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

http://en.wikipedia.org/wiki/Cavendish_experiment

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

Cavendish Experiment

http://en.wikipedia.org/wiki/Cavendish_experiment

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

Size of the angleChange is greatly exaggerated in this cartoon; it’s hard to measure (tiny)!

When: April 19, 2009Who: Teams of 3*

*Must be all male/all female

Where: DeVault Alumni Center

Find us on Facebook...Hoosiers for Jill’s House

$10/person, includes t-shirtQuestions Contact jillshouse.iu@gmail.com

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

(c) What is its potential energy at launch?(d) What is its kinetic energy at launch?

ME

• 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”]

http://www.windows.ucar.edu/tour/link=/the_universe/uts/kepler2.html&edu=elem

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)

Chapter 13 problems

Principle of equivalence

Curved Space

Hydrostatic Pressure

The magnitude of the force experienced by such a device does not dependon its orientation! It depends on the depth, g, surface pressure andarea (A). Pressure does not havea direction associated with itit 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 shapeof the container (be careful for cases wherethe 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 shapeof the container (be careful for cases wherethe 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 shapeof the container (be careful for cases wherethe 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 usedas 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 + gh

• 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-uniformFluid??

What happens to the pressure at the bottom when the salad dressing separates into oil (top) and vinegar (bottom)?

28 no answer19 same1 decrease6 increase

• 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!!

How about a non-uniformFluid??

What happens to the pressure at the bottom when the salad dressing separates into oil (top) and vinegar (bottom)?

28 no answer19 same1 decrease6 increase

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)

Small forcein

LARGE force out

Small forcein

Chapter 14 problems

Equation of Continuity(mass in must = mass out)

A1v1=A2v2Assuming that 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?