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Fyneman exercise II ch2-5

Q. The velocity of a solid object rotationg about an axis is a field . Show that a) b) where = angular velocity. Trial. a) If velocity has radial component or magnitude of angular velocity is not constant, solid object should be broken, so velocity has no radial component and magnitude of angular velocity is constant. Let’s take coordinate such that solid object is rotating in plane, axis is and use for magnitude of angular velocity. In cylinderical coordinate, we can write , . Then Divide by Radial componet of velocity is zero and , so from eq(3), (4) we get Using eq(5),(6) and (7), eq(8) can be wrriten b) Suppose rotation is counter clock wise. Then we can write Curl is Using eq(5) & (6) and (10), eq(11) can be written

Fyneman exercise II ch2-1

Q. A copper wire of radius has a concentric insulating sheath with outer radius . The wire carries an electric current that raises its temperature to while the outside of the insulation remains at near room temperature. a) What is inside the insulation? Give answer in terms of , , , . b) How big is the temperature difference if a current of 20 amps is sent through No. 10 gauage copper wire which is covered with a layer of rubber 0.2 cm thick whose thermal conductivity is ? Trial. a) b) Copper wire generate heat by Joule heating [ 1 ] Let length of copper wire , and cross-sectional area , then , where is electrical resistivity [ 2 ] . From eq(2) and (3) Radius is , Putting eq.(5) to eq(4) gives This heat trasfered throgh rubber layer, will be , where is area heat transfered, (“-” means lose energy). Suppose , then If proportionality constant is , from (7) & (8) Using eq(6), (9), mean values No...

Solenoidal & Irrotational Vector

Solenoidal If the is zero, is solenoidal , then is the curl of some vector filed . Theorem If there is a such that Irrotational If the is zero, is irrotational , then is the gradient of some scalar field . Theorem If there is a such that

Fyneman exercise II ch1-6

Q. A particle with a mass and a positive charge is at a point , , and is moving with a low velocity The charge is influenced by a negative charge fixed at the origin and by a uniform magnetic field in the direction. How large must be such that the moving particle describes a circle of radius about the stationary one? If the magnitude of the magnetic field strength were different than this, explain why the speed of the particle is a function of the radial distance only. Sketch qualitatively several cycles of the trajectory followed by the particle if it were released from the point , with zero velocity. Trial. For circular motion, For our system, if eq(1) is satisfied, it starts circular motion, then at start time radius , and From eq(1), From eq(4), From eq(5) & eq(6), condition of for circular motion is Motion is always in x-y plane, because force is always toward origin. That the force is always toward origin means is influenced only by radia...

Fyneman exercise II ch1-5

Q. In the region of space of interest there is a uniform magnetic field, such that , , . The field is constant in time, and there are no currents or electric fields in the region of space we consider. A particle of mass and positive charge is started at , , with a velocity in the direction. Sketch and describe quantitatively in terms of , , , and the path of the particle. (Assume ) Suppose, that , , but . For always small compared to , but not completely negligible, show on a sketch the qualitative behavior of the particle trajectory. (See Charpak, et al, Physical Review Letters, Vol. 6, 128 (1961) for the use of a similar field in an important expriment.) Show that the field just postulated is inconsistent with one of Maxwell’s equations if the field fills a finite volume of space and, as above, you assume there are no currents or electric fields in the volume. Trial. For , , , From Initial condition, , , Trajectory is hyperbola. If , from eq.(...

Fyneman exercise II ch1-4

Q. In a certain region of space, there is a uniform electric field of 10,000 volts per centimeter in the direction. There is in the same region a uniform magnetic filed in the direction. A beam of mu-measons with the velocity travels through this region on a straight line in the direction. a) What is the strength of the field ? (A mu-meason has a mass 210 times the electron mass and a charge equal in magnitude to the electron charge.) b) Can you tell from this experiment if the charge on the measons is + or -? Trial. Electromagnetic force a) From eq.(1) , force is . Problem said muons move on a straight line with velocity , which means no external force , . b) I can’t tell muon is + or - charge only with this experiment.

Fyneman exercise II ch1-3

Q. In copper there is one “conduction” electron for each copper atom. When a current of 10 amperes flows through a piece of No. 10 gauge copper wire, what is the average speed of the conduction electrons? How big is ? (Remember that the ratio of “magnetic” effects to “electric” effects is about this small.) Trial. , where is cross-sectional area and is velocity. Cross-sectional area of No. 10 gauge wire : (reference : American wire gauge )

Fyneman exercise II ch1-2

Q. Make a rough estimate of the work that must be done against electric force to assemble a uranium nucleus from two equal halves. What about assemble two deuterons to make a helium nucleus? Also express both answers in kilowatt hours per kg. Trial. Equation of nuclear radius from wiki( Atomic nucleus ) , where is atomic mass number and values for calculation is here . Uranium Atomic number of uranium is 92, which means nuleus has 92 protons. So, one half of it has 46 protons. From eq.(1) , Suppose we move two halves of uranium with 46 protons from to . We need about . Kilowatt-hour = So, energy in killowatt-hours per kg Helium Atomic number of uranium is 2, which means nuleus has 2 protons. So, one half of it has 1 protons. From eq.(1) , We need about . Kilowatt-hour = So, energy in killowatt-hours per kg

Electromagnetism

Field A “field” is any physical quantity which takes on different values at different points in space. Flux Circulation The laws of electromagnetism Moving charge with velocity , For any surface S (not closed) whose edge is the curve C, For any surface S bounded by the curve C,

