Thanks, Nanu. I tried to say the exact same thing about 17 posts up. I hope your explanation is more clear.
I’m sorry, Skeptico,but what you said isn’t entirely correct.
“I suggest this interactive experiment. When the cannonball goes up, it will “deflect” in flight, reaching a maximum “deflection” at apex. Once it starts falling down again, it “deflects backwards” (or if you prefer, “in reverse”), effectively correcting the original “deflection”. In other words, as it goes up, its angular velocity goes down, but when it drops again, its angular velocity goes up. In the ends, it lands at the same point it launched from.”
It’s true that the cannonball will end with it’s initial velocity, but that doesn’t mean it has caught up to the earth. Think of it this way. If you take the average angular velocity of the ball the whole time it’s in flight, it will be less than the average angular velocity of the earth during that time, hence it will miss the spot it launched from. Because the ball slows down in mid air, the earth spins ahead of it. As the ball is going up, the earth is getting ahead faster and faster. When the ball reaches it’s furthest spot, the earth is getting ahead at its fastest rate. Then, as the ball falls, the earth is getting ahead by less and less- however… the ball is not catching up. Angular Momentum stays the same, but angular kinetic energy doesn’t not. However, the effect is incredibly small, as Chronos noted, because the radius of the earth is so large compared to the vertical distance the ball will travel.
That’s exactly it. The angular momentum of the cannonball around the Earth never changes. The act of being fired straight up has no effect on the angular momentum of the cannonball around the Earth (it has the same angular momentum it would have had if it were just left in the cannon). The cannonball never dips below the surface of the Earth, although it does go above the surface. This means that the angular velocity of the cannonball sometimes falls below the angular velocity of the cannon, while it never exceeds it. The angular displacement can be found by integrating the angular velocity with respect to time, and a basic theorem from analysis tells us that the angular displacement of the cannonball from the time it is fired to the time it touches the ground is strictly less than the angular displacement of the cannon over the same time.
Nevertheless, this:
seems to be clearly wrong. It won’t come down at the “distance the Earth had moved,” whatever that could mean (I don’t suppose they mean the distance the Earth had moved relative to the cannonball ;)).
Another voice chiming in here in hopes of clarifying a little more. Skeptico has agreed that while the cannonball is far away from the earth, conservation of angular momentum dictates that its angular velocity decrease. Skeptico is also correct that when the cannonball comes back down to the ground, its angular velocity increases back to the original angular velocity it had when it was fired.
Now, given that the cannonball spends some amount of time at a lesser angular velocity than the earth, it falls behind. In order to “catch up” with the spot it was fired from, it would have to travel with angular velocity greater than the earth for at least some portion of its journey. But we know that the angular velocity is always less than or equal to that of the earth, so this does not happen.
The cannonball lands at a point west of its launch point, but with no angular velocity relative to the earth’s surface (and obviously not as far west as the earth rotated, as in the claim Galileo supposedly made).
But that’s just it - the angular momentum of the cannonball around the Earth changes CONSTANTLY. The tangential velocity says the same (assuming no air resistance, etc), but the angular momentum varies with R[sup]2[/sup] (the square of the distance from the center of the Earth). Angular momentum decreases as the ball rises, then increases as it falls, reaching its maximum angular velocity (equal to its initial angular velocity) as it hits the ground.
Analogy:
Two cars are driving side by side on a highway, at precisely 60 mph. Car J never slows. Car K slows gradually to 45 mph, then accelerates gradually back to 60. Are they still side by side? OF COURSE NOT. The one that slowed has fallen behind, and will never catch up unless either a) Car J slows down and allows K to catch up, or b) Car K exceeds its speed limit of 60 mph to catch the first one.
The fact that the angular velocity catches up to its initial value DOES NOT mean that linear displacement also does. You’re ignoring your constant of integration, so to speak.
You mean, of course, that angular velocity decreases and increases, right? Angular momentum is a constant.
Gosh, so many replies, I don’t even know where to begin…
Let’s start with Nanu:
That is correct.
That is only true if the cannonball and the Earth follow the same path (or orbit), but they don’t! You see, the cannonball’s trajectory from Earth to Pluto and back describes a parabolic-looking path, whereas Earth traces a circle (it’s really an ellipse, but circle is good enough for this discussion).
If you’re running behind a guy, there’s two ways you can catch-up: either run faster than him or take a shortcut… The cannoball is taking the shortcut, since, as you correctly point out, its angular velocity can be no greater than the angular velocity of the Earth once it reaches the original launch orbit.
RyanD004 has some interesting arguments:
That is only true if both the cannonball and the Earth follow the same path, but they don’t. The cannonball’s flight from Earth to Pluto and back is parabolic-looking, whereas Earth traces a circle (really an ellipse). (Yeah, sounds like I’m repeating the same thing I said to Nanu, doesn’t it?)
