Robittybob1

The movement along the radial line experiment - a lazy Susan - might have one at work. If we can sort this one out, then that will produce one less area of contention. Can you expand those ideas a bit for it sounds wrong. Because the two forces are present at the same time momentum is conserved (and besides momentum is always conserved). Describe your objections with a bit more flesh please. Google definition: I say there is a pair of forces, so Newton's Third Law is not broken, so why do you say it was?

Can an object's circumstances generate a force on itself? This is through the influence of the velocity of a mass and then with the application of a centripetal force (which makes the mass move in a circle). Once you have those three things , a mass, with velocity, held by a force orthogonal to that motion this produces fictitious "centrifugal force". I used the word "internal force" earlier but centrifugal forces aren't normally listed as an internal force.

That is not what is being said time and time again, but what I find is you would accept there is a centrifugal force if the object would fly off with a radial component. "If there is a force, what is doing the pushing?" You all want to see a push, but you don't get a push. But the laws of physics tell us there are pairs of forces and the force paired with centripetal force is the centrifugal force. If you let the rope slip what pulls the rope through your hands?

Ok, just to be very clear, will it be ahead or behind the radial line. Even I could accept a very slight slippage back from the radial line, for the coin has to accelerate at the larger radii and that could mean a slight difference in the coin's tangential speed and the speed of the merry go round. But this will be at the most very little. So what do you think the direction will be? If it has to slide 300mm to reach the edge how much transverse direction will there be? Just wait till I run another experiment this weekend. Weather permitting of course. Is it true there is only the need for a centrifugal force while there is a system that is rotationally accelerating the mass? As soon as it falls off the merry go round or the string breaks there is no centrifugal force. So you only get the outward force while there is a centripetal force. What pushes it out is the difference in the rotational inertia of the material being accelerated. Matter further out will have a higher rotational inertia per unit mass than parts closer in. (My terminology might be a little wrong, but it was like inertia but in a rotating frame so I guess the factor is rotational inertia. I'll attempt the math in the weekend.)

It is just simple physics, nothing extraordinary yet! Come on make a choice before the results are in please.

But what do you think will happen even before we run the experiment? Swansont has chosen a transverse line but I'm not sure what direction that is, but what about you? I have chosen the radial line. If you make a prediction I will check it out this weekend for I know where there is a merry go round.

Have you followed the whole thread?

Doesn't the solar wind strip molecules of the Earth and Mars? If this removes atmosphere off the distant planets the same would be happening on the Sun. I read the solar wind contains all the elements (a vast range at least) so where do they come from? If you were to look closely at the photos referenced earlier that was visual evidence that most incoming comets had trouble get through to the body of the Sun. The heat of the Corona (I suppose) is just too hot for the comet to survive as a body (in the cases I've seen) and they seem to be gasified and blown away from the Sun's atmosphere. Well that is my understanding of what I see. But how would this be proven?

To someone looking from a central but stationary position with respect to the merry go round spinning beneath them, they might view the coin's path as a spiral but what I'm talking about is if you were to take multiple photos of this spiral path from this view point would the coin always be on the line we drew on the merry go round? I predict it will, for it only required a centripetal force to keep it in place, and that centripetal force would have acted along that exact same "radial line" Alright, you seem to want me to prove you wrong. If I drew a radial line on the merry go round, how would I draw a transverse line from where the coin will start off from? (My two statements or questions definitely didn't deserve negative reputation points. Who is this mean person?)

I agree with you. The forces aren't from the outside, as you say, but from within, as I tried to explain OK if radial is not the right word - if I draw a radial line on the merry go round and put a coin on that line and spin it till the coin slides, I think it will follow that line to the outer edge. That would be a very easy experiment to do.

I'll have to get the guns out and try it.

Yes I watched it several times. The tension in the string is a two way force and the direction of the arrow showing the person sliding off is wrong. The "sliding off" bit is radial but the motion through the air once off is the addition of the radial rate and the tangential rate vectors. You try putting tension in the string without pulling it with two opposing forces.

Heating of molecules can rip electrons off them and make them charged and one thing bumps into another and before you know it the whole mass of gas is moving even if it was the charged ions that were accelerated first and foremost. But I stand to be corrected. Thanks.

Gosh I have just had an insight of what is going on in the tube situation. It was from reading the last three posts that has inspired this. Now think about an object in a tube starting from the center point it has one end near the center and one end further out (larger r). Now the trick is to think about the entire middle bit as a very thick string. So to begin with the two situations of a weight tied on a string inside a tube and a movable mass in the tube are the same. Now read what Swansont said, "You feel an outward force, because you have to exert the inward force on the mass that's moving in a circle" but change that to "the central part feels an outward force, because the central part has to exert the inward force on the outer mass that's moving in a circle." Now it is quite obvious in the "moving mass" situation that since the inner part is not tied down the inward force is less than the outward force and that imbalance continues as the mass slides along the tube. Whereas with the string tied to the center the two forces equal each other up to the point the string can't take the strain and snaps.

