Category: centrifuges and roller coasters

What do you feel g-forces when you ride on a roller coaster? – F

Whenever you accelerate, you feel a gravity-like sensation “pulling” you in the direction opposite your acceleration. What you feel isn’t really a force—it’s really just your own inertia trying to keep you going in a straight line at a constant speed. In other words, your inertia is trying to keep you from accelerating. For example, whenever you turn left in a roller coaster, your inertia opposes your leftward acceleration and you feel “pulled” toward the right. This “pull” of inertia is sometimes called a “fictitious force” but you should remember that it isn’t a force at all, no matter how “real” it feels. Perhaps the most striking effect of acceleration occurs during your trip around a vertical loop-the-loop. When you are arcing around the top of the loop-the-loop, you are accelerating downward so quickly that you feel an enormous “fictitious force” upward. This “fictitious force” has a stronger effect on you than the real force of gravity, so you feel as though you are being pulled upward. The result is that you feel pressed into your seat, even though your seat is actually upside-down.

In today’s lecture, you stated that a person accelerating downward OR UPWARD does not feel the effects of gravity. How do you explain the g-forces felt by astronauts at escape velocity? – TH

In the lecture, I said that a person who is falling does not feel the effects of gravity, even when they are traveling upward. But when they are falling, they are accelerating downward at a very specific rate—the acceleration due to gravity, which is 9.8 meters/second2 at the earth’s surface. When an astronaut is accelerating upward during a launch, they are not falling and they do feel weight. In fact, because they are accelerating upward, they feel particularly heavy.

Does water drain in the opposite direction in the southern hemisphere? – TL

In principle, yes, but in practice, no. To explain why, I’ll begin with the origins of directional circulations on earth. Because the earth is turning, motions along its surface are complicated. The ground at the equator is actually heading eastward at more than 1000 miles per hour. The ground north or south of the equator is also heading eastward, but not as quickly. The ground’s eastward speed gradually diminishes until, at the north and south poles, there is no eastward motion at all. As a result of this non-uniform eastward motion of the ground, objects that travel in straight lines because of their inertia end up drifting eastward or westward relative to the ground. For example, if you took an object at the equator and threw it directly northward, it would drift eastward relative to the more slowly moving ground. If someone else threw an object southward from the north pole, that object would drift westward relative to the more rapidly moving ground. In the northern hemisphere, objects approaching a center tend to deflect away from that center to form a counter-clockwise circle around it. This process is reversed in the southern hemisphere so that objects approaching a center there tend to form a clockwise circle around it. Thus hurricanes are counter-clockwise in the northern hemisphere and clockwise in the southern hemisphere.

When water drains from a basin in the northern hemisphere, it flows toward a center and should have a tendency to deflect into a counter-clockwise swirl. However, the effect is very weak in a small washbasin. The direction in which the water swirls as it drains is determined by other effects such as how the water was sloshing before you opened the drain or how symmetric the basin is. For this earth’s rotation-driven swirling effect (the Coriolis effect) to dictate the direction of a circulation the objects involved must move long distances over the earth’s surface. Even tornadoes don’t always rotate in the expected direction; they’re just not big enough to be spun consistently by the Coriolis effect.

How can one prove to students that the earth rotates. Any instructions on how to build a pendulum to show rotation or some other way? – KC

There are many indirect indications that the earth rotates, including the motions of celestial objects overhead, the earth’s winds—particularly the counter-clockwise rotation of surface winds in northern hemisphere hurricanes, and the outward bulge of the earth around its equator. But for a more direct indication, a Foucault pendulum is a good choice.

Unfortunately, a Foucault pendulum isn’t easy to interpret or build. It would be easiest to interpret if it were at the north pole, where it would swing back and forth in a fixed plane as the earth turned beneath it. To a person watching the pendulum from the ground, the pendulum’s swinging arc would appear to complete one full turn each day. However, elsewhere in the northern hemisphere, the plane of the pendulum does change and the pendulum’s swinging arc will appear to complete less than one full turn each day. Nonetheless, the fact that the arc shifts at all is an indication that the ground is accelerating and that the earth is turning.

