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Lecture 22 General Physics (PHY 2130) http://www.physics.wayne.edu/~apetrov/PHY2130/ • Solids and fluids density and pressure Pascals principle and car lift
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• Lecture 22

General Physics (PHY 2130)

http://www.physics.wayne.edu/~apetrov/PHY2130/

Solids and fluids

density and pressure Pascals principle and car lift

• Lightning Review

Last lecture: 1. Rotational dynamics

torque and angular momentum two equillibrium conditions

Review Problem: A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational inertia and her angular speed increases so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be

1. the same. 2. larger because shes rotating faster. 3. smaller because her rotational inertia is smaller.

• Example:

Given: Moments of inertia:

I1 and I2 Find: K2 =?

LIKErot 21

21 2 ==

Rotational kinetic energy is

We know that (a) angular momentum L is conserved and (b) angular velocity increases

Thus, rotational kinetic energy must increase!

• Solids and Fluids

• Question

What is a fluid?

1. A liquid 2. A gas 3. Anything that flows 4. Anything that can be made to change shape.

• States of matter: Phase Transitions

ICE WATER STEAM

Add heat

Add heat

These are three states of matter (plasma is another one)

• States of Matter

Solid

Liquid Gas Plasma

Has definite volume

Has definite shape

Molecules are held in specific location by electrical forces and vibrate about equilibrium positions

Can be modeled as springs connecting molecules

• States of Matter

Solid

Liquid Gas Plasma

Crystalline solid Atoms have an ordered structure Example is salt (red spheres are Na+

ions, blue spheres represent Cl- ions)

Amorphous Solid Atoms are arranged randomly Examples include glass

• States of Matter Solid Liquid

Gas Plasma

Has a definite volume

No definite shape

Exist at a higher temperature than solids

The molecules wander through the liquid in a random fashion

The intermolecular forces are not strong enough to keep the molecules in a fixed position

• States of Matter

Solid Liquid Gas

Plasma

Has no definite volume

Has no definite shape

Molecules are in constant random motion

The molecules exert only weak forces on each other

Average distance between molecules is large compared to the size of the molecules

• States of Matter Solid Liquid Gas Plasma

Matter heated to a very high temperature

Many of the electrons are freed from the nucleus

Result is a collection of free, electrically charged ions

Plasmas exist inside stars or experimental reactors or fluorescent light bulbs!

• Is there a concept that helps to distinguish between those states of matter?

• Density The density of a substance of uniform composition is

defined as its mass per unit volume:

some examples:

The densities of most liquids and solids vary slightly with changes in temperature and pressure

Densities of gases vary greatly with changes in temperature and pressure (and generally 1000 smaller)

Vm

=

Units

SI kg/m3

CGS g/cm3 (1 g/cm3=1000 kg/m3 )

3

2

3

34

aV

hRV

RV

cube

cylinder

sphere

=

=

=

• 14

Fluids: (liquids and gases) are materials that flow.

Fluids are easily deformable by external forces.

A liquid is incompressible. Its volume is fixed and is impossible to change.

A liquid will flow to take the shape of the container that holds it. A gas will completely fill its container.

Fluids

• 15

Pressure

Pressure arises from the collisions between the particles of a fluid with another object (container walls for example).

There is a momentum change (impulse) that is away from the container walls. There must be a force exerted on the particle by the wall.

By Newtons 3rd Law, there is a force on the wall due to the particle.

• Pressure of fluid is the ratio of the force exerted by a fluid on a submerged object or on the wall of the vessel to area

AFP

Units

SI Pascal (Pa=N/m2)

Example: 100 N over 1 m2 is P=(100 N)/(1 m2)=100 N/m2=100 Pa.

Often: 1 atmosphere (atm) = 101.3 kPa = 1.013 x 105 Pa

Pressure

• 17

Example: Someone steps on your toe, exerting a force of 500 N on an area of 1.0 cm2. What is the average pressure on that area in atmospheres?

atm 49Pa 10013.1

atm 1N/m 1

Pa 1N/m 100.5

m 101.0N 500

5226

24av

=

=

==

AFP

242

2 m 100.1cm 100

m 1cm 0.1 =

First, lets change centimeters into meters:

Then, using the definition of pressure, lets determine pressure in Pa and then change to atm:

• Pressure and Depth If a fluid is at rest in a container, all

portions of the fluid must be in static equilibrium

All points at the same depth must be at the same pressure (otherwise, the fluid would not be in equilibrium)

Three external forces act on the region of a cross-sectional area A

External forces: atmospheric, weight, normal

AghAPPAAhVMAPMgPAF

+===

==0

0

:so,:but,00 ghPP += 0

• ConcepTest 1

You are measuring the pressure at the depth of 10 cm in three different containers. Rank the values of pressure from the greatest to the smallest:

1. 1-2-3 2. 2-1-3 3. 3-2-1 4. Its the same in all three

10 cm

1 2 3

• ConcepTest 1

You are measuring the pressure at the depth of 10 cm in three different containers. Rank the values of pressure from the greatest to the smallest:

1. 1-2-3 2. 2-1-3 3. 3-2-1 4. Its the same in all three

10 cm

1 2 3

• Pressure and Depth equation

Po is normal atmospheric pressure 1.013 x 105 Pa = 14.7 lb/

in2 The pressure does not depend upon the shape of the container

ghPP o +=

Other units of pressure:

76.0 cm of mercury

One atmosphere 1 atm = 1.013 x 105 Pa

14.7 lb/in2

• Example:

Given: masses: h=100 m Find: P = ?

Find pressure at 100 m below ocean surface.

( )( )( )( )pressurecatmospheri1010

1008.910108.9

so,

6

2335

0 2

+=

+=

PamsmmkgPaP

ghPP OH

• Pascals Principle

A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

The hydraulic press is an important application of Pascals Principle

Also used in hydraulic brakes, forklifts, car lifts, etc.

2

2

1

1

AF

AFP ==

Since A2>A1, then F2>F1 !!!

• 24

11

22

2

2

1

1

AA

AF

AF

2point at 1point at

FF

PP

=

=

=

The work done pressing the smaller piston (#1) equals the work done by the larger piston (#2).

2211 dFdF =

Using an hydraulic lift reduces the amount of force needed to lift a load, but the work done is the same.

• 25

Example: Assume that a force of 500 N (about 110 lbs) is applied to the smaller piston in the previous figure. For each case, compute the force on the larger piston if the ratio of the piston areas (A2/A1) are 1, 10, and 100.

F2

1 500 N 10 5000 N 100 50,000 N

12 AA

Using Pascals Principle:

• 26

Example: In the previous example, for the case A2/A1 = 10, it was found that F2/F1 = 10. If the larger piston needs to rise by 1 m, how far must the smaller piston be depressed?

m 1021

21 == dFFd

Since the work done by both pistons is the same,

• 27

Example: Depressing the brake pedal in a car pushes on a piston with cross-sectional area 3.0 cm2. The piston applies pressure to the brake fluid, which is connected to two pistons, each with area 12.0 cm2. Each of these pistons presses a brake pad against one side of a rotor attached to one of the rotating wheels. See the figure for this problem. (a) When the force applied by the brake pedal to the small piston is 7.5 N, what is the normal force applied to each side of the rotor?

The pressure in the fluid b b .P F A=

2

b 2b

12.0 cm (7.5 N) 30 N3.0 cm

AN PA FA

= = = =

the normal force applied to each side of the rotor

Also,

ApistonpadbraketheofAreaNForceNormalPessure

Pr =

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