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Page 160 of 160 (1595 Questions and Answers)

1591. I have an active but paraplegic friend who is building an electric off-road scooter using DC motors. Those motors will have to reverse directions frequently while under load. Will they tolerate immediate reversals, or must there be a delay? — JO, Valley Springs, California
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Modern brushless DC motors are amazing devices that can handle torque reversals instantly. In fact, they can even generate electricity during those reversals!

Instant reversals of direction, however, aren't physically possible (because of inertia) and aren't actually what your friend wants anyway. I'll say more about the distinction between torque reversals and direction reversals in a minute.

In general, a motor has a spinning component called the rotor that is surrounded by a stationary component called the stator. The simplest brushless DC motor has a rotor that contains permanent magnets and a stator that consists of electromagnets. The magnetic poles on the stator and rotor can attract or repel one another, depending on whether they like or opposite poles—like poles repel; opposite poles attract.

Since the electronics powering the stator's electromagnets can choose which of the stator's poles are north and which are south, those electronics determine the forces acting on the rotor's poles and therefore the direction of torque on the rotor. To twist the rotor forward, the electronics make sure that the stator's poles are always acting to pull or push the rotor's poles in the forward direction so that the rotor experiences forward torque. To twist the rotor backward, the electronics reverses all those forces.

Just because you reverse the direction of torque on the rotor doesn't mean that the rotor will instantly reverse its direction of rotation. The rotor (along with the rider of the scooter) has inertia and it takes time for the rotor to slow to a stop and then pick up speed in the opposite direction. More specifically, a torque causes angular acceleration; it doesn't cause angular velocity. During that reversal process, the rotor is turning in one direction while it is being twisted in the other direction. The rotor is slowing down and it is losing energy, so where is that energy going? It's actually going into the electronics which can use this electricity to recharge the batteries. The "motor" is acting as a "generator" during the slowing half of the reversal!

That brushless DC motors are actually motor/generators makes them fabulous for electric vehicles of all types. They consume electric power while they are making a vehicle speed up, but they generate electric power while they are slowing a vehicle down. That's the principle behind regenerative braking—the vehicle's kinetic energy is used to recharge the batteries during braking.

With suitable electronics, your friend's electric scooter can take advantage of the elegant interplay between electric power and mechanical power that brushless DC motors make possible. Those motors can handle torque reversals easily and they can even save energy in the process. There are limits, however, to the suddenness of some of the processes because huge flows of energy necessitate large voltages and powers in the motor/generators and their electronics. The peak power and voltage ratings of all the devices come into play during the most abrupt and strenuous changes in the motion of the scooter. If your friend wants to be able to go from 0 to 60 or from 60 to 0 in the blink of eye, the motor/generators and their electronics will have to handle big voltages and powers.


1592. Suppose you have a metal sphere in vacuum and you begin putting electric charge on that sphere. Neglecting possible discharges, how much charge can the sphere store? An unlimited amount? -- BC
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By asking me to "neglect possible discharges," you're asking me to neglect what actually happens. There will be a discharge, specifically a phenomenon known as "field emission". Neglect that discharge, then yes, the sphere can in principle store an unlimited amount of charge. But on route to infinity, I will have had to ignore several other exotic discharges and then the formation of a black hole.

What will really happen is a field emission discharge. The repulsion between like charges will eventually become so strong that those charges will push one another out of the metal and into the vacuum, so that charges will begin to stream outward from the metal sphere.

Another way to describe that growing repulsion between like charges involves fields. An electric charge is surrounded by a structure in space known as an electric field. An electric field exerts forces on electric charges, so one electric charge pushes on other electric charges by way of its electric field.

As more and more like charges accumulate on the sphere, their electric fields overlap and add so that the overall electric field around the sphere becomes stronger and stronger. The charges on the sphere feel that electric field, but they are bound to the metal sphere by chemical forces and it takes energy to pluck one of them away from the metal.

Eventually, the electric field becomes so strong that it can provide the energy needed to detach a charge from the metal surface. The work done by the field as it pushes the charge away from sphere supplies the necessary energy and the charge leaves the sphere and heads out into the vacuum. The actually detachment process involves a quantum physics phenomenon known as tunneling, but that's another story.

