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Schaum's Outline of Physics for Engineering and Science, 4th Edition
by Michael E. Browne
Tough test questions? Missed lectures? Not enough time? Fortunately, there's Schaum's. More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Schaum's is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills. Schaum's Outline of Physics for Engineering and Science, Fourth Edition, features: • 788 fully solved problems • Succinct review of physics topics such as motion, energy, fluids, waves, heat, and magnetic fields • Clear, concise explanations of all physics for engineering and science concepts • Support for all the leading textbooks in physics for engineering and science • Content that is appropriate for Principles of Physics, Elements of Physics, Introductory College Physics, General Physics, Physics for Engineering courses • Access to revised Schaums.com website and new app, containing 25 problem-solving videos and more Schaum's reinforces the main concepts required in your course and offers hundreds of practice exercises to help you succeed. Use Schaum's to shorten your study time—and get your best test scores!
Book Chapter
29. Alternating Current Circuits
Consider an ideal transformer consisting of two coils, one with N1 turns (the primary) and one with N2 turns (the secondary). Assume all of the magnetic flux from one coil goes through the other. This can be accomplished by wrapping one coil on top of the other or, as is done more commonly, by wrapping both coils on an iron core (Fig. 29-1). In the latter case, the magnetic field lines tend to stay completely within the iron. The lines between the coils indicate that the transformer is filled with iron. Suppose an ac voltage V1 is applied to the primary. Since I assume no resistance in the coils, the applied voltage will equal the induced back EMF. If the flux through one coil is ΦB, then Similarly, the voltage across the secondary coil will be Divide these equations and obtain (29.1) If the transformer is ideal and has no losses, then the power into the primary is equal to the power out of the secondary, so P = I1V1 = I2V2. Thus (29.2) A transformer may be used to step up or step down a voltage or a current. PROBLEM 29.1. A transformer used to operate a neon sign steps 120 Vac up to 6000 V. The primary has 150 turns. How many turns does the secondary have? Solution PROBLEM 29.2. A filament transformer is required to provide 32 A to a 0.2-Ω load. The primary is operated off a 120-Vac line. What turns ratio is required for the transformer? Solution The secondary voltage required is V2 = I2R = (32 A)(0.2 Ω) = 6.4 V. Thus PROBLEM 29.3. In an autotransformer a single coil is used for both the primary and the secondary. In the coil shown here there are three leads attached to a coil wrapped on an iron core. Any two leads may be used as the primary, and any two may be used as the secondary. Between A and B there are 300 turns, and between B and C there are 900 turns. What output voltages are possible for a primary input voltage of 120 Vac? Solution Possible turns ratios are 300 : 900, 300 : 1200, 900 : 1200, and the reciprocals of these. Hence possible turns ratios are 1 : 3, 1 : 4, 3 : 4, 3 : 1, 4 : 1, and 4:3. 1:1 is also possible (this might be done to provide dc isolation between two circuits). Hence possible secondary voltages are 30, 40, 90, 120, 160, 360, and 480 Vac.
