|Momentum and Center of Mass|
|A Little Relativity|
|Pushing a House in Space|
|Physics and Skydiving|
|Falling in a Vacuum|
|Speed and Velocity|
|If Hens Inside a Van Fly...|
|Hunter and Monkey|
|Motion Inside a Moving Train|
|Elastic Potential Energy|
|Energy, Momentum, and Einstein's Equation|
The only force that someone in the air can use (assuming he doesn't push off another player), is air resistance, and that is too small to change the motion of their center of mass very much. You CAN do quite a bit of contortion around that center of mass, though.
If you jump straight up off a diving board, your center of mass will come back down through it. It is possible to have your body miss the board, however. Expert pole vaulters actually clear bars that are higher than their center of mass travels with their "U" shaped contortions. I don't understand how someone jumping straight up on the board could ever get their center of mass out over the edge. It sounded like you said that if the center of mass went straight up, it has to come straight down. What kind of contortion would change the location of the center of mass?
You can't change the motion of the center of mass, but by keeping the center of mass outside the body, you could theoretically miss the board (even though your center of mass would go through it). As I said, that's the way expert pole vaulters do it. Their center of mass goes below the bar (so if the board were from the bar to the ground, it would be similar to what you've asked), but every part of their body passes over the bar. What you've asked would be much harder because a pole vaulter works at the slowest point of their path and they use the space under the bar in their contortions.
The way to miss the board is, for example, to fall belly first and horizontally at the board, then, just before impact, flip your upper body, head, and arms, around under the board, and then before your legs hit, flip them around too. Another way to do it (or at least think about it) is to make your body into a rotating U, then time the rotation so that the U is pointing down when you reach the board and does half a rotation just right, so that you clear the board on the bottom side. It probably can't be done, but it IS theoretically possible.
Dr. Eric Christian
With your question, you are suggesting the following
idealized situation: one person remains on Earth for one year, while
another person takes a trip for the same amount of time to a deep
As for 1., we would have to invoke stretching of time due to the speed of the earth in its orbit (~30 km/sec) -- a problem of Special Relativity. The time on Earth would scale versus the time on the space station. Over one year, calculations show this time dilation accumulates to a mere 0.13 sec.
As for 2., we would have to invoke stretching of time due to the gravity of the earth (a problem of General Relativity). Over one year, another 0.018 sec will accumulate.
Neither change in the time elapsed for both people are worth writing home about. However, there is another wrinkle in the real world. The deep space station would have to be very far away from the sun for it to just sit there without moving on an orbit about the sun like the earth. In order to take a trip to and from this space station in a reasonable amount of time, i.e., much less than the one year, the person going here would have to travel at a speed much faster than the earth. As a consequence, time would pass slower for that person than for someone on Earth during the transit periods. Therefore, the two effects, 1. and 2., would be most likely overcompensated -- still only by seconds because we don't know yet how to boost spacecraft speed to a substantial fraction of the speed of light.
As for reading on the subject, I would suggest the
Dr. Eberhard Moebius
The question you raise is as much a study in the sociology and psychology of physics as it is a question in basic physics.
First, the basic physics question: You can move any object of any mass with the push of a finger (or a nose, or a breath) if it is in space and far from the pull of gravity. Now, that answer assumes a few things. For instance, gravity is everywhere, but it gets weaker the further you are from the source object (the sun, a star, a planet). The total gravity on any object in space will make it move, and this is why satellites orbit the Earth instead of flying off in every direction. A more complete answer would be to say the added motion due to your finger's push would be present regardless of how big the house is, if you have a sensitive instrument that can measure it.
Second, classical physics from Newton tells us that force is proportional to acceleration, not velocity or displacement. So if you continue to push, the object will continue to accelerate and eventually the speed of the house could be astronomical. It was Einstein that pu a limit on how fast the house will go and how much push it takes to get it there.
Third, Newton also tells us that for every action there is an equal and opposite reaction. This means that as you float in space and push on the brick house, you also push yourself away from the house. You are lighter than the house, so you gain more speed. You would need to brace yourself against something even more massive than the house, or you will gain as much energy moving away from the house as the house gains moving away from you.
Fourth, we always think of objects in physics as being absolutely rigid. They aren't. Here is where the sociology and psychology comes in. You have thought ahead of the problem and wondered what will happen when the force is transmitted between the many bricks. You are partly right - any motion between one brick and another that is due to the friction required to transmit the force across the structure will result in heat and reduce the acceleration of the house. The acceleration won't go to zero, but if the force is small enough, it will absorb a large fraction of the energy your finger is trying to put into house motion. It won't absorb all of it, but some. So the house accelerates even more slowly, but it does accelerate.
