Toss of a coin

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  • #1
Beanyboy
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I'm confused. According to my textbook (Hewitt), if we discount air-resistance, the ONLY force acting on a tossed coin at all points of its trajectory, is mg. For me, this begs the question: what then is causing the coin to move upward during the first half of its trajectory?
 

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  • #2
ZapperZ
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I'm confused. According to my textbook (Hewitt), if we discount air-resistance, the ONLY force acting on a tossed coin at all points of its trajectory, is mg. For me, this begs the question: what then is causing the coin to move upward during the first half of its trajectory?

Initial upward velocity.

Zz.
 
  • #3
BvU
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The fact that it already had been given an upward velocity
 
  • #4
BvU
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Inertia makes that this velocity is reduced (by g m/s per second) only 'gradually'
 
  • #5
Beanyboy
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If a force is "a push" or, "a pull", why then do we not describe the "initial upward velocity" as "a push" imparted to the coin?
 
  • #6
BvU
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Who says we don't ? The scientific term is 'momentum' : mass x velocity
 
  • #7
ZapperZ
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If a force is "a push" or, "a pull", why then do we not describe the "initial upward velocity" as "a push" imparted to the coin?

This is very difficult to decipher.

A "push" usually means there is an applied force acting on a body. Once the coin leaves your hand on the way up, there is no longer a "push" by the hand onto the coin. The ONLY force then acting on the coin is gravity. If there is no gravity, the coin will continue to move upward at a constant velocity. Due to gravity, the coin moves upward first but it is slowing down.

You need to go back to Newton's First Law and understand what it says.

Zz.
 
  • #8
Beanyboy
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Describing the "initial upward velocity" as a force seems to contradict Hewitt's assertion that the ONLY force acting on the object throughout its trajectory, is mg. We now seem to be implying there are in fact TWO forces acting on the object: "upward initial velocity" and "mg".

Thanks for the help, but still confused here.
 
  • #9
ZapperZ
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Describing the "initial upward velocity" as a force seems to contradict Hewitt's assertion that the ONLY force acting on the object throughout its trajectory, is mg. We now seem to be implying there are in fact TWO forces acting on the object: "upward initial velocity" and "mg".

Thanks for the help, but still confused here.

One more time: INITIAL UPWARD VELOCITY is NOT A FORCE!

"velocity" has units of m/s. Force has units of N. Why are you confusing these two as being the same?

Zz.
 
  • #10
Beanyboy
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This is very difficult to decipher.

A "push" usually means there is an applied force acting on a body. Once the coin leaves your hand on the way up, there is no longer a "push" by the hand onto the coin. The ONLY force then acting on the coin is gravity. If there is no gravity, the coin will continue to move upward at a constant velocity. Due to gravity, the coin moves upward first but it is slowing down.

You need to go back to Newton's First Law and understand what it says.

Zz.
One more time: INITIAL UPWARD VELOCITY is NOT A FORCE!

"velocity" has units of m/s. Force has units of N. Why are you confusing these two as being the same?

Zz.
I'm familiar with Newton's First Law and the fact that it is a special case of Newton's Second Law, since the sum of the net forces is zero.

I thank you for the distinction between the "initial applied force" and the forces acting on the coin once airborne: this helped. You might consider avoiding writing in caps - this can give the reader the impression that you are shouting.
 
  • #11
ZapperZ
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I'm familiar with Newton's First Law and the fact that it is a special case of Newton's Second Law, since the sum of the net forces is zero.

I thank you for the distinction between the "initial applied force" and the forces acting on the coin once airborne: this helped. You might consider avoiding writing in caps - this can give the reader the impression that you are shouting.

I HAD to do it because writing "initial upward velocity" didn't appear to sink in since you kept thinking that it is a force, even when the word "velocity" was in there.

However, even reading this, it appears that you're still thinking of some "initial" force, when all you should be caring about is the fact that this object has an initial velocity, no matter how it got it. You need to straighten this out before you dive into 2D projectile motion.

Zz.
 
  • #12
jtbell
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According to my textbook (Hewitt), if we discount air-resistance, the ONLY force acting on a tossed coin at all points of its trajectory, is mg. For me, this begs the question: what then is causing the coin to move upward during the first half of its trajectory?
The force that causes the coin to move upwards is the one that you exert on it with your hand. However, in this context, "trajectory" means after the coin has left your hand. The only force acting on the coin now is gravity. The force that you exerted on it has already taken place before the trajectory began. It determined the initial upward velocity.
 
  • #13
BvU
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dear Bb,

Seems you have some elementary background missing, let's try to fill that up:

http://pi.math.cornell.edu/~numb3rs/jrajchgot/508f.html

In this link, the first graph shows (only qualitatively) something similar to your coin flight: Initially it goes upward with an upward velocity. The acceleration is constant and in an opposite direction, so the upward velocity decreases linearly with time. It reaches zero at some point, meaning the extreme position (max height) is reached. Velocity changes sign and from now on is downward, in the same direction as the acceleration. Meaning its magnitude increases: falling faster and faster.

The acceleration (blue line) has been constant and negative (= downward) all the way from time zero
 
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  • #14
Beanyboy
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The force that causes the coin to move upwards is the one that you exert on it with your hand. However, in this context, "trajectory" means after the coin has left your hand. The only force acting on the coin now is gravity. The force that you exerted on it has already taken place before the trajectory began. It determined the initial upward velocity.