That is incorrect. In vaccum, there is no opposing force to slow the ball down in mid air. With an atmosphere present, air resistance (Newton’s or Stoke’s) will only manifest if the air and cannoball have different velocities. Since the cannonball has no horizontal component of velocity (as compared to the air around it), there is no air resistance on the horizontal. There will be air resistance on the vertical, but then we just launch with greater force to overcome it. In conclusion, no matter which way you cut it, the quoted statement is just wrong.
That is incorrect, since angular kinetic energy is derived from angular momentum.
As I’ve pointed out above, Chrono’s equation neglects angular momentum. It’s a nice solution to a math problem, but not to a physics problem. In our universe, Conservation of Angular Momentum is not optional, as Chrono accidentally makes it out to be.
The Weak Force has some very, VERY good arguments that I think everyone should read. I only want to append a clarification:
That is correct.
That is correct.
That is correct.
That is correct.
That is correct. (By the way, which “basic theorem” are you referring to?)
I just want to warn readers that Weak Force’s statement that the angular displacement of the cannonball is “strictly less” than the angular displacement of the Earth does not imply that the cannonball misses (or hits) the Earth. Weak Force analysis is very good, but stops short of a final conclusion, which I add below:
The angular dimension (theta) is orthogonal from the radial dimension ®. Therefore, if we take two particles in the same initial state, we can change the radial value of the first particle at will, knowing that once we return the first particle to the original radial value, we will find that the angular value of the two particles are the same. The two particles have to have the same angular value, since they are at the same r, and we did not perturb the angular dimension of the first particle at all. The fact that the two particles have differing angular displacements is not surprising, since they did not remain at the same r throughout the time evolution.
Now, we turn to ntucker’s concerns:
That is correct.
That is incorrect. For the third time: that is only true if both the cannonball and the Earth follow the same path, but they don’t. The cannonball’s flight from Earth to Pluto and back is parabolic-looking, whereas Earth traces a circle (really an ellipse).
Imagine a car chase through the streets of San Francisco (have you ever seen the 70’s movie “Bullet”?). The bad guy is going at a constant 100MPH. The cop is not as good of a driver, so sometimes he drops below 100MPH, but never goes above it. Can the cop catch up if he is always at a lesser speed? Yes, if the cop takes a shortcut through
Lombard Street.
No such thing happens. I still say cannonball lands at launch spot.
I’m not certain whether Joe_Cool is with or against me… At any rate:
That is incorrect. Where are you getting this free energy? The cannonball is not attached to lateral rocket propellers.
[quote]
The tangential velocity says the same (assuming no air resistance, etc), but the angular momentum
varies with R2 (the square of the distance from the center of the Earth).
[\quote]
That is correct. (By the way, you won’t have lateral air resistance even with an atmosphere, so that’s not an issue).
[quote]
Angular momentum decreases as the ball rises, then increases as it falls, reaching its maximum angular velocity (equal to its
initial angular velocity) as it hits the ground.
[\quote]
That is incorrect. The angular velocity is what decreases as the ball rises, then increases as it falls. The angular momentum is the quantity (angular velocity)x(radius)^2, and it remains constant due to the Conservation of Angular Momentum.
Or c) Car K takes a shortcut; a shorter path than Car J.
Or d) Car K veers off the road and back again, in a parabola shape. Since Car J was running in circles, they meet at the same spot again.
That is correct. The Weak Force didn’t “ignore” it, since he basically implied what you said. Nobody is arguing with you on that point, buddy.
Skeptico.
Skeptico. It has nothing to do with air resistance. Angular momentum is the cross product of the linear momentum and the radius vector. Stated differently, it is the product of the mass, the angular velocity, and the square of the radius. I think you’re been ignoring the radius term. For angular momentum to be conserved, as the radius term increases, since the mass doesn’t change, the angular velocity must decrease. The Earth will spin ahead of the cannonball. Do you see that this is a necessary consequence of the conservation of angular momentum?
When the Earth is spinning, an observer on the surface is no longer in an inertial reference frame. I think what you’re saying would be true in an inertial reference frame (maybe I don’t understand what you’re saying completely), but not necessarily in a non-inertial frame.
The Weak Force, you’re absolutely correct, thanks for catching my error, I’ve had to type far too many replies on this topic (I was thinking wayyy too near-earth when I typed that, I think). To clarify what I’m trying to say: Yes, the earth will spin ahead of the cannonball during the ascent flight, but on descent, the cannonball will catch-up (not by speeding up ahead of Earth, but by way of differing route) and land on the exact same spot it was launched from.
When the Earth is spinning, the observer on the surface is no longer an inertial reference frame (ie. Newton’s laws AS IS no longer hold). To fix the situation, we introduce an imaginary “force” component, aka centripedal acceleration. That effectively makes Newton’s laws work again on rotating frames, and all is good (ie. it’s inertial again).
Doesn’t this actually demonstrate that the cannonball will not land at the launch point?