The problem with building a Foucault pendulum is that it must retain its swinging energy for hours or even days and that it must not be perturbed by activities around it. It must have a very dense, massive pendulum bob supported on a strong, thin cable and that cable must be attached to a rigid support overhead. The longer the cable is, the longer it will take the bob to complete each swing and the more slowly the pendulum will move. Slow movements are important to minimize air resistance. If I were building a Foucault pendulum, I’d find a tall empty shaft somewhere, away from any moving air, and I’d attached a lead-filled metal ball (weighing at least 100 pounds but probably more) to the top of the shaft with a thin steel cable. I’d make sure that nothing rubbed and that the top of the cable never moved. (Over the long haul, there is the issue of damage to the top of the cable because of flexure…it will eventually break here. Wrapping the cable around a drum so that there is no specific bending point helps.) Then I’d pull the pendulum away from its equilibrium position and let it start swinging slowly back and forth. Over the course of several hours, its swing would decrease, but not before we would notice that its arc had turned significantly away from the original arc because of the earth’s rotation.

What is the “optimal” shape for a pinewood derby car — I’m guessing some sort of short, flat, thin rectangle. – BP

The car’s biggest obstacle is air resistance, which in this case is a force known as “pressure drag.” The pressure drag force is proportional to the size of the turbulent wake the car creates in the air as it passes through the air. Streamlining is important to minimizing this wake. The thinner and shorter you can make the car, the smaller its wake will be. The ideal shape would be an airfoil, like those used in airplane wings and bodies. These carefully tapered shapes barely disturb the air at all and experience very little pressure drag. If you design your car to resemble a wingless commercial jet airliner, you will be doing pretty well.

What is the “optimal” weight distribution for a pinewood derby car — in front/behind, above/below the center of gravity? – BP

I’ll assume that the car starts on a slope and coasts downhill to a level finish. If that’s the case, then you want to put the car’s center of gravity as far back in the car as you can get it. That way, the center of gravity will start as high as possible in the tilted car and will finish as low as possible in the level car. During a race, the car obtains its kinetic energy (its energy of motion) from its gravitational potential energy. The farther the car’s center of gravity descends during the race, the more gravitational potential energy will be converted to kinetic energy and the faster the car will go.

When you spin an object around a fixed point, a sling for example, does the object at the end build up energy that causes it to shoot out quickly when released?

Yes. As you whip the object around on a string, you are doing work on it. You do this by making subtle movements with your hand, exerting forces that aren’t exactly toward the center of the circle. When you do this, the object begins to travel faster and faster, so its kinetic energy increases. Traveling in a circle doesn’t change this kinetic energy because kinetic energy is proportional to speed squared, and doesn’t depend on direction. Finally, when you let go of the string, the object stops circling and begins to travel in a straight line. It carries with it all the kinetic energy you gave it by whipping it about.

Why is the outward force in a loop-the-loop a “fictitious” force? Why isn’t it a “real” force?

A real force causes acceleration. If the outward “fictitious” force on a circling object were “real,” that object wouldn’t circle. It would accelerate outward. When you swing an object around on a string, you feel the object pulling outward on the string. But it isn’t itself being pulled outward by anything. What you’re feeling is the object’s inertia trying to make it travel straight. The inward force you’re exerting on it isn’t opposing some real force, it’s causing the object to accelerate inward.

Can you explain the term centripetal?

Centripetal means “directed toward a center.” A centripetal force is a force that’s directed toward a center. For example, a ball swinging around in a circle at the end of a string is experiencing a force toward the center of the circle—a centripetal force. Because the ball accelerates in the direction of the force, it accelerates centripetally. And because it experiences a fictitious force in the direction opposite its acceleration, it experiences an outward fictitious force away from the center of the circle. That fictitious force is called centrifugal “force.” However, you should always recognize that this outward “force” is not a force at all, but an effect caused by the ball’s inertia—its tendency to travel in a straight line.

If all the kids on the merry-go-round are clustered around its center while it is spinning at a constant angular velocity, then if all the kids were to “cautiously” move away from its pivot to the outer edges (while still spinning), would that cause the merry-go-round to slow down faster than if they had remained in the center?

Yes. When the kids move away from the center, the merry-go-round will slow down. If they then return to the center, the merry-go-round will speed up!