29. Alternating Current Circuits
Figure 29-1 
Book Chapter
29. Alternating-Current Circuits
The frequency of an alternating current is the number of complete back-and-forth cycles it goes through each second (Fig. 29-1). As in the case of harmonic motion, the unit of frequency is the hertz (Hz), where 1 Hz = 1 cycle/s. An alternating emf of frequency f whose maximum value is V max varies with time according to the formula The quantity ω = 2π f is the angular frequency of the emf in radians per second. Similarly, an alternating current (ac) of frequency f whose maximum value is I max varies with time according to the formula
29. Alternating-Current Circuits
Figure 29-1 
Book Chapter
11. Angular Momentum
The angular momentum with respect to the origin of a particle with position r and momentum p = mv is (11.1) If the angle between r and p is θ, then the magnitude of L is (11.2) If r and p lie in the xy plane, L is along the z axis (Fig. 11-1). The time rate of change of the angular momentum is (11.3) The cross product of a vector with itself is always zero, so From Newton’s second law, F = dp/dt. Thus (11.4) If we think of a rigid body as a collection of particles of mass mi, the z component of the angular momentum can be expressed in terms of the moment of inertia, (11.5) This expression is analogous to p = mv, where I is like m and ω is like v. Apply Eq. 11.4 to a rigid body. All terms involving internal forces cancel, so (11.6) Equation 11.6 is analogous to Fext = dp/dt for translational motion. If no external torque acts on a system, the angular momentum of the system remains constant. (11.7) This is the law of conservation of angular momentum. PROBLEM 11.1. In an interesting lecture demonstration, a student sits in a swivel chair. She has moment of inertia Is about a vertical axis. She holds vertical the axis of a bicycle wheel of moment of inertia IB ≪ Is spinning with large angular velocity ω1. The wheel is spinning counterclockwise, as viewed from above. Now she rotates the wheel axis by 180°. What happens? Solution No external torque acts (friction is negligible) so the angular momentum of the system remains constant. Initially L1 = Iwω1, directed upward. After the wheel axis is inverted, the angular momentum vector of the wheel will point downward, so the chair will rotate counterclockwise with angular velocity ω2 so that the total angular momentum remains unchanged in magnitude and points upward. PROBLEM 11.2. A disk is mounted with its axis vertical. It has radius R and mass M. It is initially at rest. A bullet of mass m and velocity v is fired horizontally and tangential to the disk. It lodges in the perimeter of the disk. What angular velocity will the disk acquire? Solution Angular momentum is conserved. The initial angular momentum is just the angular momentum of the bullet, L1 = mvR. After the collision PROBLEM 11.3. The force of gravity on the Earth due to the Sun exerts negligible torque on the Earth (assuming the Earth to be spherical), since this force is directed along the line joining the centers of the two bodies. The Earth travels in a slightly elliptical orbit around the Sun. When it is nearest the Sun (the perihelion position), it is 1.47 × 108km from the Sun and traveling 30.3 km/s. The Earth’s farthest distance from the Sun (aphelion) is 1.52 × 108km. How fast is the Earth moving at aphelion? Solution No torque acts, so the Earth’s angular momentum is constant.
11. Angular Momentum
Figure 11-1 
Book Chapter
34. Atomic Physics
Under certain conditions moving bodies exhibit wave properties. The quantity whose variations constitute the matter waves (or de Broglie waves) of a moving body is known as its wave function ψ (Greek letter psi). The likelihood of finding the body at a particular time and place is proportional to the value of ψ 2 at that time and place. A large value of ψ 2 signifies a high probability of finding the body; a small value of ψ 2 signifies a low probability of finding the body. The matter waves associated with a moving body are in the form of a group of waves that travels with the same velocity as the body. The wavelength of the matter waves of a body of mass m and velocity v is Find the de Broglie wavelengths of (a) a 46-g golf ball moving at 30 m/s, and (b) an electron moving at 10 7 m/s. An electron microscope uses a beam of fast electrons that are focused by electric and magnetic fields to produce an enlarged image of a thin specimen on a screen or photographic plate. Find the resolving power of an electron microscope which uses 15-keV electrons by assuming that this is equal to the electron wavelength. The kinetic energy of the electrons is Since The electron wavelength is therefore
34. Atomic Physics
34.1.1. SOLVED PROBLEM 34.1
The wavelength of the golf ball is so small compared with its dimensions that we would not expect to find any wave aspects in its behavior.
