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

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

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.

In an interesting lecture demonstration, a student sits in a swivel chair. She has moment of inertia I_s about a vertical axis. She holds vertical the axis of a bicycle wheel of moment of inertia I_B ≪ I_s 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 = I_wω₁, 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.

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

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.

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 λ_max is the wavelength at which the radiation intensity is a maximum.

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.

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 = 4πR²S_e. Equating these, 4πr²σT⁴ = 4πR²S_e:

so

r = 3.9 × 10⁸ m

about half as big as the Sun.

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)

Calculate the capacitance of two plates of area 4 cm² separated by 1 mm.

Solution

Note: 1 pF = 1 picofarad = 10⁻¹² F.

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)

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.

Find the magnitudes of D and P for a dielectric material in which E = 0.15 MV/m and χ_e = 4.25.

Since ϵr=χe+1=5.25,

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 a_c. 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.

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

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 F_c. Thus

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 F_up = F_down, 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.

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, F_f = mg and N = mrω².

Thus μmrω² = mg:

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

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 F_c = mrω² = F_f · F_f = μmg, so

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 F_f = μmg, so the braking acceleration is F_f/m = –μg. From Eq. 3.9, v2=v02+2ax=0 when stopped. Thus the stopping distance is

For turning, friction provides the centripetal force, so

Thus x <>r, and it is better to brake than to turn.

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.

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 F_c = ma_c 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 T1=23T2. The tension is greater farther out, and it increases as the square of the frequency of rotation. Always wear eye protection when working around equipment or machinery with rotating parts.

Often no. 12 copper wire (0.205-cm diameter) is used in residential construction. What is the drift velocity of electrons in such a wire when it carries 12 A? Copper has a density of 8.95 g/cm³ and an atomic mass of 63.5 g/mol. Each copper atom provides one free electron for conductivity.

Solution

The electrons drift very slowly!