Lastly, why don't we realize this is true from our own Earthly experience? It's that pesky friction again. The friction between a house and the ground is so much that moving it takes big equipment. Then again, put that house on hard metal rollers on a hard metal field and you could get tha house moving! It's all a matter of reducing friction. House, and lighthouses for that matter, are moved on wooden rails and rubber tires in part for this reason.
So who is right? I guess your friend is right, but you had a big piece of the next puzzle already in hand.
Charles W. Smith
This is a case where the real life situation is quite a bit more complicated than the simple physical models we like to use to describe it, as your skydiver friend knows from experience.
In the simple picture, our skydiver steps off a 1000 ft tall building without a parachute, curls into a ball and waits for the ground to arrive. Neglecting wind and air resistance, it's a straightforward falling object problem and our friend will expire in just under 7.9 seconds at a final vertical speed of about 250 ft/sec (172 miles/hour). The acceleration is straight down due to the force of gravity. Having stepped, not jumped off the building, there is no significant horizontal speed at the beginning and no wind to encourage one en route, so the landing site (crash site?) will be directly below where they stepped off.
The real story is more complicated. The air that the skydiver is falling into pushes back, slowing him/her down. You can think of it simply as an increasing wind or probably more accurately as a fluid problem, like falling through thin molasses. This slows the person down, making the flight longer and the final velocity lower. So far this hasn't changed the outcome, our poor friend is still squashed directly below at the bottom of the building.
What changes the picture is that the skydiver's body is not shaped like a bowling ball (we assume...) but appears to the air more like an airfoil, like the wing of an airplane. In the neutral "box" position with arms and legs extended to the sides and head slightly down, a skydiver falls straight down in a stable position with the air slipping by evenly on all sides. Raise one arm or leg and more air gets deflected out that side and the body moves the other way. Move your arms back a little and flatten the legs and you slide forward. Controlled turns and slides happen with combinations of these movements. The physics involved is now more related to flying than falling.
So how far can you get? It's a pretty complicated problem and it will depend on the body position, movements, and many other factors. Rather than attempt that, consider one more very rough estimate to get a sense of what *might* be possible. Our skydiver dropping like a rock took 8 seconds to fall 1000 feet with an average speed of about 125 ft/sec (half of the final speed of 250 feet/sec). If using body movements could achieve an average horizontal speed of only 10% of the average vertical speed (12.5 ft/sec), the skydiver could move sideways over 100 feet before hitting the ground. Then remember that the air resistance that is allowing him or her to move sideways is also slowing them down. Also, 10% isn't that much, if they could generate a little more horizontal speed they could conceivably move much farther in the same time.
Add a parachute and the potential becomes even higher. Modern square (actually they're rectangular) parachutes are airfoils and create lift, allowing for tip-toe perfect standing landings at very low vertical speed. These allow very long flight times and large horizontal velocities so you could imagine moving much farther away once the chute comes open.
Happy (safe) skydiving!
Dr. Jeff George
I hope this answer relieves some of your confusion!
In a vacuum both bodies would fall at an identical rate and hit the ground at the same time. But in our atmosphere the air affects the motion. (Think of the air as a fluid the bodies are moving through.) Falling bodies accelerate until the air drag (force acting on the bodies due to air resistance) balances the gravitational force. At that point, the body stops accelerating and moves at what is called "terminal velocity". The air drag starts out at zero and builds up as a body falls.
Moreover, the air drag depends not only on surface area, but on velocity and length (it's a complicated dependence: for example, at high velocities the air drag is proportional to velocity squared, while at low velocities, it is proportional to velocity).
Anyway, the box filled with lead will hit the ground first because the air drag will rapidly build up and retard the motion of the empty box, while the lead-filled box continues to accelerate up to a higher terminal velocity.
Dr. Louis Barbier
The acceleration due to gravity depends only upon the mass of the object you're accelerating toward, so it is constant and the same for both heavy and light objects.
Dr. Eric Christian
Speed and velocity are two different things. Velocity is a vector quantity (i.e. it has a magnitude and a direction), whereas speed is a scalar quantity. In fact, speed is the magnitude of the velocity vector. In other words, velocity has both a speed and a direction.
Examples: Two people are riding in separate cars, both going 40 mph. If they are going in the same direction, they have the same speed AND the same velocity (vector). If they are going in different directions - say one is traveling south and the other is traveling east - they still have the same speed, but they have different velocities.