The dimensions of atoms are comparable with this figure—the radius of the hydrogen atom, for instance, is 5.3 × 10 −11 m. It is therefore not surprising that the wave character of moving electrons is the key to understanding atomic structure and behavior.34.1.2. SOLVED PROBLEM 34.2
, the velocity of the electrons is
Book Chapter
35. Atoms and Photons
The story begins around 1900 when attempts were made to understand the radiation emitted by hot blackbodies. It was well known that when an object is heated, it radiates energy. If a horseshoe is held in a flame, it first radiates dull red light, and as it gets hotter and hotter, the intensity increases and the light becomes brighter red and then yellow. This behavior is shown in Fig. 35-1. Classical theory predicted that the intensity of radiation should continue to increase indefinitely at shorter wavelengths, but instead the curves start turning downward around 400 nm (the “ultraviolet catastrophe”). Max Planck was able to reproduce the experimental curves with a theory based on the radical assumption that radiation of frequency f and wavelength λ is emitted only in discrete packets, which he called photons. He postulated that the energy of a photon is (35.1) Here h is Planck’s constant: h = 6.63 × 10−34 J · s. On this basis he calculated the intensity of blackbody radiation to be (35.2) kB is Boltzmann’s constant. Planck envisioned a collection of radiators whose energy was E = nhf = n(hc/λ), where n = integer. When the system emits a photon, it drops down one energy level, and when it absorbs a photon, it moves up one energy level. Such discrete energy levels are quantized, and the packets of radiation are called photons or quanta. PROBLEM 35.1. Calculate the energy in joules and electronvolts for the following photon wavelengths: 1 × 10−10 m (x ray), 200 nm (ultraviolet), 500 nm (yellow light near the peak of the Sun’s spectrum), 10 μm (infrared radiated by your body), and 200 m (AM radio station). Solution Substituting numerical values in E = hc/λ and using 1 eV = 1.6 × 10−19 J, E = 2.0 × 10−15 J = 12.4keV (x rays), E = 9.9 × 10−19 J = 6.2 eV (ultraviolet) E = 4.0 × 10−19 J = 2.5 eV (yellow light), E = 9.9 × 10−28 J = 6.2 × 10−9 eV (radio waves) Note that it usually takes a few electronvolts to break chemical bonds, which is why UV light is much more likely to cause skin cancer and tanning than is visible light. X rays and gamma rays are even more dangerous. From Eq. 35.1 we can deduce the following useful relation for blackbody radiation: (35.3) T is the absolute temperature, and λmax is the wavelength at which the radiation intensity is a maximum. PROBLEM 35.2. The peak of the Sun’s spectrum is at 484 nm. Estimate the temperature of the Sun’s surface. Human skin is at about 33°C (306 K). At what wavelength is the peak of its radiation, assuming it is a blackbody? Solution We can integrate Eq. 35.1 to obtain the total energy radiated by a blackbody. The energy radiated per meter squared per second is (35.4) Here the Stefan-Boltzmann constant is σ = 5.67 × 10−8 W/m2 · K4. PROBLEM 35.3. Suppose that from a star’s radiation spectrum its surface temperature is determined to be 5400 K. Astronomical observation determines its distance from Earth to be 5.2 × 1018 m, and the starlight reaching us has intensity 1.4 × 10−4 W/m2. From this data, estimate the size of the star. Solution The surface area of a star of radius r is 4πr2, so the total power radiated is P = 4πr2S, where S = σT4. By the time the light reaches us, it is spread over a sphere of radius R, so P = 4πR2Se. Equating these, 4πr2σT4 = 4πR2Se: so r = 3.9 × 108 m about half as big as the Sun.
