Download Instantaneous momentum principle

vine 藤
Tarzan 泰山
Instantaneous momentum
principle
Newton’s Second Law of Motion
(Momentum Principle)
 
dp
 Fnet
dt
instantaneous
rate of change of
momentum
变化率
We now know at least
four different way of
writing this law…
 
dp
 Fnet Rules for identifying forces
dt
(1) Choose a system.
(2) Distance forces: Show all forces due to
objects in the surroundings that interact
with the system at a distance
(gravitationally or electrically).
(3) Contact forces: Show all forces due to
objects in the surroundings that are
touching the system.
Your turn…
What objects exert significant forces on
the red block?
1) Earth, String 1, String 2
2) Earth, String 1, String 2, Hand
3) Earth, String 1, String 2, Hand,
Ceiling
4) Earth, Hand, Ceiling
Your turn…
What is the unit vector in the
direction of the force?
F1 and F2 are magnitudes of
forces.
1) < cosq, cos(90+q), 0 >
2) < cosq, cos(90–q), 0 >
3) < cos(90+q), cos(q), 0 >
4) < cos(90–q), cos(q), 0 >
Your turn…
1) 0 = – F1*cos(q) + F2
2) 0 = F1 – F2
F1 and F2 are magnitudes of
forces.
Which equation correctly states
that dpx/dt = Fnetx?
3) 0 = F1 + F2 – mg
4) 0 = F1*cos(90+q) + F2
Your turn…
1) 0 = F1 – mg
2) 0 = F1 + F2 – mg
F1 and F2 are magnitudes of
forces.
Which equation correctly states
that dpy/dt = Fnety?
3) 0 = –F1*cos(q) + F2
4) 0 = F1*cos(q) – mg
 
dp
Rules for finding the
 Fnet
rate of change of momentum
dt
(1) Draw two arrows:

- one for pi , the momentum a short time
before the moment of interest;

- one for p f , the momentum a short time
after the moment of interest
 

(2) Find p  p f  pi

(3) The direction of dp dt is the same as the

direction of p .
Your turn…
A ball hangs from the bottom of a vertical
spring. You pull the ball downwards and
release it, and the ball oscillates up and
down.
At the bottom of each oscillation, where the
ball’s instantaneous momentum is zero,
what is the direction of
?
Your turn…
A ball hangs from the bottom of a vertical
spring. You pull the ball downwards and
release it, and the ball oscillates up and
down.
At the bottom of each oscillation, where the
ball’s instantaneous momentum is zero,
what is the direction of
?
Comet 彗星
Your turn…
Which of the red arrows
labeled 1, 2, or 3
represents the vector
,
the change in the
momentum of the comet?
Comet simulation…
Change of momentum can be split into two
parts:
(1) Parallel to the momentum
(change of speed)
(2) Perpendicular to the momentum
(change of direction)


d
p
 
   Fnet | |
 dt | |


 dp 
   Fnet 
 dt  
Changes speed
Changes direction
Ball and mallet video…
kissing circle
Bigger force, smaller radius.
Bigger momentum, bigger
radius.
Your turn…
The Moon travels in a nearly
circular orbit around the Earth, at
nearly constant speed.
When the Moon is at location A,
which components of
are zero?
1) The parallel component
2) The perpendicular
component
3) Both the parallel and
perpendicular components
4) Neither component
Analytical solution for the rate of
change of direction:


 v
 dp 
   p
R
 dt  
Here’s the proof…
radius of
“kissing circle”

dp d 
  p pˆ 
dt dt

dp
 dpˆ

pˆ  p
(product rule)
dt
dt
Parallel to
momentum

 dp 
 
 dt | |
Perpendicular to
momentum

 dp 
 
 dt  
After a time Δt, particle moves to position 2.
p̂2

v t
p̂1
R θ R
R
p̂
θ
1
θ
1
These are similar triangles.
R
After a time Δt, particle moves to position 2.
p̂2

v t
p̂1
R θ R
R
θ
R
p̂
1 θ1
These are similar triangles.
These are similar triangles, so:
pˆ
1
pˆ
limit Δt -> 0



v t

v t
R

v
t
R

v
dpˆ

dt
R
R
θ
R
p̂
1 θ1
These are similar triangles, so:
pˆ
1
pˆ
limit Δt -> 0



v t
R

v
t
R

v
dpˆ

dt
R

 dpˆ
 dp 
   p
dt
 dt | |

 v
 p
R
Example…
What is the speed of a
satellite (卫星) in a
circular orbit (轨道) of
radius R around the
Earth?
Back to Tarzan…
Why does the vine break?
Your turn…
Tarzan swings from a vine.
At the bottom of the swing, what
is the direction of his
?
Your turn…
Tarzan swings from a vine.
At the bottom of the swing, what is
the direction of the net force on
Tarzan?
Your turn…
Tarzan swings from a vine.
At the bottom of the swing, what
objects exert forces on Tarzan?
1) Earth, vine
2) centrifugal force (离心力)
only
3) Earth and centrifugal
force
4) Earth, vine, centrifugal
force
Your turn…
At the bottom of the swing, how
1) F vine > F Earth
does the magnitude of the force on
Tarzan by the vine compare to the
2) F vine = F Earth
magnitude of the force on Tarzan by
the Earth?
3) F vine < F Earth
4) Not enough info.
Your turn…
- Tarzan's mass: 100 kg.
- Length of vine: 5 m.
- Tarzan's speed: 13 m/s.
What is the tension in the vine at
this instant?
1) 980 N
2) 3380 N
3) 2400 N
4) 4360 N
QUIZ
ks
m
+x
L
1.) < 1.65, 0, 0 > kg m/s
2.) < –0.264, 0, 0 > kg m/s
A block with mass m = 0.5 kg is
3.) < –0.114, 0, 0 > kg m/s
attached to a spring with stiffness
4.) < –0.228, 0, 0 > kg m/s
ks = 45 N/m and relaxed length
L0 = 0.15 m. The current length of
5.) < –0.186, 0, 0 > kg m/s
spring is L = 0.11 m and the current
velocity of block is < –0.30, 0, 0> m/s.
Friction is negligible.
What is the momentum of the block
at a time 0.02 seconds later?
(Approximate the force as constant
over this time interval.)