1. Physics I LECTURE 21 12/2/09

2. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
a) (5pts) What is the speed of the mass when it leaves the spring at point A?
b) (5pts) What is the speed of the mass at point B?
c) (5pts) What is the speed of the mass at point C?
d) (5pts) What is the speed of the mass at point D?
e) (5pts) How high above point D does the spring travel?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

3. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
a) (5pts) What is the speed of the mass when it leaves the spring at point A?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

4. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
b) (5pts) What is the speed of the mass at point B?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

5. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
c) (5pts) What is the speed of the mass at point C?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

6. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
d) (5pts) What is the speed of the mass at point D?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

7. Exam Prep Question
A mass of 5kg sits on the surface shown below. A spring with k=3000N/m is compressed from equilibrium by 30cm and placed next to the mass. The spring is released, and the mass shoots forward.
e) (5pts) How high above point D does the spring travel?
30cm, μk=0 A B C D
1m, μk=0
3m, μk=0.3
μk=0
h=0.5m

8. Review Example
What is the vector dot product of the two vectors:

9. Outline Vector Cross Products Conservation of Angular Momentum
Weightlessness
Kepler’s Laws
Work by Constant Force
Scalar Product of Vectors
Work done by varying Force
Work-Energy Theorem
Conservative, non-conservative Forces
Potential Energy
Mechanical Energy
Conservation of Energy
Dissipative Forces
Gravitational Potential Revisited
Power
Momentum and Force
Conservation of Momentum
Collisions
Impulse
Conservation of Momentum and Energy
Elastic and Inelastic Collisions2D, 3D Collisions
Center of Mass and translational motion
Angular quantities
Vector nature of angular quantities
Constant angular acceleration
Torque
Rotational Inertia
Moments of Inertia
Angular Momentum
Vector Cross Products
Conservation of Angular Momentum
What do we know?
Units
Kinematic equations
Freely falling objects
Vectors
Kinematics + Vectors = Vector Kinematics
Relative motion
Projectile motion
Uniform circular motion
Newton’s Laws
Force of Gravity/Normal Force
Free Body Diagrams
Problem solving
Uniform Circular Motion
Newton’s Law of Universal Gravitation

10. Review of Lecture 20 Introduced concept of Angular Momentum
Conservation of Angular Momentum
With no external torques acting on a system, the angular momentum of the system is conserved.
Vector Cross products

11. Review of Angular Motion
We know equations of motion for angular motion
We know torques cause angular acceleration
Objects can have rotational kinetic energy
And angular momentum
So why the cross product?

12. Torque and the Cross Product
When we first introduced torque as the product of the radius and the perpendicular component of the Force, we were only interested in the magnitude of the torque!
Magnitude given by RFsinθ…same as cross product
FROM LECTURE 19

13. Torque and the Cross Product
However, we since learned that
And we know that angular acceleration points in direction of axis of rotation… so Torque must as well!
Torque is cross product of R,F
FROM LECTURE 19

14. Angular Momentum of a Particle
We have already defined angular momentum as
But this definition is for objects rotating with some angular velocity and moment of inertia around an axis of rotation.
More general, alternate, definition:

15. Equivalence of our two definitions
Suppose we have a mass rotating around an axis
Use cross product y x
m=2kg 2m

16. Equivalence of our two definitions
Suppose we have a mass rotating around an axis
Use y x
m=2kg 2m

17. So why use cross product?
Cross products are messy…why would we ever use them, instead of the simpler
Because the cross product allows us to determine the angular momentum of, or torque on, objects which are not necessarily moving with constant, or even circular motion!

18. Example
Calculate the angular momentum (about the origin) of the rock of mass m dropped from rest off the cliff. d
(0,0)

19. Example
What Torque is exerted (about the origin) on the rock? d
(0,0)

20. Relationship of torque to angular momentum
When we discussed linear momentum, we revised Newton’s 2nd Law to state
Similarly, we can write an expression for net torque in terms of angular momentum
Double check with our falling rock:

21. Newton’s 2nd Law: Angular Form
The vector sum of all of the torques acting on a particle, object, or system, is equal to the time rate of change of the angular momentum of the particle, object or system.

22. Conservation of Angular Momentum
We used the revised expression for conservation of linear momentum to argue that if there is no net external force on a system or object, then the momentum of the system or object is conserved.
Similarly:
If the net external torque acting on a system of object is zero, then the angular momentum of that object will remain constant.
L constant!

23. Conservation of Momentum

24. Does this make sense? What is happening? What do we need to know?
System= Container + Skinner

25. Moments of Inertia Skinner=point mass Shipping Container Mass=3,500kg

26. I Container Determine surface mass density σ Divide into 6 slabs
Top & Bottom:

27. Sides
Use parallel axis theorem
h=6.1m

28. Conservation of Momentum
Skinner seems to be making one rotation every 2 seconds.

29. Finally
In the clip, it takes about 20 seconds to turn the container 90 degrees.

30. Conservation of Angular Momentum
In Class Demo