35. Atoms and Photons
Figure 35-1 
Book Chapter
23. Capacitance
Consider two conductors, on one of which is charge +Q and on the other −Q. The capacitance of the conductors is defined as (23.1) We measure capacitance in farads (F): Note that the capacitance of two conductors does not depend on whether or not they are charged, just as the capacity of a 5-gal bucket does not depend on whether or not it contains water. To find the capacitance of two conductors, use the following procedure: Place charge +Q on one conductor and −Q on the other. Calculate the electric field in the space between the conductors. Usually this requires using Gauss' law. Determine the potential difference between the conductors by integrating: Choose any convenient path from conductor A to conductor B. It is only the magnitude of V that counts, so I dropped the minus sign in front of the integral. Since E depends on Q, you will obtain an expression of the form V = (constant) Q, where the constant is 1/C. PROBLEM 23.1. Calculate the capacitance per unit area of two large parallel plates of area A separated by a distance d. Solution Using Gauss' law, I found that E = σ/ϵ0. Thus and (23.2) PROBLEM 23.2. Calculate the capacitance of two plates of area 4 cm2 separated by 1 mm. Solution Note: 1 pF = 1 picofarad = 10−12 F. PROBLEM 23.3. Calculate the capacitance per unit length of a long, straight coaxial line made of two concentric cylinders of radii R1 and R2. Solution Using Gauss' law I found that so Thus (23.3) PROBLEM 23.4. Calculate the capacitance of two concentric spheres of radii R1 and R2. From Gauss' law I found that Thus (23.4) When R2 → ∞, C → 4πϵ0R1. This describes the capacitance of an isolated sphere. Whenever we speak of the capacitance of an isolated conductor, we think of the second conductor as being at infinity.
23. Capacitance
Book Chapter
26. Capacitance
Click here for the Inductance, Capacitance, Resonance and Time Constants - Basic Calculations spreadsheet calculator. A capacitor is a system that stores energy in the form of an electric field. In its simplest form, a capacitor consists of a pair of parallel metal plates separated by air or other insulating material. The potential difference V between the plates of a capacitor is directly proportional to the charge Q on either of them, so the ratio Q/V is always the same for a particular capacitor. This ratio is called the capacitance C of the capacitor: The unit of capacitance is the farad (F), where 1 farad = 1 coulomb/volt. Since the farad is too large for practical purposes, the microfarad and picofarad are commonly used, where A charge of 10 −6 C on each plate of 1-μF capacitor will produce a potential difference of V = Q/C = 1 V between the plates.
26. Capacitance
Book Chapter
6. Circular Motion
Whenever a moving object turns, its velocity changes direction. Since acceleration is a measure of the rate of change of velocity, an object that turns is accelerating. This kind of acceleration, called centripetal or radial acceleration, is related to υ, the speed, and r, the radius of the curvature of the turn, by ac = υ2/r. This relationship was obtained in Section 4.4. There we saw that there were two kinds of acceleration. Tangential acceleration (I called this “speeding up or slowing down” acceleration) measures the rate of change of speed. Radial, or centripetal, acceleration (“turning” acceleration) measures the rate of change of velocity associated with changing direction. Radial acceleration is directed perpendicular to the velocity vector and points toward the center of the arc on which the object is moving. If an object turns left, it accelerates left. If it turns right, it accelerates right. From Newton’s second law of motion we saw that in order for an object to accelerate, it must be subject to a net force. This is illustrated in Fig. 6-1. The amount of force needed to cause an object with speed υ to curve along an arc of radius r is thus (6.1) Observe that Fc is not a kind of force. Many different kinds of forces can be used to make an object turn. For example, gravity causes the Moon to curve and travel a circular path around the Earth. The tension in a rope causes a tether ball to travel in a circle. The normal force exerted by a banked curve on a highway causes a car to travel a circular path. The friction force between a car’s tire and the roadway causes the car to turn. When you draw a force diagram, do not draw in a force labeled Fc. Centripetal force is a way of using a force, not a kind of force. An object does not have to travel in a complete circle to experience centripetal acceleration. However, frequently objects do travel around and around, as is the case with a spinning wheel or compact disk. Suppose that an object makes f rev/s. The frequency of revolution is f. One revolution per second is called 1 hertz (a dumb