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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 =