This statement is not completely correct (at least not obviously so) for two reasons: one (as The Weak Force has stated) is that we have not specified our reference frame–in the earth’s rotating coordinate system (relative to space–the isotropy of which is the real source of ang momentum conservation, not Newton’s laws), momentum conservation cannot be simply stated as it would in an inertial frame. Furthermore, while the r direction and the theta direction are orthogonal, their derivatives are not independent. This is a consequence of working in a curvilinear coordinate system.
Since the board still seems skeptic (and that’s a good thing!) of my Conservation of Angular Mometum argument, and Orthogonality argument, let me try a third explanation:
Newton destroyed the idea of absolute space. Newton showed us that motion is relative. When you’re standing still, you’re not really standing still, since the earth is rotation. Ok fine, stop the earth on it’s tracks. That does no good since the Sun itself is rotating around our galatic cluster, and so on and so forth…
Since every inertial frame is identical to any other inertial frame, we can look at our problem as follows. Fix Earth’s inertial frame, and have whole universe rotate about it instead. Mind you, the earth still rotates around it’s own axis, but it doesnt’ rotate around the Sun (about instead the Sun rotates the Earth).
Now launch the cannonball and see if you can visualize it’s path…
(If the board is still not convinced, I’ll sit down and work out the math… but will anyone read through my maths???)
Hi Don Roberto
No, since the one particle is going around in a circle, and the other other has a parabolic-looking trajectory.
Humm… I will investigate this point further.
The problem is that Earth is not an inertial frame, because of the centripetal forces that allow its rotation.
Alright, it’s on now. I dusted off the physics book. Ok, unfortunately I lost my equation sheets (I love those things), so I’m mainly going off on memory here, but there are a few blatent flaws in your arguements. Number one: yes there is a force that slows the cannonball down. It’s called gravity. If gravity doesn’t accelerate rotating objects then the planets should all be flying off into space right about now. Ok,heres one key concept. Angular velocity is normal velocity times the radius of the angular path.
I’m sorry… I’m having trouble with this. I’m just a bit rusty. Looking for an analogy. Well, to start off, imagine you are spinning a yoyo on a string in a circle. If, at any moment, you cut that string, the yo yo will fly off with a velocity tangent to it’s location in the circle you were ‘swinging’. This ‘velocity’ represnts the non-angular value of the yoyo’s velocity while it was spinning. No, imagine you’re really tall,and you get a yoyo string twice as long. You will find its harder to make the yoyo spin as fast. Why? Because it’s inertia changes. Inertia… contrary to the way they seem to use the term on tv, it angular resistance to motion. Whereas mass is just regular resistance to motion. As you try to rotate something,and it gets further from you, it’s harder to do. Test this is you like, just get a stick in your back yard. (a big stick) Hold it in the center- spin. Then hold it at the end, try rotating in a circle again.
Ok,another thing. You keep saying “the earth and the ball don’t take the same path.” Thats precisely why the Earth gets ahead. The cannonball is taking a longer path. When you spin the yoyo on the longer string it still goes 360 degrees, but the radius is larger so the circumference, or path, of the yoyo is larger than the one with the small string. When you shoot the cannonball up, it has to travel a larger circle than the earth.
Ok,got it. Here we go. The key is that as the cannonball gets further away, it’s Inertia increases. Inertia is equal to mass of the object timesits radius squared. Now, i’m treating the cannonball like a point particle because thats basically how it acts (especially if were talking about travelling to pluto and back). Thank god Galileo picked a spherical object. So, if inertia gets bigger, and energy isn’t coming out of the seams of the universe, the ball has to slow down. Now, all of this could be a little wrong. I’m getting rusty on my physics, but I am possitive that the cannonball does not land in the same spot. Chronos knows it… it becomes obvious because you take anular momentum into account… how can’t be ignoring it.
by the way… "Newton destroyed the idea of absolute space. Newton showed us that motion is relative. " … that was Einstein idea, not newton. Hence Einstein’s general theory of ‘relativity.’ 
I really wish that little edit button at the bottom of the screen worked. I got frustrated from being logged of so many times while writing that, that I didn’t proofread. Hopefully you get the idea.
Numerical example:
r[sub]e[/sub] = 6.3778x10[sup]6[/sup] metres
period of rotation = 86,164 seconds
w[sub]e[/sub] = 7.2921x10[sup]-5[/sup] radians/second
tangential velocity at equator v[sub]e[/sub]= r[sub]e[/sub]w[sub]e[/sub] = 483.74 metres/second.
Now suppose the cannon ball reaches a maximum altitude of 1,000 kilometres.
r[sub]ball[/sub] = 7.3778x10[sup]6[/sup] metres
w[sub]ball[/sub] = v[sub]e[/sub]/r[sub]ball[/sub] = 6.5567x10[sup]-5[/sup] radians/second
The cannonball has a significantly reduced angular velocity compared to the earth. As it falls, its angular velocity increases until it again matches that of the earth.
However, because the cannonball has spent time moving at a lower angular velocity, while the earth has not, the cannonball cannot catch up.
Just as in Joe_Cool’s cars on the highway analogy.