But Musicat, surely that arc wouldn’t remain the same no matter what velocity the golf ball was projected?
I think it’s possible you are beyond help, but try this thought experiment that was mentioned above: you are on a train traveling 100MPH and drop a baseball from six feet above the floor. Meanwhile someone standing next to the train drops a baseball from six feet above the ground. How long does it take each ball to hit the floor/ground?
The picture does show two arcs, one of them is just really steep. Despite the different shape of the arcs the ball falls at the same rate. So yes, the speed will affect the shape of the arc, but the time to “fall” to the ground is identical.
The whole point of that image is to show that the arc changes due to the speed the golf ball was projected, but the balls fall down at the exact same speed. The first ball has zero forward velocity. The second has a forward velocity of, say, X. If you shot the ball with a forward velocity of 2X the arc would be flatter, with the ball landing approximately twice as far to the right. But all three balls would land at exactly the same time, minus small aerodynamic affects.
I’d like to add that the spinning of the bullet does cause *horizontal *drift. If the bullet is fired from a right twist barrel, the bullet will drift to the right. If the bullet is fired from a left twist barrel, the bullet will drift to the left.
While this is true, the OP is assuming the barrel is perfectly horizontal and the earth is not curved. Even then, the dropped bullet will hit the ground before the fired bullet. The fired bullet is “nose high” throughout its trajectory, meaning that the angle between the center line of the bullet and the trajectory (i.e. the yaw angle) is not zero… the center line of the bullet is angled slightly higher than the trajectory. This provides lift. In addition, the spinning of the bullet helps to ensure this yaw angle doesn’t get too large. Without the spin, the yaw angle could get so large that the bullet tumbles in flight.
Why? How can a spinning bullet emerge nose-high, from a barrel that’s parallel to the ground?
Presumably, “nose high” is relative to the direction of travel, not the horizontal, and the yaw angle would change through the flight of the bullet. See the link and discussion in ZenBeam’s post 30.
Just a nit-pick: Things “yaw” around a vertical axis; that is, they turn left or right while moving forward. When something moves about a horizontal axis, and the nose points up or down, that’s called “pitch”.
When your bullet, being axisymmetric and spinning, has turned considerably away from horizontal, it may be hard to tell what’s pitch and what’s yaw. But as long as it remains substantially parallel to the ground, the definitions are clear.
**Crafter_Man ** said:
**Askance ** said:
Strictly speaking, you are correct, the bullet leaves the barrel exactly aligned with the trajectory. However, as gravity works, the bullet’s path begins to curve downward. However, the spin stabilization resists the bullet from turning nose down. Thus the nose is “high” from the trajectory, and thus we get the “nose high” description that Crafter_Man mentions.
Crafter_Man, originally I said I didn’t think spinning would cause any aerodynamic lift, but because of the gyroscopic effect resisting turning, I now concede the “nose high” profile will make the aerodynamic cross-section nonuniform, and thus there may be lift.
ivan astikov, all of this discussion on the “nose high” and drop time from these experiments are discussing aerodynamic effects on the bullet. They are complications that inhibit your understanding of the basic physics principles that everyone is trying to explain to you. It would be better for you to ignore the discussions relating to spin and such until you grasp the fundamentals.
While the links provided by Jragon are appropriate, it does take a bit of time to learn the material. It isn’t that complicated, but does take unlearning your intuitive understanding. Going through the reading would help, but you appear unwilling to spend the time.
The first issue you seem unable to grasp is the fundamental of vectors, that is that motion can be separated into component parts that are independent. Let me try an example.
Suppose you wish to go to a mall that is located 1/2 mile east of you. That is short enough you can walk it, and since you have no car (in my hypothetical), you think it is okay.
But there is a rectangular lake between you and the mall. In order to go around the lake, you must first hike 5 miles north, then the 1/2 mile east, then 5 miles south. The 10 miles traveling north and south make your proposed journey a lengthy, time consuming, and potentially painful venture. But the 1/2 mile East is still 1/2 mile east. You could travel 20 miles north, 1/2 mile east, then 20 miles south and still reach the same mall, and you will have still traveled the same 1/2 mile east. That is the separability of vectors. Perpendicular axes are independent.
Velocity works the same way. Velocity is not just speed, but requires direction. Speed is scalar, i.e. direction independent. Velocity is a vector and requires direction to be complete. And any vector can be equivalently represented by breaking it into two components on perpendicular axes. Oops, now you need to know trigonometry. 
Acceleration (rate of change of velocity), for physics purposes, is also a vector and requires direction. We can therefore state that motion requires direction to be complete. Not just how fast is it going and how quickly is it getting to that speed, but which way is it going.
