Transcript Momentum of a System

Exam 2 Review
8.02
W08D1
Announcements
Test Two Next Week Thursday Oct 27
7:30-9:30 Section Room Assignments on
Announcements Page
Test Two Topics: Circular Motion, Energy, Momentum,
and Collisions
Circular Motion: Vector Description
Position
r
r(t)  R rφ(t)
Angular Speed
  d / dt
Velocity
r
φ  R (t) (t)
φ
v(t)  v (t) (t)
Speed
r
v  v(t)
Angular Acceleration
  d / dt  d 2 / dt 2
Acceleration
r
φ
a(t)  ar (t) rφ a (t)φ   v rφ(t)  R (t)
ar  v  R 2  v 2 / R
Modeling the Motion: Newton’s
Second Law
• Define system, choose coordinate system.
• Draw free body force diagrams.
• Newton’s Second Law for each direction.
• Example: x-direction
ˆi : F total
x
• Example: Circular motion
rˆ : F
total
r
d 2x
m 2 .
dt
v2
 m .
R
Strategy:
Applying Newton’s Second Law for
Circular Motion
• Always has a component of acceleration pointing radially inward
• May or may not have tangential component of acceleration
• Draw Free Body Diagram for all forces
• mv2/r is not a force but mass times acceleration and does not
appear on force diagram
• Choose a sign convention for radial direction and check that signs
for forces and acceleration are consistent
Concept Question: Tension
and Circular Motion
A stone attached to a string is
whirled in a vertical plane. Let T1,
T2, T3, and T4 be the tensions at
locations 1, 2, 3, and 4 required
for the stone to have the same
speed v0 at these four locations.
Then
1. T3 > T2 > T1 = T4
2. T1 = T2 = T3 = T4
3. T1 > T2 = T4 > T3
4. none of the above
Dot Product
A scalar quantity
Magnitude:
A  B  A B cos 
The dot product can be positive, zero, or negative
Two types of projections: the dot product is the parallel
component of one vector with respect to the second vector
times the magnitude of the second vector
A  B  A (cos  ) B  A B
A  B  A (cos  ) B  A B
Review: Potential Energy
Difference
Definition: Potential Energy Difference between the
points A and B associated with a conservative force Fc
is the negative of the work done by the conservative
force in moving the body along any path connecting
the points A and B.
B
U    Fc  dr   Wc
A
Review: Examples of Potential
Energy with Choice of Zero
Point
(1) Constant Gravity:
U ( y)  mgy
U ( y  0)  0
(2) Inverse Square Gravity
Gm1m2
U (r) = 
r
U (r0  )  0
(3) Spring Force
U (x)  (1 / 2)kx 2
U (x  0)  0
Work-Energy Theorem:
Conservative Forces
The work done by the total force in moving an
object from A to B is equal to the change in kinetic
energy
W
total
zf
 F
z0
total
1 2 1 2
 dr  mv f  mv0  K
2
2
When the only forces acting on the object are
conservative forces
F total  Fc
then the change in potential energy is
U  W  K
Therefore
U  K  0
Change in Energy for Conservative
and Non-conservative Forces
Force decomposition:
r r
r
F  Fc  Fnc
Work done is change in kinetic energy:
B
r r B r
r
r
W   F  d r   (Fc  Fnc )  dr  U  Wnc  K
A
A
Mechanical energy change:
K  U  E mech  Wnc
Strategy: Using Multiple Ideas
Energy principle: No non-conservative work
K  U  E mech  Wnc  0
For circular motion, you will also need to
Newton’s Second Law in the radial direction
because no work is done in that direction hence
the energy law does not completely reproduce
the equations you would get from Newton’s
Second Law
2
rφ: Fr   m
Constraint Condition:
vf
R
N  0 at    f
Modeling the Motion Energy
Concepts
Change in Mechanical Energy:
Identify non-conservative forces.
final
Calculate non-conservative work
Wnc   Fnc  dr .
initial
Choose initial and final states and draw energy diagrams.
Choose zero point P for potential energy for each
interaction in which potential energy difference is welldefined.
Identify initial and final mechanical energy
Apply Energy Law.
Wnc  K  U  E mech
Bead on Track
A small bead of mass m is constrained
to move along a frictionless track. At
the top of the circular portion of the
track of radius R, the bead is pushed
with an unknown speed v0. The bead
comes momentarily to rest after
compressing a spring (spring constant
k) a distance xf. What is the direction
and magnitude of the normal force of
the track on the bead at the point A, at
a height R from the base of the track?
Express your answer in terms of m, k,
R, g, and xf.
Block Sliding off Hemisphere
A small point like object of
mass m rests on top of a
sphere of radius R. The object
is released from the top of the
sphere with a negligible speed
and it slowly starts to slide. Find
an expression for the angle θf
with respect to the vertical at
which the object just loses
contact with the sphere. There
is a non-uniform friction force
with magnitude f=f0sinθ acting
on the object.
Table Problem: Potential Energy
Diagram
A body of mass m is moving along the xaxis. Its potential energy is given by the
function U(x) = b(x2-a2) 2 where b = 2 J/m4
and a = 1 m .
a) On the graph directly underneath a graph of
U vs. x, sketch the force F vs. x.
b) What is an analytic expression for F(x)?
Momentum and Impulse: Single
Particle
• Momentum
p  mv
SI units
[kg  m  s ]  [N  s]
• Change in momentum
-1
p  mv
tf
• Impulse
I   Fdt
ti
• SI units
[N  s]
External Force and Momentum
Change
The momentum of a system of N particles is defined as the sum of the
individual momenta of the particles
i N
r
r
r
psys   pi  msys Vcm
i1
Force changes the momentum of the system
r
r
dpsys
r
r
dp i
F   Fi  