way of labeling something, but we’re stuck with it). Later we will extend this idea of frequency to anything that varies periodically, whether or not it moves in a circle. Thus the electricity in your house varies at 60 times per second, or 60 Hz. The AM radio station in my town broadcasts radio waves at a frequency of 1400 kHz. (The announcer always says, “KRPL at 1400 on your AM dial.” He means 1,400,000 variations per second in the radio wave electric field.) In one revolution an object travels a distance 2 πr. If it makes f rev/s, the distance traveled in 1 s is 2 πrf. The distance traveled per second is the speed υ, so (6.2) One revolution is 2 radians (rad). The number of radians swept out per second is called the angular frequency or angular velocity, in radians per second (both terms are used). This symbol ω that looks like a small w is a lowercase Greek omega. It is measured in radians per second. Thus, (6.3) and (6.4) In terms of ω the centripetal force can be written (6.5) When dealing with objects rotating at constant frequency, it is easiest to use Eq. 6.5. When an object is simply turning, such as a jet plane pulling out of a dive, use Eq. 6.1. PROBLEM 6.1. A ball of mass 0.15 kg slides with negligible friction on a horizontal plane. The ball is attached to a pivot by means of a string 0.60 m long. The ball moves around a circle at 10 rev/s. What is the tension in the string? Solution The ball stays in the horizontal plane, so the upward normal force exerted by the table balances the downward force of gravity (the weight of the ball). In the horizontal plane only one force acts on the ball, the tension of the string. Thus a net force is acting on the ball, and it is not in equilibrium. It is accelerating with acceleration ac. The force required to cause this acceleration is Note that the units for ω and f are s−1. Radians and revolutions are not “units” as such. PROBLEM 6.2. In a popular carnival ride a person sits in a chair attached by means of a cable to a tall central post. The pole is spun, causing the rider to travel in a horizontal circle, with the cable making an angle θ with the vertical pole. A contraption like this is called a conical pendulum. Suppose the rider and chair have mass of 150 kg. If the cable length is 8 m, at what frequency should the chair rotate if the cable is to make an angle of 60° with vertical? What is the tension in the cable? Solution Always begin by drawing the force diagram, as shown here. The rider is not moving up or down, so the vertical forces are in balance. The component of the cable tension T directed horizontally, that is, toward the center of rotation, is a net force causing a centripetal acceleration toward the center. The magnitude of this required net force is given by Eq. 6.5: T sin θ = mrω2, where r = L sin θ. Thus T sin θ = mL sin θω2, T = mLω2, and so PROBLEM 6.3. On a level roadway the coefficient of friction between the tires of a car and the asphalt is 0.80. What is the maximum speed at which a car can round a turn of radius 25 m if the car is not to slip? Solution The force of friction provides the needed turning force Fc. Thus PROBLEM 6.4. A car traveling on a freeway goes around a curve of radius r at speed υ. The roadway is banked to provide the necessary inward centripetal force in order for the car to stay in its lane. At what angle should the roadway be banked if the car is not to utilize friction to make the turn? Solution Draw the force diagram. The car is not accelerating in the vertical direction, so Fup = Fdown, and N cos θ = mg. The horizontal component of the normal force N provides the needed centripetal force: Divide: Caution: You might be tempted to resolve the weight and the normal force into components parallel and perpendicular to the road surface. You might then imagine that the components perpendicular to the surface are in balance. This is wrong because the road surface is not truly directed perpendicular to the plane of the paper. Our drawing is somewhat misleading in this respect. The normal force is actually larger than the car’s weight since it must support the weight and also provide an inward force to make the car turn. PROBLEM 6.5. In the “Human Fly” carnival ride a bunch of people stand with their backs to the wall of a cylindrical room. Once everyone is in place, the room begins to spin. The inward normal force exerted by the wall on the back of each person provides the needed centripetal force to ensure that each person travels in a circle of diameter equal to the diameter of the room. Once the room is spinning rapidly, the floor drops out from beneath the people. Friction between the wall and each person’s back “glues” each one to the wall, although with some effort they can squirm and move about (like human flies on a wall). Personally, this is not my cup of tea. I get motion