For the bullet scenario (flat terrain, no air), we are defining two perpendicular axes - horizontal and vertical. All motion can be defined in these two axes. We are looking 2 dimensional; the same can be done with three dimensions (horizontal being defined into length and width or North and East or X and Y), but it is an unneccessary complication to grasp the principle.
The dropped bullet has zero motion in the horizontal plane. Gravity begins to act as soon as it is released, and the bullet accelerates downward at 9.8 m/s[sup]2[/sup], until it hits the ground.
The bullet fired from the gun has some huge horizontal velocity (say 300 m/s). Therefore, it will sail sideways. Without any forces acting horizontally (and we said no air), it will go on forever in that direction at that speed. But as soon as it leaves the barrel, there is another force: gravity. Immediately gravity begins to act, but like the case of the dropped bullet, gravity is not a velocity. It does not give an instant speed. Instead, gravity is an acceleration. The fired bullet immediately begins to drop at 9.8 m/s[sup]2[/sup], exactly like the dropped bullet. However, it is moving sideways at the 300 m/s velocity. Therefore, the path of the fired bullet will be a parabolic curve, while the dropped bullet will have a straight line path. But because the motions are independent, the vertical motion for both is identical. Ergo, they both fall exactly the same and hit at the same time. One just hits a long ways to the side of the other one.
From the column:
ivan astikov said:
The word “tend” used above is a descriptive term used to represent inertia, which was described by Isaac Newton as the principle that objects resist changes to their motion. This is an observed property that no one can explain why (the mechanism), but he described how (the behavior). Newtons three laws:
- Objects in motion tend to stay in motion.
- Objects at rest tend to stay at rest.
- For every action there is an equal and opposite reaction.
For our purposes, the first 2 laws are a description of inertia. Take it as a fundamental law of the universe that motion does not change (speed or direction) without a Force being applied.
So why do planets travel in arcs instead of straight lines? They are being affected by a force - gravity.
The word ‘tend’ also implies there is an exception to the rule. Is there? And how does gravity - a force capable only of accelerating a mass to a maximum of 100 m per sec - overcome a force that is propelling an object at 300 m per sec, as though it is motionless? Also, why is there a terminal velocity at all? Forgive me wearing my ignorance on my sleeve.
You read too much into a single word. Cecil is trying to explain the concept as concisely as possible for a lay audience.
The idea behind Newton’s law here is that a particle moving at a certain speed in a certain direction will stay moving at that speed and in that direction, as long as there are no other forces on the particle.
However, in the question at hand, there are other forces on the particle (or the bullet, rather): gravity and aerodynamic drag. Those other forces change the speed and direction of motion, in a way which nicely follows Newton’s Laws.
By parenthetically referring to Newton’s first law, Cecil is making a point in context, which is that, in the absence of gravity (and drag, I suppose), the bullet would continue in a straight line, and the Earth would gradually curve away from it.
First of all, how on Earth do you conclude that gravity is “a force capable only of accelerating a mass to a maximum of 100 m per sec”? That is completely and fundamentally wrong, and I haven’t the foggiest notion of why you would think that is true.
Second of all, once the bullet leaves the rifle barrel, there is no “force that is propelling” it. It simply is travelling at that velocity. In any case, gravity produces an acceleration, which is not the same thing as velocity.
And thirdly and most importantly: you need to go back and read the portion of Irishman’s post that deals with vectors. If you don’t understand parts of it, ask. I’m sure Irishman would be happy to explain; if he’s not, I will.
However, in a nutshell: gravity acts vertically, the velocity of the bullet is primarily horizontal. They are not in the same direction; they do not oppose each other. Have you ever pushed a car? Why do you suppose one person can roll a car forward, but cannot lift it in the air?
Why are you bringing terminal velocity into this discussion? I can and would be happy to explain the answer to this question, but it has nothing at all to do with the rest of this thread.
See this?
Isn’t 9.8 m/s squared, around about a 100 m/s, or is it you that is wrong?
Look, I’m having a chat with someone who can do graphs, and I’ll get back to you regarding the rest. And you can forget the ‘terminal velocity’.
Gravity is an acceleration; its units are meters per second per second. In other words, 9.8m/(s[sup]2[/sup]), not (9.8m/s)[sup]2[/sup].
As an aside, are you clear on the difference between velocity and acceleration? That’s fundamentally important, but also a concept that’s not infrequently misunderstood.
So, how many m per sec is terminal velocity* then? Isn’t that what we are talking about, in essence?
Tbh, no. If it can be explained without too much math, I’m all ears.
- I knew there was a reason I mentioned it earlier.
Depends on the object. Terminal velocity is when the drag caused by air resistance exactly equals the force of gravity so the object stops accelerating.
No. Terminal velocity doesn’t come into it at all. The question in its simplest form assumes a vacuum, in which there is no terminal velocity; in addition, a bullet dropping from shoulder height doesn’t have time to achieve terminal velocity.