The amount of charge the sphere can store before field emission begins depends on the radius of the sphere and on whether the charge is positive or negative. The smaller that radius, the faster the electric field increases and the sooner field emission starts. It's also easier to field-emit negative charges (as electrons) than it is to field-emit positive charges (as ions), so a given sphere will be able to hold more positive charge than negative charge.


1593. Why can't the Japanese stop the chain reaction in the Fukushima Daiichi nuclear reactors? — FE
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The Japanese did stop the chain reactions in the Fukushima Daiichi reactors, even before the tsunami struck the plant. The problem that they're having now is not the continued fissioning of uranium, but rather the intense radioactivity of the uranium daughter nuclei that were created while the chain reactions were underway. Those radioactive fission fragments are spontaneously decaying now and there is nothing that can stop that natural decay now. All they can do now is to try to contain those radioactive nuclei, keep them from overheating, and wait for them to decay into stable pieces.

The uranium atom has the largest naturally occurring nucleus in nature. It contains 92 protons, each of which is positively charged, and those 92 like charges repel one another ferociously. Although the nuclear force acts to bind protons together when they touch, the repulsion of 92 protons alone would be too much for the nuclear force—the protons would fly apart in almost no time.

To dilute the electrostatic repulsion of those protons, each uranium nucleus contains a large number of uncharged neutrons. Like protons, neutrons experience the attractive nuclear force. But unlike protons, neutrons don't experience the repulsive electrostatic force. Two neutron-rich combinations of protons and neutrons form extremely long-lived uranium nuclei: uranium-235 (92 protons, 143 neutrons) and uranium-238 (92 protons, 146 neutrons). Each uranium nucleus attracts an entourage of 92 electrons to form a stable atom and, since the electrons are responsible for the chemistry of an atom, uranium-235 and uranium-238 are chemically indistinguishable.

When the thermal fission reactors of the Fukushima Daiichi plant were in operation, fission chain reactions were shattering the uranium-235 nuclei into fragments. Uranium-238 is more difficult to shatter and doesn't participate much in the reactor's operation. On occasion, however, a uranium-238 nucleus captures a neutron in the reactor and transforms sequentially into neptunium-239 and then plutonium-239. The presence of plutonium-239 in the used fuel rods is one of the problems following the accident.

The main problem, however, is that the shattered fission fragment nuclei in the used reactor fuel are overly neutron-rich, a feature inherited from the neutron-rich uranium-235 nuclei themselves. Midsize nuclei, such as iodine (with 53 protons), cesium (with 55 protons), and strontium (with 38 protons), don't need as many neutrons to dilute out the repulsions between their protons. While fission of uranium-235 can produce daughter nuclei with 53 protons, 55 protons, or 38 protons, those fission-fragment versions of iodine, cesium, and strontium nuclei have too many neutrons and are therefore unstable—they undergo radioactive decay. Their eventual decay has nothing to do with chain reactions and it cannot be prevented.

How quickly these radioactive fission fragment nuclei decay depends on exactly how many protons and neutrons they have. Three of the most common and dangerous nuclei present in the used fuel rods are iodine-131 (8 days half-life), cesium-137 (30 year half-life), and strontium-90 (29 year half-life). Plutonium-239 (24,200 year half-life) is also present in those rods. When these radioactive nuclei are absorbed into the body and then undergo spontaneous radioactive decay, they damage molecules and therefore pose a cancer risk. Our bodies can't distinguish the radioactive versions of these chemical elements from the nonradioactive ones, so all we can do to minimize our risk is to avoid exposure to them or to encourage our bodies to excrete them by saturating our bodies with stable versions.


1594. Are smart meters bad for people’s health? Is this actually not knowable at this time? -- ED
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If by smart meters you mean the devices that monitor power usage and possibly adjust power consumption to periodically, then I don't see how they can affect health. Their communications with the smart grid are of no consequence to human health and having the power adjusted on household devices is unlikely to be a health issue (unless they cut off your power during a blizzard or a deadly heat wave).

