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

#### 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 doe…

#### 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

#### 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…

#### 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

#### 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

#### 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.

## If all the kids on the merry-go-round are clustered around its center while it i…

#### 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!

## If the fictitious force you experience on a loop-the-loop isn’t greater than you…

#### If the fictitious force you experience on a loop-the-loop isn’t greater than your weight, will you fall?

Yes. If you go over a loop-the-loop too slowly, so that you don’t accelerate downward quickly enough, you will leave the track and fall. That’s why some roller coasters strap you in carefully before taking you upside-down slowly. Without the supports, you would fall out of the car.

## If you feel fictitious force upward on a loop the loop, how can that fictitious …

#### If you feel fictitious force upward on a loop the loop, how can that fictitious force make objects fall upward? Is fictitious force fictional or real?

As you travel over the top of the loop the loop, you observe the world from an inverted perspective. The sky is below you and the ground is above you. If you were to take a coin out of your pocket and release it, you would see it fall toward your seat. From that observation, and the feeling of being pressed into your seat, you might think that gravity is suddenly pulling you toward the sky. It isn’t. Gravity is still pulling you toward the ground, but you are in a car that is accelerating rapidly toward the ground. As a result, the car is having to push you toward the ground with a force on the seat of your pants. You feel pressed into your seat because the car is pushing you downward hard. When you release the coin, it seems to fall toward the sky, but it’s really just falling more slowly than you are. With the car pushing you downward, you’re accelerating toward the ground faster than the coin and you overtake it on the way down. It drifts toward the seat of the car because the car seat accelerates toward it. As you can see, the only forces around are the force of gravity and support forces from the car. There is no outward or upward force here. The fictitious force is truly fictional; a way of talking about the strange pull you feel toward the outside of the loop.

## When a ball swings in a horizontal circle at the end of a string, what’s the for…

#### When a ball swings in a horizontal circle at the end of a string, what’s the force on the ball pulling it straight? What’s the force pulling it out?

Let’s neglect gravity, which isn’t important in this horizontal motion problem. When a ball swings in a circle at the end of a string, there is only one force on it and that force is inward (toward the center of the circle). We call such a force a centripetal force, meaning toward the center. There are many kinds of centripetal forces and the string’s force is one of them. As for the ball’s tendency to travel in a straight line, that’s just the ball’s inertia. With no forces acting on it, it will obey Newton’s first law and travel in a straight line. There is no real force pulling the ball outward. But a person riding on the ball will feel pulled outward. We call this feeling a fictitious force. Fictitious forces always appear in the direction opposite an acceleration. In this case (an object traveling in a circle) the outward fictitious force is called centrifugal “force.” But remember that it’s not a real force; it’s just the object’s inertia trying to make it go in a straight line.