1. Physics: Principles with Applications, 6th edition
Lecture PowerPoints
Chapters 7, 8
Physics: Principles with Applications, 6th edition
Giancoli
© 2005 Pearson Prentice Hall
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2. Momentum Center of Mass
Up to this point, we have assumed that all object move as if they were particles
The center of mass of an object is that point within the object that acts like a particle when the object is moving

3. Momentum Center of Mass
“One point in a moving object which would act like a particle subjected to the same net force”
To determine center of mass (CM)
Place two objects of masses mA and mB along x-axis
Defined at xCM = (mAxA + mBxB) / M , where M is total mass
Motion of CM of a system of particles is directly related to net force acting on system as a whole
Linear momentum of a system of particles is product of mass M and velocity of CM

4. Momentum Center of Mass

5. Momentum Center of Mass
For center of mass in two dimensions
Calculate xCM using x values
Calculate yCM using y values
Total mass is located at coordinates xCM, yCM

6. Momentum Momentum and Relation to Force
An object’s momentum is product of an object’s mass and its velocity
P = mv
If you only see work “momentum”, it means linear momentum
Velocity is a vector, so momentum is a vector
Direction is the direction of the velocity vector
Units for momentum are kg m/s (no special name)
The rate of change of momentum of an object is equal to the net force applied to it
ΣF = Δp / Δt

7. Momentum Momentum and Relation to Force
Example 7-1: Force of a tennis serve (see notes)
If a tennis ball leaves a racket at 55m/s, and the 0.060kg ball is in contact with the racket for 4 x 10-3s, what is the average force on the ball?

8. Momentum Momentum and Relation to Force

9. Momentum Conservation of Momentum
Take two billiard balls, headed in opposite directions, which collide
momentum before = momentum after
mA vA + mB vB = mA v’A + mB v’B

10. Momentum Conservation of Momentum
The Law of Conservation of Momentum: “The total momentum of an isolated system of objects remains constant”
System is a set of objects we choose
An isolated system is one without external forces
If net external forces not zero, then momentum won’t be conserved
In this case, redefine the system to include the objects exerting these forces
Then momentum will be conserved

11. Momentum Conservation of Momentum
Example 7-4: Rifle recoil (see notes)
Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.020kg bullet at 620m/s

12. Momentum Conservation of Momentum

13. Momentum Collisions and Impulse
Since F = Δp / Δt, F Δt = Δp
F Δt has a name: Impulse
Impulse is often represented by the letter J
Since Δp = m Δv, can also use F Δt = m Δv
Units are kg m/s or N s
Total change in momentum is equal to the impulse
Useful for forces that act over a very short period of time, even though force is not constant

14. Momentum Collisions and Impulse

15. Momentum Collisions and Impulse
Example 7-6: Bend your knees (see notes)
a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. b) Estimate average force landing with stiff legs (happens over 1 cm) and c) with bent legs (happens over 50 cm)

16. Momentum Collisions and Impulse
a) v = 7.7 m/s just as he hits the ground
F Δt = m Δv ; 70 kg x (0 m/s – 7.7 m/s) = -540 N s
Force acts upward (opposite downward momentum)
b) Person decelerates from 7.7 m/s to 0 m/s over 0.01 m
Average velocity is (7.7 m/s + 0 m/s) / 2 = 3.9 m/s
Since v = d / Δt, Δt = d / v = m / 3.9 m/s = 2.6x10-3 s
Average F = -540 N s / 2.6x10-3 s = 2.1 x 105 N
c) Same as b except distance for deceleration is 0.50 m
Δt = d / v = 5 x 10-1 m / 3.9 m/s = 0.13 s
Average F = -540 N s / 0.13 s = 4.2 x 103 N

17. Momentum Collisions and Impulse
Rock climbers use nylon ropes on steep cliffs
If a rock climber loses grip, will begin to fall
Momentum will ultimately be halted by means of the rope
The ropes are made of nylon or similar stretchy material
If the rope stretches when pulled taut by the falling climber's mass, force applied on climber over longer time.
Longer time for breaking momentum means less force exerted on the falling climber.
Source: Momentum Real-World Applications