Dr. Louis Barbier
If the van doors are closed, no. All the forces balance on the inside. If the hens have their wings sticking out the windows, that's another story.
Dr. Eric Christian
A circle is 2-dimensional motion. With one dimensional motion, in order to get zero average velocity, you'll need both positive and negative instantaneous velocities, and in transitioning from one to the other, you must cross zero. The case of the circle can be broken up into two 1-D motions (X & Y) each of which will be zero at some point.
Dr. Eric Christian
This problem is known as the "Hunter and Monkey" problem. It works because the motions in the horizontal and vertical directions are separable, i.e. they don't affect each other. Also, the acceleration felt by both objects in the vertical direction is the same. I don't know of a "law" that puts this forth, however.
Dr. Eric Christian
I think that the ball will fall in a straight line, because there's no horizontal acceleration on the ball (the train is at a constant speed), so the only acceleration applied to the ball is gravity. My uncle's answer is that it will fall behind because when we cut the wire, equilibrium is broken and the ball will start to decelerate horizontally because it's no longer "touching" the wagon. I've tried to convince him otherwise, stating that intertia will keep the ball moving at the same speed as the train, but he refuses to believe me. Who is right?
For the most part, you are exactly right. But make sure that you state it correctly. It is true that the only force acting on the ball is gravity, but be sure to note that the ball will fall down in a straight line with respect to the point of view of someone in the train. The ball's inertia will cause it to continue moving at the same speed as the train once it is no longer in contact with the train. So, I hereby grant you nephew bragging privileges!
Thank you for this rather challenging question. It provides the opportunity to tackle an issue that is rather often subject to misconceptions.
The forces experienced in a rotating frame of reference can be rather unintuitive. Therefore, I would like to use a carousel as an illustration for the forces that play a role during rotation.
1) Let us start with the "simplest" force that is only present in a rotating frame: the centripetal force. An object is NOT subject to a force, if and only if it is either at rest or moves at CONSTANT pace WITHOUT CHANGING DIRECTION. Not so, when the object is fixed to a rotating carousel. The carousel may rotate at constant speed, and so will move an object fixed to that platform. However, while moving, the object changes direction at all times, violating the condition (of Newton's first Law of Motion) for not being subjected to any force. During the entire ride there is this inward pulling (centripetal) force on every passenger on the carousel. However, this is the only force felt on the carousel, when you stay put.
2) Now let's take this one step further: The carousel starts rotating, i.e. it accelerates from a stop to the normal speed of the ride. Any object or passenger on the carousel will now feel another force, that of the accelerating carousel, or being pushed to speed up for the ride. This force is the same that you feel when your car accelerates. You still stay put, but now you feel two forces: the acceleration of the carousel for the ride AND the (centripetal) push towards the center on the way around.
3) Now finally, let's take a look what happens when you elect to walk inward on the carousel to reach that other horse over there, while the ride is in full swing. Because the horse further inward travels on a smaller circle, its speed during the ride is slower than that of a horse further outward. When you walk radially inward the carousel will slow you down to the speed of that other horse. As during the braking of your car, you will feel another force: This is called the Coriolis force. From the sequence of our steps 1) through 3) it hopefully becomes apparent, why:
You can read more about the Coriolis force, particularly how it might affect the human body during spaceflight, in the Science@NASA story Spinning Brains.
Dr. Eberhard Moebius
Elastic potential energy is the energy stored in the deformation of an elastic body. I would not think of air as an elastic body, normally this is used for solids (springs, rubber, etc.). I would think your project was just fine, but there may be some technicalities hidden in the rules that I am not aware of.
Dr. Louis Barbier
Thank you very much for this timely question at the end of the 100th anniversary of Einstein's wonder year. You are addressing possibly the most famous physics equation.
Let me try to give you the answer in two ways, both with some formulae, because you asked the question this way and I suspect you can make sense of them. I will give you an answer first that is more descriptive, while still containing the relations that you want explained, and then, because I suspect that you will be longing for more, an answer that contains some more mathematical expressions to show how EKin = 1/2 mv2 and E = mc2 are still connected.
I sense you are referring to the kinetic energy (or the energy of motion) of an object when you ask the question, why the expression doesn't say 1/2 mc2, correct? Kinetic energy is only the additional energy that an object attains when it is moving at a certain speed v (or velocity; I prefer speed here because the term velocity taken strictly also defines the direction of motion, whereas speed only means the magnitude). Widely known as the relation for the kinetic energy (EKin) is: EKin = 1/2 mv2. However, this formula is only accurate as long as the speed v of the object is slow compared with the speed of light c.