sickness, but kids love it. What minimum coefficient of friction is needed if the people are not to slip downward, assuming the room diameter is 8.0 m and the room spins at 18 rev/min? Solution For no slipping down, Ff = mg and N = mrω2. Thus μmrω2 = mg: PROBLEM 6.6. An F14 jet fighter traveling 260 m/s (about 580 mi/h) pulls out of a vertical dive by turning upward along a circu-lar arc of radius 2.4 km. What acceleration does the pilot experience? Express the result as a multiple of g. If the pilot’s weight is 560 N, what force does the seat exert on him? (This is his “apparent weight.”) Even with a pressurized suit, the maximum acceleration a person can experience without suffering brain hemorrhaging resulting in a blackout is about 11g. Thus a pilot must avoid turning too sharply so that he or she won’t black out and the wings won’t break off the airplane. Solution Here g is the acceleration due to gravity. At the lowest point in the dive, gravity acts downward on the pilot with force W, and the seat pushes up with a normal force N, so so PROBLEM 6.7. A woman stands a distance of 2.40 m from the axis of a rotating merry-go-round platform. The coefficient of friction between her shoes and the platform surface is 0.60. What is the maximum number of revolutions per minute the merry-go-round can make if she is not to start slipping outward? Solution Friction provides the needed inward centripetal force, so Fc = mrω2 = Ff · Ff = μmg, so PROBLEM 6.8. Suppose you are driving at speed υ0 and find yourself heading straight for a brick wall that intersects the line of your path at 90°. Assuming that the coefficients of friction for stopping and for turning are the same, are your chances of avoiding a crash better if you continue straight ahead while braking or if you simply turn along a circular path at a constant speed? Solution The braking force is Ff = μmg, so the braking acceleration is Ff/m = –μg. From Eq. 3.9, For turning, friction provides the centripetal force, so Thus x <>r, and it is better to brake than to turn. PROBLEM 6.9. The first space habitats built by humans will most likely be cylindrical in shape. It is envisioned that such a space habitat will be rotated about the axis of the cylinder in order to simulate the effect of gravity here on Earth. Thus an inhabitant will feel the floor pushing on his or her feet with a force of N = mrω2 in order to cause him or her to move along a circular path as the space cylinder rotates. He or she will then experience a sort of “artificial gravity” pulling downward to counter the upward force of the floor. (Downward will be radially out.) Preliminary NASA designs have been developed for a cylinder about 6.4 km in diameter and 32 km in length. Later much larger structures could be built. At what rate would such a structure have to rotate in order to simulate the same acceleration due to gravity found here on Earth? Solution We require mg = mrω2 so Note that your apparent weight mrω2 decreases as r becomes smaller, that is, as you approach the axis of the cylinder. This could have important applications. For example, it is very difficult to hospitalize severely burned patients since they must lie on open wounds. Near the center of the space habitat, a person would feel “weightless” and could simply float above a bed with very little support. PROBLEM 6.10. The inward centripetal force required to cause any object to move in a circle can be very large when rigid objects are rotated at high frequency, as is the case in most machines. Spinning gears and wheels can be subject to huge forces that can cause them to fracture and cause serious damage. My father, a machinist, lost an eye when a grinding wheel he was using fractured and sent fragments in all directions. Some of the pieces struck his face. Experimental cars have been designed that are propelled by the energy stored in spinning flywheels (as an alternative to using internal combustion engines), but a limiting factor in the use of such machines is their ability not to fracture when rotated at high speed. To gain an understanding of the forces involved when objects rotate, consider the following simple model. A very light rod of length L is rotated in a horizontal plane with one end fixed. At one end is attached a mass m1, and at the center of the rod is attached a mass m2. Determine the tension T1 in the portion of the rod between m1 and m2 and the tension in the rod between m2 and the axis of rotation. Solution The forces acting on m1 and m2 are drawn here. Note that the tension T1 in the outer portion of the rod pulls inward on m1 and outward on m2. There must be a net inward centripetal force on each mass, so T2 > T1. Applying Fc = mac to each mass, T1 = m1r1(2πf)2. T2 – T1 = m2r2 (2 πf)2, where r1 = L, and r2 = 1/2L. Thus, and If m1 = m2, we see that
6. Circular Motion
Figure 6-1 