My point of the thread was to show the Sun was different that other cold gravitational bodies. The extreme heating that an object is subjected to as it goes through the atmosphere of the Sun reduces most objects to molecules which are then blown away in the Solar Wind. Really massive objects might still be able to get through to the body of the Sun. There would be limits on how much could be evaporated in the limited time of an impact. Those events would be rare.

What is clear from my experiment and Mike's one is that as the objects go along the tubes this is a speed which will make the object leave the tube at an angle to the tangential velocity component. I'll have to read through what you have written again later, but my experiment was done horizontally so I could ignore the force of gravity which is confounding Mike's object's trajectory. Mike run your experiment again horizontally. Trouble becomes how do you film it then? For the best vantage point is from above it looking down.

That is right its angular momentum won't change after the bullet has left the barrel. But as I have been asking is, what happens while it is in the barrel? "Mythbusters: Is it possible to Curve a Bullet part 2" 1:38 in the video we get a slow motion shot of the gun and as the puff of smoke comes out it looks like it is arrested in rotation. Do you agree?

Yes a "cross product"! I do wish I was better at math. OK r is the radius but as I look at it the radius is going to increase as the bullet travels the length of the barrel. Now is a man's arm powerful enough to add the required angular momentum to the bullet in 1/1200th of a second? A man's reflexes would not be able to anticipate the force required so does the gun momentarily halt in its rotation? For that is what I was seeing in the Mythbuster video on whether a bullet curved in flight. P is the symbol for momentum but we are only interested in the momentum in the tangential direction and that will be increasing as the radius increases too won't it?

OK I wasn't thinking of the rotation of the bullet along its path. Ignore that effect. Swansont -- What I find is I don't understand your way of asking questions sorry. "What about the angular momentum?" To me that is an ambiguous question. "What is it that you think is happening here?" I'm not sure what aspect you are referring to by "here". Off to work. Later! But I'm wanting to know what is happening while it is still in the barrel. At the moment don't mention what happens once it has left the barrel. What happens while it is still in the barrel. "barrel was 100 metres long and did a 1000 revs a second it will still come out straight" agreed but can you turn a gun at 1000 Hz while a bullet is going along the barrel? And if you can what effect would it have on the torques required.

If the target and the shooter were going the same speed parallel to each other, ignoring wind resistance the bullet would go straight across. If the target was stationary the gun would need to track the target. If the bullet starts off at say 0.5 meter from the point of rotation and the barrel is say 1 meter long is the swinging action of the gun transmitted to the bullet? I'll need an estimate of the mass of a bullet and we should give an estimate of the angular speed and see how much angular momentum is required.

There is no way the shooter can affect the trajectory once it has left the gun. I was talking about the angular momentum while it is still inside the barrel. If the bullet goes at 1200 m/sec it might be in the barrel for 1/1200th of a second if the barrel was a meter long. OK that is not long to affect the path of the bullet but motion is motion.

They talk about a bullet getting extra velocity when fired along the direction of travel, and likewise slowed when fired to the rear. But what happens to a bullet fired from a gun either moving sideways or rotated (swung sideways) rapidly. I saw a YT where the Mythbusters proved a bullet does not curve in flight after being fired from a gun being swung around rapidly. I had a feeling if a gun is being swung at the same time as being fired, the gun momentarily stops, or slows at least, for there is no way to add the angular momentum to the bullet in the short time it is within the barrel. Any ideas?

I'll look out for it.

It is not just a matter of being told "you are wrong" but the proof of it that counts. If I am on the wrong track, I'd like someone to show me where I'm wrong and not just told to go back to school. It shows a deficiency in the rebuttal as well as those who are "in error".

Centrifugal pumps would be designed to maximize the pressure and the throughput. I have seen their shape and they are definitely designed to thrust the liquids outward, drawing new liquid into the central inlet. Now in the design of my juice extractor using centrifugal force I reversed the rotation of a normal centrifugal pump, and the engineers who were working with me took a bet with me for they thought that if the direction of rotation of a centrifugal pump was reversed it would reverse the direction of the fluid going through it, but they were wrong. OK it wasn't as efficient but it pumps regardless of the spin direction of the rotor. In the experiment I tried to keep the tube continually radial so not to accentuated the "pumping action". OK I wasn't working with a fluid but a solid mass. Today I followed up with some lessons on Newton's Second Law of motion and there were examples that motion vectors were added. If the liquid is rushing along the vanes to the outer edge, that motion has to be added to the tangential motion and together you get an exit direction that is predominantly tangential but not perfectly tangential. The question still stands as to what "pushes" the fluid through a centrifugal pump, but the fact is that it does. It takes a lot of energy to achieve the effect. Once the fluid is given extra energy, its angular momentum forbids the fluid moving inward, only allowing an outward motion. The same is happening with a solid object because the centripetal force is the friction and back pressure so as long as there is somewhere for the liquid to go the object's inertia that friction is soon exceeded.