Velocity is speed (with a vector (direction) component). Acceleration is a change in velocity; getting faster or slower (or changing direction). Acceleration requires a force. Once that force goes away, the acceleration stops and the velocity remains constant. The fired bullet, once it leaves the barrel (really, once it leaved the cartridge), no longer has a force acting on it in the horizontal vector. The only force acting on it is gravity, which acts on the vertical vector, in the exact same way as the dropped bullet, accelerating them downward. Again this is all assuming that this occurs in a vacuum where there is no air resistance.
The term 9.8m/s[sup]2[/sup] means that for every second that passes, the velocity increases by 9.8m/s. So after one second, a dropped object is travelling 9.8m/s, after two seconds 19.6m/s, etc. Absent another force (like air resistance or hitting the ground) there is no limit to this increase in velocity.
As just said, terminal velocity is only relevant when you are dropping an object in an atmosphere. The only acceleration acting on the object is gravity. In a vacuum, nothing would slow down the object and it would continue increasing in speed until it hit the earth (or whatever massive object was the source of the gravity). Near the earth, it would be traveling at 9.8 m/s at the end of the first second, 18.6 m/s at the end of the second second, 29.4 m/s at the end of the third second, and so on.
If you introduce air, however, the object doesn’t fall a 9.8 m/s^2. At the end of the first second it will be going somewhat slower. And this trend continues the faster the object travels. Eventually, the force of 9.8 m/s^2 is exactly matched by the drag of the object moving through the air, and that speed is the object’s terminal velocity. It’s different for every object because drag depends on the shape, density, and orientation of the object. It will be different in different mediums (such as air versus water) and at different distances from the center of mass of the gravity source.
A lot of physics is handled in the world of no atmosphere and no friction because it’s simpler to understand the concepts. When you add in those complications it becomes more engineering then physics.
**ivan astikov ** said:
“Tend” is the phrasing Newton chose to signify that objects have a certain natural state with regards to motion. That natural state is that the motion will not change on its own, but requires some action in order to change. Newton defined the action as a force, and defined force, and gave a nice mathematical relationship, usually simplified as Force = mass x acceleration.
Read “tend” to mean “objects will do this unless something else interferes”.
Ahh, I see a lack of clarity in my presentation. Sometimes it is difficult to remember just how basic the information has to be to be clear.
I think zut and fachverwirrt have clarified, but I will repeat. Velocity is how fast something is going (and in which direction), acceleration is how quickly it is changing velocity (and in which direction is it changing). Here is the relationship.
distance is linear, meters (m)
velocity is rate of change of distance, meters / second (m/s)
acceleration is rate of change of velocity, meters/second / second (m/s /s = m/(s*s) = m/(s[sup]2[/sup]))
Terminal velocity is how fast something falls through air. It is a measure of the velocity at which the force from air resistance equals the force from gravity, gravity pulling down and air pushing up. In a vacuum, there is no air, so there is no terminal velocity. The object just keeps getting faster and faster until it hits something or reaches the speed of light.
When a bullet is fired from a gun, it is accelerated by the propellent in the gun burning and pushing the bullet down the barrel. As the bullet leaves the barrel, the gas no longer pushes on the bullet and expands outward, so the velocity that the bullet has upon exiting the barrel is the fastest it will go. Without other forces, it will remain at that speed until it hits something (which is another force). Air resistance is a force which will slow the bullet down, but that slowing is not very effective, so the bullet keeps going a long ways (i.e. miles). Gravity is also a force, which acts to make the bullet fall to the ground.
If shot horizontally, the gravity will pull the bullet into hitting the ground long before the air resistance could possibly slow the bullet to a stop. The air resistance is essentially negligible for the comparison purposes.
It isn’t, as has been pointed out, so I’ll skip this for now.
Actually, you already know the difference between velocity and acceleration, you just don’t know you know.
Velocity is a change in position with time.
Acceleration is a change in velocity with time.
Think of driving somewhere in car. Suppose you start at home, drive 50 miles, and it takes you an hour. That’s 50 miles per hour, right? That’s velocity: a change in position (from home to 50 miles away) with time (one hour). Likewise, if you travel 25 miles and it takes you 30 minutes, that’s still 50 mph, right? And, of course you can change the units: meters per second is also a velocity, which describes how many meters you would travel in one second.
Now, think of driving in a car again. Suppose you’re traveling at 30 mph. You desice that’s too slow, and you hit the accelerator, and change your speed to 50 miles per hour. It takes you 10 seconds. That’s acceleration: a change in velocity with time. In this case, you changed 20 mph in 10 seconds, or 2 miles per hour per second. That’s a weird unit (normally we’d use meters per second per second, or feet per second per second), but it’s an acceleration.
Yes - the exception is when a force acts upon the objects so as to defy their tendency.