1. Chapter 12 problems

2. Chapter 12 problems

3. 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.

4. Cavendish Experiment
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”

5. Cavendish Experiment Size of the angle
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”
Size of the angle
Change is greatly exaggerated in this cartoon; it’s hard to measure (tiny)!

6. jills-house.org/events.php for a registration form
Kick off Little 5 Week with Fun, Competition -AND-
an Extreme Inflatable Obstacle Course…
When: April 19, 2009
Who: Teams of 3*
*Must be all male/all female
Where: DeVault Alumni Center
jills-house.org/events.php for a registration form
$10/person, includes t-shirt
Questions Contact
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7. Chapter 13 problems

8. Chapter 13 problems (c) What is its potential energy at launch?
(d) What is its kinetic energy at launch?

9. Chapter 13 problems ME (c) What is its potential energy at launch?
(d) What is its kinetic energy at launch?

10. 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= 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”]

11. 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)

12. Chapter 13 problems

14. Principle of equivalence

15. Curved Space

16. 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!!

17. Pascal’s Vases (from UIUC)

18. 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).

19. 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?

20. 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

21. P = Po + rgh 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

22. 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.

23. 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!!

24. +L
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!

25. Pascal’s Principle (hydraulic systems)
LARGE force out
Small force in
Small force in

26. Chapter 14 problems

27. Equation of Continuity (mass in must = mass out)
A1v1=A2v2
Assuming that r is constant
(i.e. an incompressible fluid)

28. E.G. with the Equation of Continuity (mass in must = mass out)
What is the flow through the unmarked pipe?