dt
i1
i1 dt
i N
i N
Force equals external force, internal forces cancel in pairs
r r
F  Fext
Newton’s Second and Third Laws for a system of particles: The
external force is equal to the change inr momentum of the system
r
Fext 
r
dpsys
dt

d(msys Vcm )
dt
r
 msys A cm
Strategy: Momentum of a System
1. Choose system
2. Identify initial and final states
3. Identify any external forces in order to determine whether
any component of the momentum of the system is
constant or not
i) If there is a non-zero total external force:
r
dpsys
r total
Fext 
dt
ii) If the total external force is zero then momentum is
constant
p sys,0  p sys,f
Problem Solving Strategies:
Momentum Flow Diagram
•
Identify the objects that comprise the system
•
Identify your choice if reference frame with an appropriate
choice of positive directions and unit vectors
•
Identify your initial and final states of the system
•
Construct a momentum flow diagram as follow:
Draw two pictures; one for the initial state and the other for
the final state. In each picture: choose symbols for the mass
and velocity of each object in your system, for both the initial
and final states. Draw an arrow representing the
momentum. (Decide whether you are using components or
magnitudes for your velocity symbols.)
Modeling: Instantaneous
Interactions
• Decide whether or not an interaction is instantaneous.
• External impulse changes the momentum of the
system.
I[t , t  tcol ] 
t tcol

Fext dt  (Fext ) ave tcol  p sys
t
• If the collision time is approximately zero,
tcol
0
then the change in momentum is approximately zero.
p system  0
Collision Theory: Energy
Types of Collisions
Elastic:
K 0sys  K sys
f
1
1
1
1
2
2
2
m1v1,0  m2 v2,0      m1v1, f  m2v2, f 2    
2
2
2
2
Inelastic:
K0sys  K sys
f
Completely Inelastic: Only one body emerges.
Superelastic:
K0sys  K sys
f
Elastic Collision: 1-Dim
Conservation of Momentum and
Relative Velocity
Momentum
Relative Velocity
m1vx,1,i  m2vx,2,i  m1vx,1, f  m2vx,2, f
v1,x,i  v2,x,i  v2,x, f  v1,x, f
Table Problem: One
Dimensional Elastic Collision:
Relative Velocity
Consider the elastic collision of two carts; cart 1 has
mass m1 and moves with initial speed v0. Cart 2 has
mass m2 = 4 m1 and is moving in the opposite direction
with initial speed v0 . Immediately after the collision, cart
1 has final speed v1,f and cart 2 has final speed v2,f. Find
the final velocities of the carts as a function of the initial
speed v0 .