Transcript P221_2009_week11

Chapter 12 problems
Chapter 12 problems
•
•
•
The first experimental determination of the universal Gravitational
Constant (G), which appears in Newton’s law for gravitational force was
derived (some 100 years after the fact) from experiments by Henry
Cavendish using equipment he had inherited (and modified) from Rev.
John Michell. When Cavendish published his results in the Philosophical
Transactions of the Royal Society of London, his article actually had the
title “Experiments to determine the Density of the Earth”. Explain how one
might make a connection between the determination of G and the
determination of this quantity. [27 no answer; about half had a rough
idea, but didn’t nail it succinctly, the other half were confused]
Gravity is related to mass thoguh the equation F=Gm1m2/r^2 and
mass is related to density by the equation Density=mass/volume
[True, but what does knowing G really give you?]
(13-11) g =GME/RE since we know g=9.80m/s2; knowing G gives
us ME., and RE had been well-known since the time of the greeks.
Cavendish Experiment
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”
http://en.wikipedia.org/wiki/Cavendish_experiment
Cavendish Experiment
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”
http://en.wikipedia.org/wiki/Cavendish_experiment
Size of the angle
Change is greatly
exaggerated in this
cartoon; it’s hard
to measure (tiny)!
When:
Who:
April 19, 2009
Teams of 3*
*Must be all male/all female
Where:
DeVault Alumni
Center
jills-house.org/events.php
for a registration form
Find us on Facebook...Hoosiers for
$10/person, includes t-shirt
Questions Contact
[email protected]
Chapter 13
problems
Chapter 13 problems
(c) What is its potential energy at launch?
(d) What is its kinetic energy at launch?
Chapter 13 problems
ME
(c) What is its potential energy at launch?
(d) What is its kinetic energy at launch?
•
The Schwarzschild radius of an astronomical object is approximately
equal to be that radius for which a sphere of the mass in question has
an escape velocity equal to the speed of light. Estimate the
Schwarzschild radius for our Sun. If you could compress the mass of
our Sun into a sphere of this radius, it would form a black hole.
•
(19 correct; 5 made errors, some way off, some silly;
26 no answer; 2 were confused). Basically, for the
most part the class got this.
•
Schwarzchild radius=2949.62 m; v^2=(2GM)/R; v=3*10^8 m/s,
G=6.67*10^-11, M=1.99*10^30 kg. [right idea, but an ESTIMATE with
6 sig figs??]
Using the escape speed formula, I found the Schwarzschild radius for
the Sun to be approx 2950m. My only concern with my answer is
whether or not the gravitational constant in the equation should be
6.67x10^-11 or is there a separate gravitational constant specifically
for the Sun. [YES, the same G works for everything: “Universal”]
•
Kepler’s second Law
(equal areas in equal times)
This is equivalent to
saying that the
angular momentum
of the planet must be
conserved
throughout the orbit.
l = m(r x v)
http://www.windows.ucar.edu/tour/link=/the_universe/uts/kepler2.html&edu=elem
Chapter 13 problems
Principle of equivalence
Curved Space
Hydrostatic Pressure
The magnitude of the force experienced
by such a device does not depend
on its orientation! It depends on
the depth, g, surface pressure and
area (DA). Pressure does not have
a direction associated with it
it is in all directions at once!!
Pascal’s Vases (from UIUC)
http://demo.physics.uiuc.edu/lectdemo/scripts/demo_descript.idc?DemoID=229
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!! See the
next slide).
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!!).
Why is the pressure
at the bottom of
these two containers
the same?
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!!).
The walls in the first
container provide the same
forces provided by the
extra fluid in the second
container
Manometer as a P gauge
The height difference can be used
as a measure of the pressure difference
(assuming that the density of the
liquid is known). Hence we have
Pressures measured in “inches of Hg”
or “mm of Hg” (i.e. Torr).
P = Po + rgh
How about a non-uniform
Fluid??
28 no answer
19 same
1 decrease
6 increase
What happens to the pressure at the bottom when the salad
dressing separates into oil (top) and vinegar (bottom)?
• The pressure on the bottom of the vessel remains the same
because the total weight of the fluid above it has not changed even
though the lighter fluid has moved to the top. [THE TOTAL WEIGHT
IS NOT THE RELEVANT issue; think of along pipe in a barrel]
• The pressure on the bottom of the vessel remains the same
according to Pascal's Principle, which states that the pressure is
transmitted uniformly to all portions of the fluid. [BUT THIS ONLY
HOLDS FOR HOMOGENEOUS FLUIDS; as we shall soon see]
• THIS IS NOT AN EASY QUESTION, BUT IT IS A GOOD ONE! You
have to think about WHY the pressure in the homogeneous case
does not depend on the shape of the container.
How about a non-uniform
Fluid??
28 no answer
19 same
1 decrease
6 increase
What happens to the pressure at the bottom when the salad
dressing separates into oil (top) and vinegar (bottom)?
• The pressure on the bottom of the vessel remains the same
because the total weight of the fluid above it has not changed even
though the lighter fluid has moved to the top. [THE TOTAL WEIGHT
IS NOT THE RELEVANT issue; think of along pipe in a barrel]
• The pressure on the bottom of the vessel remains the same
according to Pascal's Principle, which states that the pressure is
transmitted uniformly to all portions of the fluid. [BUT THIS ONLY
HOLDS FOR HOMOGENEOUS FLUIDS; as we shall soon see]
• THIS IS NOT AN EASY QUESTION, BUT IT IS A GOOD ONE! You
have to think about WHY the pressure in the homogeneous case
does not depend on the shape of the container.
• The downward force from the slanted portion of the vessel is
reduced because the density of the fluid at the top has
decrease after separation!!-> PRESSURE AT BOTTOM WILL
DECREASE!!
The mass (and
weight) of water
above the bottom is
much less on the
left than for that of a
cylinder of height
H+L and constant
diameter D, but the
pressure at the
bottom is the same
for both vessels!
+L
Pascal’s Principle (hydraulic
systems)
LARGE
force out
Small force
in
Small force
in
Chapter 14
problems
Equation of Continuity
(mass in must = mass out)
A1v1=A2v2
Assuming that r is constant
(i.e. an incompressible fluid)
E.G. with the Equation of Continuity
(mass in must = mass out)
What is the flow through the unmarked pipe?