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


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


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)


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


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


Example: The simple pendulum.

y

y=0

y=h1


E = K + U = constant and so K is maximum when U is a minimum.



v

h1

y

y=h1

y=0


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



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


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


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 ?


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


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


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


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


What is the normal force on the skate boarder?

R=10 m

..

m = 25 kg

ExampleSkateboard

..

R=10 m

30°.

.

N

mg60°


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°


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


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.


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


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!


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


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


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


F-s relation for a foot arch:

For

ce (

N)

Displacement (mm)


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


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


Energy diagrams In general:

Ene

rgy

K

y

U

Emech

Ene

rgy

K

s

U

Emech

Spring/Mass systemBall falling


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


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.


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

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Physics 207: Lecture 13, Pg 1 Lecture 13 Goals: Assignment: l HW6 due Wednesday, Feb. 11 l For...

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