How can a balloon support the air around it (pressure wise) and still rise?

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How can a balloon support the air around it (pressure wise) and still rise?

If you could have filled a balloon with nothing at all, it would float very nicely because it would have had no weight and the only force on it would have been the buoyant force upward. But an empty balloon will be crushed by the surrounding air, which will push inward on its surface with enormous forces. To keep the balloon from crushing, you must fill it with gas. Since this gas will weight the balloon down, you should choose the lightest gases around: hot air or helium. In the case of hot air, a relatively small number of air molecules create the pressure needed to keep the balloon inflated. With helium atoms, lots of helium atoms are needed to create the pressure but helium atoms are very light and their total weight is less than that of an equal volume of air. Thus the upward force on the helium filled balloon is more than its weight and floats upward.

Have you heard about the Egyptian Lighthouse (one of the seven wonders of the an…

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Have you heard about the Egyptian Lighthouse (one of the seven wonders of the ancient worlds) that was found in the sea? Well, people are proposing lifting the 70-ton sections of the lighthouse with balloons. Can a balloon do this? What special aspect of the balloon will lift 70-ton bricks? What kind of balloon would be used?

I would guess that they intend to use balloons in the water. The blocks are at the bottom of the sea, so they must be lifted up to a boat. A balloon experiences an upward net force that is equal to the difference between the upward buoyant force on it and its downward weight. If the balloon displaces air, then the buoyant force on it is rather small and it would have to be extraordinarily big to displace enough air to lift a 70-ton block. But if it displaces water, then the buoyant force on it is much greater. To displace 70 tons, it would only have to displace about 65 cubic meters of water. That’s not hard at all. The balloon could be made of heavy reinforced canvas and still work just fine underwater.

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

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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.

When you spin an object around a fixed point, a sling for example, does the obje…

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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.

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

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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.

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

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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.

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

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

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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!

Can you explain the term centripetal?

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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.

Would a baseball bat do more damage on a person if the point of contact was the …

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Would a baseball bat do more damage on a person if the point of contact was the very end of the bat (torque=force x lever arm) or at the sweet spot? (assuming the bat was swung with a constant angular momentum)

The sweet spot. Hitting someone with the bat is very similar to hitting a ball. When you hit a ball with the sweet spot of the bat, the bat slows down and begins to rotate slowly. The slowing is good because it means that some of its kinetic energy has been transferred to the ball. The rotation is bad, because it means that the bat has put energy into rotation (spinning objects have kinetic energy). If the ball hit the bat’s center of mass, the bat wouldn’t rotate and the transfer of energy would be better; except for one new problem: the bat would begin to vibrate and that vibration would use energy. By hitting the ball on the sweet spot, you keep the bat from vibrating and wasting some of its energy. The transfer of energy and momentum to the ball is maximized. The same occurs when hitting any other object, including a person.