Problems

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

16. Static Equilibrium

17. Rotational Statics: ladder

18. Rotational Statics: ladder
Rotation axis
Ladder slips

19. Tipping Box

20. Question 1

21. 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

22. 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

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

25. 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

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

29. Homework Problem 31

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

31. 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 33

32. 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 34

33. 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 35

34. 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

35. 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 37

36. Applying T
ACM = w2d
Centripetal acceleration d w y x Mg 38

37. We will now work out the general case
Ladder HW problem:
We will now work out the general case

38. 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

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

40. Maximum Angle with man
Fwall
L/2 Mg d mg q y x N

41. 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

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

44. 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 46

45. 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. 47

46. 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)

47. 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.

48. 78.54%
98.18%