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Transcript of Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignment: l HW6 due Wednesday, Feb. 11 l For...
Physics 207: Lecture 13, Pg 1
Lecture 13Goals:Goals:
Assignment: Assignment: HW6 due Wednesday, Feb. 11HW6 due Wednesday, Feb. 11 For Thursday: Read all of Chapter 11For Thursday: Read all of Chapter 11
• Chapter 10Chapter 10 Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Develop and exploit conservation of energy principle
in problem solving
• Chapter 11Chapter 11 Understand the relationship between force, displacement and work
Physics 207: Lecture 13, Pg 2
Energy
-mg y= ½ m (vy2 - vy0
2 )
-mg yf – yi) = ½ m ( vyf2 -vyi
2 )
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf
2 + mgyf
We now define mgy as the “gravitational potential energy”
A relationship between y-displacement and change in the
y-speed
Physics 207: Lecture 13, Pg 3
Energy
Notice that if we only consider gravity as the external force then
the x and z velocities remain constant
To ½ m vyi2 + mgyi = ½ m vyf
2 + mgyf
Add ½ m vxi2 + ½ m vzi
2 and ½ m vxf2 + ½ m vzf
2
½ m vi2 + mgyi = ½ m vf
2 + mgyf
where vi2 = vxi
2 +vyi2 + vzi
2
½ m v2 terms are defined to be kinetic energies
(A scalar quantity of motion)
Physics 207: Lecture 13, Pg 4
Energy If only “conservative” forces are present, the total energy If only “conservative” forces are present, the total energy
((sum of potential, U, and kinetic energies, K) of a system) of a system is is conservedconserved
For an object in a gravitational “field”
Emech = K + U
K and U may change, but Emech = K + U remains a fixed value.
Emech = K + U = constant
Emech is called “mechanical energy”
K ≡ ½ mv2 U ≡ mgy
½ m vyi2 + mgyi = ½ m vyf
2 + mgyf
Physics 207: Lecture 13, Pg 5
Example of a conservative system: The simple pendulum.
Suppose we release a mass m from rest a distance h1 above its lowest possible point.
What is the maximum speed of the mass and where does this happen ?
To what height h2 does it rise on the other side ?
v
h1 h2
m
Physics 207: Lecture 13, Pg 6
Example: The simple pendulum.
y
y=0
y=h1
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum.
Physics 207: Lecture 13, Pg 7
Example: The simple pendulum.
v
h1
y
y=h1
y=0
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing
Physics 207: Lecture 13, Pg 8
Example: The simple pendulum.
y
y=h1=h2
y=0
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2
Physics 207: Lecture 13, Pg 9
ExampleThe Loop-the-Loop … again
To complete the loop the loop, how high do we have to let the release the car?
Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R)
Use fact that E = U + K = constant !
(A) 2R (B) 3R (C) 5/2 R (D) 23/2 R
h ?
R
Car has mass mRecall that “g” is the source of
the centripetal acceleration and N just goes to zero is the limiting case.
Also recall the minimum speed at the top is gRv
Ub=mgh
U=mg2R
y=0
Physics 207: Lecture 13, Pg 10
ExampleThe Loop-the-Loop … again
Use E = K + U = constant mgh + 0 = mg 2R + ½ mv2
mgh = mg 2R + ½ mgR = 5/2 mgR
h = 5/2 R
R
gRv
h ?
Physics 207: Lecture 13, Pg 11
What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest?
Assume we can treat the skateboarder as a “point” Assume zero of gravitational U is at bottom of the hill
R=10 m
..
m = 25 kg
ExampleSkateboard
..
R=10 m
30°
y=0
Physics 207: Lecture 13, Pg 12
What speed will the skateboarder reach halfway down the hill if there is no friction and the skateboarder starts at rest?
Assume we can treat the skateboarder as “point” Assume zero of gravitational U is at bottom of the hill
R=10 m
..
m = 25 kg
ExampleSkateboard
..
R=10 m
30°
Use E = K + U = constant
Ebefore = Eafter
0 + m g R = ½ mv2 + mgR (1-sin 30°)
mgR/2 = ½ mv2
gR = v2 v= (gR)½
v = (10 x 10)½ = 10 m/s
Physics 207: Lecture 13, Pg 13
Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless
1. The ball is dropped
2. The ball slides down a straight incline
3. The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare?
(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell
h
1 32
Physics 207: Lecture 13, Pg 14
Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless
1. The ball is dropped
2. The ball slides down a straight incline
3. The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare?
(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell
h
1 32
Physics 207: Lecture 13, Pg 15
What is the normal force on the skate boarder?