The guiding principle for relativity is that the speed of light c is always the same, no matter how fast we are traveling, whether light is observed from the perspective of the traveler or from the one left behind. As a consequence you will never catch up with a light beam, no matter how fast you get, and therefore c + v = c, which sounds like nonsense, but comes out perfectly once you realize that time and space dimensions are seen differently by the traveler compared with the one left behind. Therefore, if you launch a rocket with speed v from another rocket that already flies at high speed v, the newly launched rocket will now travel slower than 2v as seen from an Earthbound observer.
Yet the momentum of the two rockets must still add up. To fulfill this "conservation of momentum" Einstein found that the only way out was for the fast rockets to increase their mass.
Where does this mass come from? It is the kinetic energy of the rocket that shows up as mass, disguised in Einstein's most famous formula for the energy E = mvc2, where mv is the mass of an object moving with speed v. This means that even at rest, and this realization was new, the rocket already has an intrinsic (or rest) energy Eo = mo c2 (mo is rest mass) that is equivalent to the mass that you find by weighing the rocket.
Now you can see that the pure motion energy, or kinetic energy, must be:
EKin = E - Eo = mvc2 - moc2
where E is the total energy of the moving object and Eo is its energy at rest. When Einstein calculated the kinetic energy from this relation for very slow speeds v, he obtained EKin = 1/2 mv2 as the limiting approximation, showing that his new theory is perfectly compatible with Newton's relations for slow speeds.
The whole point of Einstein's formula is that energy is equivalent to mass, meaning one can be turned into the other. Nuclear bombs and nuclear reactors work that way; the Sun along with all the other stars generate their immense amount of energy that way. The stars' trick is to fuse four hydrogen nuclei into one helium nucleus, where the mass of the latter is slightly less that that of the four hydrogen nuclei. The fraction of the mass that has disappeared surfaces simply as energy, according to Einstein's formula, as has been tested in many ways. (You can find the explanations of the basic concepts along with light reading on Einstein's life and the development of his ideas in the book "Einstein's Cosmos" by Michio Kaku, a book which I currently enjoy reading.)
Let me try to explain to you where this comes from and how the kinetic energy that you learned in freshman physics still comes out of Einstein's relativity theory.
To satisfy the requirement from Maxwell's theory of electro-magnetism, that the speed of light c is always the same, Einstein had to abandon the notion of absolute time and space. Time runs slower according to:
tv = t *(1- v2/c2)1/2
(tv is the time measured when moving at speed v, while t is the time as seen when at rest.)
and lengths shrink according to
xv = x *(1- v2/c2)1/2
(xv is a distance measured when moving at speed v, while x is the distance measured when at rest.)
To arrive at these relations is a nice geometrical game, but would take awhile to explain, likewise the translation of these relations into a new way to add velocities. You can find a detailed yet geometrically intuitive explanation of the relations in the book "Spacetime Physics" by Edwin F. Taylor and John A. Wheeler.
Let me move on. Now v + v is not simply 2v anymore after invoking the relations for slowed time and shortened length scales. However, momentum must be conserved when you add the momentum p = mv, for an object that you speed up to v in a rocket that already goes at speed v, i.e. p + p = 2p. Therefore, Einstein concluded that the mass mv of an object moving with speed v is now larger than its mass at rest mo according to: mv = mo/(1- v2/c2)1/2.
As you can see, the mass approaches infinite mass when v approaches c. In other words, any further acceleration goes into increasing mass rather than speed, thus really making c the cosmic speed limit. This enormous increase in mass is indeed observed with very high accuracy when physicists accelerate electrons and nuclei to speeds close to c. Then it becomes very hard to bend the path of these particles with magnets because of their ever-increasing mass.
Let us now compute the kinetic energy for the case v << c. Using Einstein's famous formula for energy and mass we conclude
EKin = E - Eo = mv c2 - mo c2
where we fill in mv as noted above. For v << c or v2/c2 << 1 we can make use of the approximations that 1/(1- v2/c2)1/2 is approximately equal to (1 + v2/c2)1/2 which is approximately equal to 1+ 1/2 v2/c2.
Now we get:
E_kin = mv c2 - mo c2 = (mv - mo)c2 which is approximately equal to [mo(1+ 1/2 v2/c2) - mo]c2 = mo 1/2 v2 And this is exactly the well-known formula for kinetic energy from freshman physics, just limited for slow speeds.
Dr. Eberhard Moebius
This file was last modified: June 14, 2012