The radiated power from all of these wireless communications devices is so small that we have yet to find mechanisms whereby they could cause significant or lasting injury to human tissue. If there is any such mechanism, the effects are so weak that the risk associated with it are dwarfed by much more significant risks of wireless communication: the damage to traditional community, the decline of ordinary human interaction, and the surge in distracted driving.


1595. If a penny fell from the Empire State Building, could it actually punch a hole in the sidewalk?
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A famous urban legend states that a penny dropped from the top of the Empire State Building will punch a hole in the sidewalk below. Given the height of the building and the hardness of the penny, that seems like a reasonable possibility. Whether it's true or not is a matter that can be determined scientifically. Before we do that, though, let's get some background.

Falling rocks can be dangerous and, the farther they fall, the more dangerous they become. Falling raindrops, snowflakes, and leaves, however, are harmless no matter how far they fall. The distinction between those two possibilities has nothing to do with gravity, which causes all falling objects to accelerate downward at the same rate. The difference is entirely due to air resistance.

Air resistance—technically known as drag—is the downwind force an object experiences as air moves passed it. Whenever an object moves through the air, the two invariably push on one another and they exchange momentum. The object acts to drag the air along with it and the air acts to drag the object along with it, action and reaction. Those two aerodynamic forces affect the motions of the object and air, and are what distinguish falling snowflakes from falling rocks.

Two types of drag force affect falling objects: viscous drag and pressure drag. Viscous drag is the friction-like effect of having the air rub across the surface of the object. Though important to smoke and dust particles in the air, viscous drag is too weak to affect larger objects significantly.

In contrast, pressure drag is strongly affects most large objects moving through the air. It occurs when airflow traveling around the object breaks away from the object's surface before reaching the back of the object. That separated airflow leaves a turbulent wake behind the object—a pocket of air that the object is effectively dragging along with it. The wider this turbulent wake, the more air the object is dragging and the more severe the pressure drag force.

The airflow separation occurs as the airflow is attempting to travel from the sides of the object to the back of the object. At the sides, the pressure in the airflow is especially low due as it bends to arc around the sides. Bernoulli's equation is frequently invoked to help explain the low air pressure near the sides of the object. As this low-pressure air continues toward the back of the object, where the pressure is much greater, the airflow is moving into rising pressure and is pushed backward. It is decelerating.

Because of inertia, the airflow could be expected to reach the back of the object anyway. However, the air nearest the object's surface—boundary layer air—rubs on that surface and slows down. This boundary layer doesn't quite make it to the back of the object. Instead, it stops moving and consequently forms a wedge that shaves much of the airflow off of the back of the object. A turbulent wake forms and the object begins to drag that wake along with it. The airflow and object are then pushing on one another with the forces of pressure drag.

Those pressure drag forces depend on the amount of air in the wake and the speed at which the object is dragging the wake through the passing air. In general, the drag force on the object is proportional to the cross sectional area of its wake and the square of its speed through the air. The broader its wake and the faster it moves, the bigger the drag force it experiences.

We're ready to drop the penny. When we first release it at the top of the Empire State Building, it begins to accelerate downward at 9.8 meters-per-second2—the acceleration due to gravity—and starts to move downward. If no other force appeared, the penny would move according to the equations of motion for constant downward acceleration, taught in most introductory physics classes. It would continue to accelerate downward at 9.8 meters-per-second2, meaning that its downward velocity would increase steadily until the moment it hit sidewalk. At that point, it would be traveling downward at approximately 209 mph (336 km/h) and it would do some damage to the sidewalk.

That analysis, however, ignores pressure drag. Once the penny is moving downward through the air, it experiences an upward pressure drag force that affects its motion. Instead of accelerating downward in response to its weight alone, the penny now accelerates in response to the sum of two force: its downward weight and the upward drag force. The faster the penny descends through the air, the stronger the drag force becomes and the more that upward force cancels the penny's downward weight. At a certain downward velocity, the upward drag force on the penny exactly cancels the penny's weight and the penny no longer accelerates. Instead, it descends steadily at a constant velocity, its terminal velocity, no matter how much farther drops.