18. Momentum Conservation of Energy and Momentum in Collisions
If two (hard) objects collide, their KE is the same before and after the collision; called an elastic collision
Few perfect elastic collisions, since energy ‘lost’ as sound and heat
But collision of two billiard balls very close to being perfectly elastic, so treated as such

19. Momentum Elastic Collisions in One Dimension
Head-on collision two small objects along one line
KE and momentum conserved, so collision elastic
After algebraic manipulation (not shown)
Relative speed of objects same magnitude but opposite direction after collision

20. Momentum Elastic Collisions in One Dimension
Example 7-7: Pool or billiards (see notes)
Billiard ball A of mass mA moving with speed vA collides head-on with billiard ball B of equal mass (mB = mA) at rest (vB=0). What are speeds of balls after collision?
During collision, ball A comes to a stop; ball B moves in same direction at ball A’s initial velocity

21. Momentum Collisions in Two or Three Dimensions
Need to find x, y components of each path
Momentum is conserved in x direction and in y direction

22. Momentum Conservation of Energy and Momentum in Collisions
If two objects collide, and KE is not conserved, called an inelastic collision
Even though KE not conserved, total energy is conserved

23. Momentum Inelastic Collisions
In inelastic collisions, KE is not conserved
If objects stick after collision, is said to be completely inelastic
Example 7-9: Railroad cars (see notes)
Railroad car A mA=10,000 kg) traveling at vA = 24.0 m/s strikes an identical car B (mB = mA) at rest (vB = 0 m/s). Calculate how much of initial KE is converted to other forms.

24. Momentum Inelastic Collisions
Before collision, only car A moving, so KE = ½ m(vA)2 = ½ 10000kg x (24 m/s)2 = 2.88 x 106 J
After collision, both cars move together at 12 m/s,* KE= ½ (20000kg) (12 m/s)2 = 1.44 x 106 J
So energy converted to other forms = 2.88 x 106 J – 1.44 x 106 J = 1.44 x 106 J
* By conservation of momentum

25. Momentum
Multimedia

26. Torque and Angular Momentum
Objects, such as wheels rotate around an axis of rotation
It is easier to measure rotation in radians (rad) instead of degrees
1 radian = angle subtended by arc with length = radius
Θ = l / r
If l = r, Θ = 1 rad
Note:
counterclockwise rotation is +
clockwise rotation is -

27. Torque and Angular Momentum
Rotational motion
Average angular velocity
Analogous to linear velocity
ω = Δθ/Δt
Units are rad/s
Angular velocity is a vector quantity

28. Torque and Angular Momentum
Rotational motion
Average angular velocity
Direction of angular velocity according to right-hand rule

29. Torque and Angular Momentum
Angular acceleration
Analogous to linear acceleration
α = Δω/Δt
Units are rad/s2

30. Torque and Angular Momentum
Applying the same force at different points, rA and rB
If rA = 3•rB, FB must be 3•FA
Perpendicular distance from axis of rotation to line along which force acts is called the lever arm or moment arm
Force times lever arm = moment of force about the axis = torque (τ)
τ = r┴F = rF┴ (┴ means perpendicular)
τ = Frsinθ
Longer lever arm means less force must be applied means more torque

31. Torque and Angular Momentum
Both force and radius are vectors
Torque is the vector product of multiplying force x radius: τ = F d sin ϴ
Direction of torque determined by right-hand rule

32. Torque and Angular Momentum Moment of Inertia
For a mass in a circle of radius r around an axis of rotation, F = mrα, where m is mass, r is radius and α is angular acceleration
Multiply both sides by r to get τ = mr2α
mr2 represents rotational inertia and is called moment of inertia (I)
Consider moments of inertia for all particles making up a rigid rotating object:
Στ = Iα
Analogous to F = ma

33. Torque and Angular Momentum
Moments of inertia for some various objects of uniform composition

34. Torque and Angular Momentum
Angular momentum (L):
L = Iω, where I is moment of inertia and ω is angular velocity about axis of rotation
Under some circumstances, angular momentum is conserved
“The total angular momentum of a rotating object remains constant if the net torque acting on it is zero”
In this case, Iω is constant

35. Torque and Angular Momentum
Angular momentum (L)
Is a vector quantity
Direction of angular momentum determined by right-hand rule

36. Torque and Angular Momentum