Transcript Chapter9-10.temp

PH 105-003/4 ----Monday, Oct. 8, 2007
Homework: PS8 due Wednesday:
conservation of energy
Chapter 9: momentum
Review Wednesday & Friday:
define p = mv
Why? Then dp/dt=0 for isolated system
Dp =  F dt = impulse
Elastic collision: KE conserved
Completely inelastic collision: stick together
(v1f = v2f = vf) – solve C of Mom. eqn. for vf.
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PH 105
Elastic Collisions in 1D
Equations:
m1v1i + m2v2i = m1v1f + m2v2f (Cons . of
px)
½m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
(Conservation of E)
Known: v1i & v2i
Unknown: v1f & v2f – can solve 2 equations.
But quadratic!
(Wednesday, Oct. 10)
Trick (only works for 1D elastic collisions):
can show relative velocity reverses,
(v1f - v2f) = - (v1i - v2i)
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PH 105
Ungraded question (related to PS 7, Chapter 8)
A mass m strikes a spring with spring constant k. There is
a friction force of magnitude
f = mKmg. In a process in which the spring moves a
distance d, conservation of mechanical energy
E = K + U can be written as
A. Ef = Ei
B. Ef = Ei + f d
C. Ef = Ei - f d
Answer is (c) -- the mass must lose
energy to friction.
Example:
Consider two equal masses m1 = m2 = m,
m1 moving to the right at velocity v and m2
stationary. Find the velocities after an elastic
collision.
Solution: We know conservation of momentum,
m1v1i + m2v2i = m1v1f + m2v2f
and (ONLY because this is 1D elastic)
relative velocity reverses, (v1f - v2f) = - (v1i - v2i)
....
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PH 105
Clicker question:
A small mass moving at +8 m/s collides
elastically with a stationary larger mass,
which moves away at +2 m/s. What is
the velocity of the smaller mass after the
collision?
-6
0.1
Clicker question:
In the previous question, how
much heavier was m2? That is,
find the ratio m2/m1.
7
0.1
PH 105-003/4 ----Monday, Oct. 15, 2007
Homework: PS9 (short) due Wednesday:
conservation of momentum
Rest of Chapter 9: Center of Mass
Why? If a system isn’t just a point mass, but we want to treat it
as one (use F=ma, for example)
what is its position x?
Answer: use xcm
where (for a simple system of 2 masses m1 and m2)
xcm
m1 x1  m2 x2

m1  m2
m1
x
x1
Check: x1=x2 case, m1=0 case, …
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m2
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PH 105
xcm
x2
Example:
Consider a system of two masses
m1
m2
x
0
xcm
6m
If they are equal (m1 = m2) then
m1 x1  m2 x2 m1 x1  m1 x2 x1  x2 0  6 m
xcm 



 3m
2
m1  m2
m1  m1
2
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PH 105
Clicker question (not recorded):
In the system of two masses, suppose
m1=1 kg, m2=2 kg. Find the center of
mass coordinate xcm.
m1
m2
x
4
0.1
0
6m
A. 9.7 Deformable bodies (skip)
9.8 Rockets:
Serway derives
vf
vi=0
mi
 ln(
)
v0
mf
mi
exhaust v0
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mf
vf
Clicker question (didn’t have time):
Use the rocket velocity equation to
calculate the final velocity vf
(in m/s) , given that mf = 1000 kg,
mi = 7389 kg, and the rocket
exhaust velocity v0 = 200 m/s,
400
20
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vf
mi
 ln(
)
v0
mf
A. PH 105-003/4 ----Wednesday, Oct. 17,
2007
Please turn in filled-out evaluations in box at front.
Talk Thursday: “Energy for All in the 21st Century ”
Homework: PS9 (short) due tonight:
conservation of momentum
Test next Friday
Recoil problems: rocket, gun, hose, ….
Linear vs. Rotational motion: x <--> q, etc.
Torque:
F=ma  t = I a
where I = S m r2
Kinetic energy: ½ m v2  ½ I w2
Clicker Question: Two cylindrical
drums (on axles through their
centers) have string wound around
them. One string has an object of
mass m attached to it, the other is
pulled on by a constant force T = mg.
Which drum will turn faster?
A
T
= mg
A. The one with constant tension mg.
B. The one with the mass m attached
C. They will turn at the same rate.
B
m
Clicker Q [not done]: Find the acceleration of the
falling mass in the case mdisk = m = 2 kg, r = 0.2 m.
r
6.7
0.4
m
A. PH 105-003/4 ---- Friday, Oct. 19, 2007
Homework (PS 9, Ch 9&10) is on WebAssign:
Test next Friday
Did:
Linear vs. Rotational motion: x <--> q, etc.
F=ma  t = I a, where I = S m r2
Torque: t = F r
More generally, t = F r sin f
Line of action
Calculating I for various shapes (make table)
Parallel Axis Theorem
Kinetic energy: ½ m v2  ½ I w2
A. PH 105-003/4 ---- Monday, Oct. 22, 2007
Homework (PS 9, Ch 9&10) is due Wednesday 11PM
Test this Friday
Review:
F=ma  t = I a, where I = S m r2
Torque: t = F d = F r sin f (d = r sin f = moment arm)
Line of action
Calculating I for various shapes (made table)
Parallel Axis Theorem: I = Icm + Mr2
A. Example: An airplane has wingspan 20 m, length
20 m, and mass 20,000 kg. It snags a weather
balloon at the end of its wing, which exerts a drag
force 105 N on the wing. Will the plane crash?
Assume it crashes if it rotates > 1 radian in 1
second (the pilot’s reaction time).
To answer this, answer
(a) what is the torque about the center of mass?
(aa) where is the center of mass?
model: 2 rods, each 10,000 kg
(ab) what is the torque?
(b) What is the moment of inertia?
(c) What is the angular acceleration?
(d) How far does it turn in 1 second?
Clicker question:
What is the moment of inertia of a rod of length L
about a point on the rod, L/8 from the center?
For simplicity, use m = 1 kg and L = 20 m, and give
the answer in kg m2.
Hint: for a rod of length L rotating about its center of
mass, I/mL2 = 1/12.
39.6
1
A.
Monday, continued
Kinetic energy:
Ktotal = Ktrans + Krot = ½ m vcm2 + ½ I w2
Rolling motion -- calculate v at bottom of incline
Conservation of energy: mgh = ½ mv2 + ½ I w2
& v = w r give
v2 = 2gh/(1+I/MR2)
A. PH 105-003/4 ---- Wednesday, Oct. 24,
2007
Homework (PS 9, Ch 9&10) is due tonight 11PM
Test this Friday
Review:
Parallel Axis Theorem: I = Icm + Mr2
Kinetic energy:
Ktotal = Ktrans + Krot = ½ m vcm2 + ½ I w2
Rolling motion -- calculate v at bottom of incline
Conservation of energy: mgh = ½ mv2 + ½ I w2
& v = w r give
v2 = 2gh/(1+I/MR2)
vsphere>vdisk > vhoop
Angular momentum