Shankar exercise-2.1.2

Q. Write the Lagrangian and get the equation of motion for the coupled-mass problem. Trial. coupled-mass-spring-system Lagrangian We have two coordinate systems which origins are and respectively. Lagrange’s equation for Therefore, Equation of motion is Lagrange’s equation for Therefore, Equation of motion is

Shankar exercise-2.1.3

Q. A particle of mass moves in three dimensions under a potential . Write its and find the equation of motion. Trial. Lagrangian . But, potentail depends only on , , Line element is From eq.(1) & eq.(2), Lagrange’s equations for For From eq.(3), Putting the above result to Eq.(4) gives For From eq.(3), For From eq.(3), Equations of motion Classical mechanics For From eq.(5) For From eq.(7.1) For From eq.(9.1)

Lagrangian mechanics-momentum&force

Lagrangian . Canonical momentum conjugate to Generalized force conjugate to Lagrange’s Equation From Principles of Quantum Mechanics, 2nd Edition by R. Shankar . Lagrangian for classical mechanics , where = Kinetic Energy, = Potential Energy.

Fyneman exercise I ch39-6

Q. A cylinder with a leakless, frictionless piston contains of a monatomic gas ( ) at gauge pressure 1 atmosphere ( ). The gas is slowly compressed at constant temperature until the final volume is only . How much work must be done to accomplish this compression? A. means and Constant value is Work is

Fyneman exercise I ch39-5

Q. An atmosphere in which the pressure and density as a function of height satisfy the relation is called an adiabatic atmosphere. a) Show that the temperature of such an atmosphere decreases linearly with height, and find the constant of proportionality. This temperature gradient is called the adiabatic lapse rate . Evaluate this temperature gradient for the earth’s atmosphere. b) Use an argument based on energy considerations to show that an atmosphere having less or more temperature gradient than the adiabatic lapse rate will be stable or unstable against convection, respectively. A. a) From , For Ideal gas, Put from eq.(1) to eq.(2) Using eq.(1) of previous problem eq.(3) becomes RHS of eq.(4) is constant, which means temperature decrease linearly, because is greater than 0. For earth’s atmosphere, symbol meaning value Earth’s Heat capacity ratio 1.4 ...

Fyneman exercise I ch39-4

Q. A) Imagine a tall vertical column of gaseous or liquid fluid whose density varies with height. Show that the pressure as a function of height follows the differential equation . B) Solve this differential equation for the case of a gaseous atmosphere of molecular weight , in which the temperature is constant as a function of . A. A) Imagine the volume of unit area and infinitesimal height . Number of molecules = At equilibrium the force of gravity in the volume plus pressure difference is zero. , where is mass of one molecule. For unit mass B) For ideal gas , where is number of molecules in volume . Then, for unit volume , where is number of molecules in unit volume, which means density. From eq.(1) We need . Substituting to eq.(2) gives , where . Reference molecule weight Avogadro constant

Fyneman exercise I ch39-2

Q. A bicycle pump is being used to inflate a tire to a pressure of gauge pressure, starting with air at atmospheric pressure, at . If for air, at what temperature centigrade is the air as it leaves the pump? Neglect heat loses to the walls of the pump. A. Note : Gauge pressure Using eq.(1)

Fyneman exercise I ch10-2

Q. Two equally massive gliders, moving in a level air trough at equal and opposite velocities, and , collide almost elastically, and rebound with slightly smaller speeds. They lose a fraction of their kinetic energy in the collision. If these same gliders collide with one of them initially at rest, with what speed will the second glider move after the collision? (This small residual speed may easily be measured in terms of the final speed of the originally stationary glider, and thus the elasticity of the spring bumpers may be determined.) NOTE : If A. Let initial speed of glider-1 , final speed of the other glider-2 initially at rest , and final speed of glider-1 . Suppose is opposite . Kinetic Energy Momentum From eq.(1) & eq.(2) Eq.(3) is not possible, because , , and are all positive. We can conclude two glider moves same direction after collision. Therefore eq.(2) should be From eq.(1) & (4) ...

Fyneman exercise I ch10-1

Q. Two gliders are free to move in a horizontal air through. One is stationary and the other collides with it perfectly elastically. They rebound with equal and opposite velocities. What is the ratio of their masses? A. Let moving glider’s mass , velocity and stationary one’s mass , velocity . Initially and total momentum is Kinetic energy is Finally, and so total momentum is Kinetic energy is Since elastic collision, momentum and kinetic energy are conserved, which means and . From eq.(5) From eq.(6) Let and put from eq.(7) to eq.(8) gives Therefore, ratio or . Since , ratio

Fyneman exercise I ch39-1

Q. If an ideal gas is compressed adiabatically, we have found (Eq. 39-14) that . On the other hand, under all conditions, . Combine these to deduce how and , or and , are connected during an adiabatic compression. A. P & T Putting to gives V & T Putting to gives

Fyneman exercise II ch1-1

Q. Electric and gravitational forces. a) What would a proton’s mass be if the gravitational force between two protons at rest were to equal the electric force? How does this compare with its actual mass? b) What would be the electric force between two dimes placed at opposite ends of a 10 meter lecture table if their nuclear and electronic charges were unbalanced by about 1%? Can you think of some object whose “weight” equals that force? A. symbol meaning value or Elementary charge coulombs Vacuum permittivity F/m G Gravitational constant Proton mass kg a) Let proton’s mass , gravitational force , electric force and distance between two protons . Gravitational force equals electric force, Using above numbers in table, we get kg Electric force is very very strong than gravitational force! b) Mass of dime = . Suppose dime is composed only ...