R=10 m
..
m = 25 kg
ExampleSkateboard
..
R=10 m
30°.
.
N
mg60°
Physics 207: Lecture 13, Pg 16
Now what is the normal force on the skate boarder?
R=10 m
..
m = 25 kg
ExampleSkateboard
..
R=10 m
30° Fr = mar = m v2 / R
= N – mg cos 60°
N = m v2 /R + mg cos 60°
N = 25 100 / 10 + 25 10 (0.87)
N = 250 + 220 =470 Newtons
..
N
mg60°
Physics 207: Lecture 13, Pg 17
Elastic vs. Inelastic Collisions
A collision is said to be elastic when energy as well as momentum is conserved before and after the collision.
Kbefore = Kafter
Carts colliding with a perfect spring, billiard balls, etc.
vvi
Physics 207: Lecture 13, Pg 18
Elastic vs. Inelastic Collisions
A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved.
Kbefore Kafter
Car crashes, collisions where objects stick together, etc.
Physics 207: Lecture 13, Pg 19
Inelastic collision in 1-D: Example 1
A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V.
What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved?
vV
before after
x
Physics 207: Lecture 13, Pg 20
Inelastic collision in 1-D: Example 1
What is the momentum of the bullet with speed v ?
What is the initial energy of the system ?
What is the final energy of the system ?
Is momentum conserved (yes)? Is energy conserved? Examine Ebefore-Eafter
vV
before after
x
vm
v2
1 vv
2
1 2mm
V)(2
1 2Mm
V)( 0 M v Mmm
)(1v2
1 vv)(
2
1 v
2
1 V]V)[(
2
1 v
2
1 222
Mm
mm
Mm
mmmMmm
No!
Physics 207: Lecture 13, Pg 22
Variable force devices: Hooke’s Law Springs
Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms)
In this spring, the magnitude of the force increases as the spring is further compressed (a displacement).
Hooke’s Law,
Fs = - k s
s is the amount the spring is stretched or compressed from it resting position.
F
s
Rest or equilibrium position
Physics 207: Lecture 13, Pg 23
Exercise 2Hooke’s Law
50 kg
9 m
8 m
What is the spring constant “k” ?
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
Physics 207: Lecture 13, Pg 24
Exercise 2Hooke’s Law
50 kg
9 m
8 m
What is the spring constant “k” ?
F = 0 = Fs – mg = k s - mg
Use k = mg/s = 500 N / 1.0 m
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
Fspring
Physics 207: Lecture 13, Pg 25
F-s relation for a foot arch:
For
ce (
N)
Displacement (mm)
Physics 207: Lecture 13, Pg 26
Force vs. Energy for a Hooke’s Law spring
F = - k (x – xequilibrium) F = ma = m dv/dt
= m (dv/dx dx/dt)
= m dv/dx v
= mv dv/dx So - k (x – xequilibrium) dx = mv dv Let u = x – xeq. & du = dx
m
f
i
f
i
v
v
x
xdvmvduku
f
i
f
i
v
v
x
x mvku | | 2212
21
2212
212
212
21 ifif mvmvkxkx
2212
212
212
21 ffii mvkxmvkx
Physics 207: Lecture 13, Pg 27
Energy for a Hooke’s Law spring
Associate ½ kx2 with the “potential energy” of the spring m
2212
212
212
21 ffii mvkxmvkx
fsfisiU K UK Hooke’s Law springs are conservative so the
mechanical energy is constant
Physics 207: Lecture 13, Pg 28
Energy diagrams In general:
Ene
rgy
K
y
U
Emech
Ene
rgy
K
s
U
Emech
Spring/Mass systemBall falling
Physics 207: Lecture 13, Pg 30
Equilibrium
Example Spring: Fx = 0 => dU / dx = 0 for x=xeq
The spring is in equilibrium position
In general: dU / dx = 0 for ANY function establishes equilibrium
stable equilibrium unstable equilibrium
U U
Physics 207: Lecture 13, Pg 31
Comment on Energy Conservation
We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Mechanical energy is lost:
Heat (friction)Bending of metal and deformation
Kinetic energy is not conserved by these non-conservative forces occurring during the collision !
Momentum along a specific direction is conserved when there are no external forces acting in this direction. In general, easier to satisfy conservation of momentum
than energy conservation.
Physics 207: Lecture 13, Pg 32
Lecture 13
Assignment: Assignment: HW6 due Wednesday 2/11HW6 due Wednesday 2/11 For Monday: Read all of chapter 11For Monday: Read all of chapter 11