The penny's terminal velocity depends primarily on two things: its weight and the cross sectional area of its wake. A heavy object that leaves a narrow wake will have a large terminal velocity, while a light object that leaves a broad wake will have a small terminal velocity. Big rocks are in the first category; raindrops, snowflakes, and leaves are in the second. Where does a penny belong?

It turns out that a penny is more like a leaf than a rock. The penny tumbles as it falls and produces a broad turbulent wake. For its weight, it drags an awful lot of air behind it. As a result, it reaches terminal velocity at only about 25 mph (40 km/h). To prove that, I studied pennies fluttering about in a small vertical wind tunnel.

Whether the penny descends through stationary air or the penny hovers in rising air, the physics is the same. Of course, it's much more convenient in the laboratory to observe the hovering penny interacting with rising air. Using a fan and plastic pipe, I created a rising stream of air and inserted a penny into that airflow.

At low air speeds, the penny experiences too little upward drag force to cancel its weight. The penny therefore accelerated downward and dropped to the bottom of the wind tunnel. At high air speeds, the penny experienced such a strong upward drag force that it blew out of the wind tunnel. When the air speed was just right, the penny hovered in the wind tunnel. The air speed was then approximately 25 mph (40 km/h). That is the terminal velocity of a penny.

The penny tumbles in the rising air. It is aerodynamically unstable, meaning that it cannot maintain a fixed orientation in the passing airstream. Because the aerodynamic forces act mostly on the upstream side of the penny, they tend to twist that side of the penny downstream. Whichever side of the penny is upstream at one moment soon becomes the downstream side, and the penny tumbles. As a result of this tumbling, the penny disturbs a wide swath of air and leaves a broad turbulent wake. It experiences severe pressure drag and has a low terminal velocity.

The penny is an example of an aerodynamically blunt object—one in which the low-pressure air arcing around its sides runs into the rapidly increasing pressure behind it and separates catastrophically to form a vast wake. The opposite possibility is an aerodynamically streamlined object—one in which the increasing pressure beyond the object's sides is so gradual that the airflow never separates and no turbulent wake forms. A penny isn't streamlined, but a ballpoint pen could be.

Almost any ballpoint pen is less blunt than a penny and some pens are approximately streamlined. Moreover, pens weigh more than pennies and that fact alone favors a higher terminal velocity. With a larger downward force (weight) and a smaller upward force (drag), the pen accelerates to a much greater terminal velocity than the penny. If it is so streamlined that it leaves virtually no wake, like the aerofoil shapes typical of airplane components, it will have an extraordinarily large terminal velocity—perhaps several hundred miles per hour.

Some pens tumble, however, and that spoils their ability to slice through the air. To avoid tumbling, a pen must "weathervane"—it must experience most of its aerodynamic forces on its downstream side, behind its center of mass. Arrows and small rockets have fletching or fins to ensure that they travel point first through the air. A ballpoint pen can achieve that same point-first flight if its shape and center of mass are properly arranged.

Almost any ballpoint pen dropped into my wind tunnel plummeted to the bottom. I was unable to make the air rise fast enough to observe hovering behavior in those pens. Whether they would tend to tumble in the open air was difficult to determine because of the tunnel's narrowness. Nonetheless, it's clear that a heavy, streamlined, and properly weighted pen dropped from the Empire State Building would still be accelerating downward when it reached the sidewalk. Its speed would be close to 209 mph at that point and it would indeed damage the sidewalk.

As a final test of the penny's low terminal velocity, I built a radio-controlled penny dropper and floated it several hundred feet in the air with a helium-filled weather balloon. On command, the dropper released penny after penny and I tried to catch them as they fluttered to the ground. Alas, I never managed to catch one properly in my hands. It was a somewhat windy day and the ground at the local park was uneven, but that's hardly an excuse—I'm simply not good at catching things in my hands. Several of the pennies did bounce off my hands and one even bounced off my head. It was fun and I was more in danger of twisting my ankle than of getting pierced by a penny. The pennies descended so slowly that they didn't hurt at all. Tourist below the Empire State Building have nothing fear from falling pennies. Watch out, however, for some of the more streamlined objects that might make that descent.


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