physics 207: lecture 14, pg 1 lecture 14goals: assignment: l l hw6 due tuesday oct. 26 th l l for...

Physics 207: Lecture 14, Pg 1

Lecture 14Goals:Goals:

Assignment: Assignment: HW6 due Tuesday Oct. 26th For Monday: Read Ch. 11

• Chapter 10 Chapter 10 Understand the relationship between motion and energy Define Potential Energy in a Hooke’s Law spring Exploit conservation of energy principle in problem solving Understand spring potential energies Use energy diagrams energy diagrams


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 = U as the “gravitational potential energy”

A relationship between

y- displacement and change in the y-speed squared


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 = K terms are defined to be kinetic energies

(A scalar quantity of motion)


Reading Quiz

What is the SI unit of energy

A erg

B dyne

C N s

D calorie

E Joule


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!


Elastic vs. Inelastic Collisions

A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision.

How, if no net Force then momentum will be conserved.

Kbefore Kafter

E.g. car crashes on ice: Collisions where objects stick together

A collision is said to be elastic when both energy & momentum are conserved before and after the collision.

Kbefore = Kafter

Carts colliding with a perfect spring, billiard balls, etc.


Kinetic & Potential energies

Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of masses

Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy

Mechanical energy, EMech, is defined to be the sum of U and K.


Energy

If only “conservative” forces are present, the total mechanical If only “conservative” forces are present, the total mechanical energy 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


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


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)

Exploit the fact that E = U + K = constant ! (frictionless)

(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=0U=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 ?


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 u

s is the amount the spring is stretched or compressed from it resting position.

F

u

Rest or equilibrium position


Exercise Hooke’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 u - mg

Use k = mg/u = 500 N / 1.0 m

(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m

mg

Fspring


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

u

udvmvduku

f

i

f

i

v

v

u

u mvku | | 2212

21

2212

212

212

21 ifif mvmvkuku

2212

212

212

21 ffii mvkumvku


Energy for a Hooke’s Law spring

Associate ½ ku2 with the “potential energy” of the spring m

2212

212

212

21 ffii mvkumvku

fsfisiU K UK Ideal 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

u = x - xeq

U

Emech

Spring/Mass systemBall falling

00


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


Recap

Assignment: Assignment: HW6 due Tuesday Oct. 26th For Monday: Read Ch. 11

physics 207: lecture 14, pg 1 lecture 14goals: assignment: l l hw6 due tuesday oct. 26 th l l for...

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