This is very helpful. I was aware that at the crest of its motion, the sum of the forces would be zero. I was asking myself: What's the other force that's being negated by the weight of the coin?".Hewitt's text threw me, but thanks to you, I now see that "trajectory" means after the coin has left the hand.

Thanks again for the clarifying.
 
  • #15
sophiecentaur
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I'm familiar with Newton's First Law and the fact that it is a special case of Newton's Second Law, since the sum of the net forces is zero.
The net sum of the forces is NOT zero. If they were, it would carry on upwards and leave the Earth. There is a very non-zero force acting which is the weight force on the object. Gravity is the only force acting, once it has left the arm of the thrower and it applies for all the lifetime of the object - before and after the throw. Before and after it's journey, the weight force is balanced by the table it was resting on and the ground where it lands.

Your confusion was explained, way back, by Galileo and others and the 'reality' is not intuitive so you were in good company.
 
  • #16
ZapperZ
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This is very helpful. I was aware that at the crest of its motion, the sum of the forces would be zero.

Again, this is why I said earlier that I still think you have not understood this scenario. This is still wrong.

Throughout the trajectory, i.e. AFTER it has left the tossing hand, there is only ONE constant force - gravitational force. This is true at every point along the motion of the coin, even at the "crest".

So no, I do not believe you've understood what Newton's First Law is. You may have read it, and think you know, but everything that you've said here seems to indicate that you do not know what the words mean physically.

Zz.
 
  • #17
Beanyboy
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The net sum of the forces is NOT zero. If they were, it would carry on upwards and leave the Earth. There is a very non-zero force acting which is the weight force on the object. Gravity is the only force acting, once it has left the arm of the thrower and it applies for all the lifetime of the object - before and after the throw. Before and after it's journey, the weight force is balanced by the table it was resting on and the ground where it lands.

Your confusion was explained, way back, by Galileo and others and the 'reality' is not intuitive so you were in good company.
The thick plottens! Thanks for drawing attention to my mistake. "Of all the intellectual hurdles which the human mind has confronted and has overcome in the past fifteen hundred years, the one which seems to me to have been the most amazing in character and the most stupendous in the scope of its consequences is the one relating to the problem of motion.... " (Butterfield)

What then is the acceleration of the coin at the top of its trajectory?
 
  • #18
russ_watters
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What then is the acceleration of the coin at the top of its trajectory?
You tell us. I'll give you a hint: you've already said it, here in this thread!
 
  • #19
Beanyboy
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You tell us. You have all the information in front of you, here in this thread!

Well, I would have assumed it was zero. But, I assumed that this zero was the sum of two forces. Since now I'm being told there's only one force in operation while the coin is in motion, I'm left wondering how does that become zero?,

I used a = f/m. Since the net forces in the numerator come to zero, this yields zero acceleration.
 
  • #20
russ_watters
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Well, would have assumed it was zero. But, I assumed that this zero was the sum of two forces. Since now I'm being told there's only one force in operation while the coin is in motion, I'm left wondering how does that become zero?
You gave us the correct answer in your very first post and the rest of the thread has been you repeatedly not wanting to accept the answer that you already know. So please: tell us the answer you already gave us -- and then accept it!
 
  • #21
Beanyboy
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I'm trying to understand something that's confusing to me. If you think you can help me, then I'm grateful to you. If you think I'm deliberately being willfully stupid, or unable/unwilling to accept an answer, then I'm sorry you feel that way - I assure you, I'm not.

As a learner one must humble oneself to reach out and expose one's lack of understanding. Seeming "stupid" is not something that gives me any pleasure. Please try to be a bit more empathetic.
 
  • #22
BvU
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I was aware that at the crest of its motion, the sum of the forces would be zero.
It is not. The velocity itself is zero, but the change in velocity (the acceleration ##g##) is as it was and remains as it was: 9.81 m/s2 in a downward direction.

Before launch, when the coin is at rest on the launchers thumb, the sum of forces is zero. No acceleration (Fnet = Fgravity+ Fthumb = 0 = ma ##\Rightarrow## a = 0). No change in velocity, no velocity, and therefore a constant position.

During launch the thumb exerts a hefty upward force > mg that accelerates the coin upwards, so the coin shoots up.

After launch the only one who exerts a force is mother earth, pulling down on the coin with a constant force mg = ma so the acceleration (= the change in velocity per unit time) is downwards.

I am sincerely convinced you are not stupid. It's just that you have difficulty distinguishing position ##x##, velocity ##v = {dx\over dt} ## and acceleration ##a = {dv\over dt} = {d^2 x\over dt^2}##
 
  • #23
Beanyboy
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It's very kind of you to take time out to explain. I'm extremely grateful to you.

I thought I was aware of the distinction between velocity and acceleration, both qualitatively and quantitatively, but it seems my thinking is muddled.

It's going to take me a while to digest your answer and try to figure out the nature of my misunderstanding. Expect a more detailed reply within 24 hours.
 
  • #24
Dale
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Describing the "initial upward velocity" as a force
The initial upwards velocity is not a force. The units are different.
 
  • #25
Vanadium 50
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Here's something that will help. Draw three graphs - one of the coin's position vs. time, one of the coin's velocity vs. time and one with the coin's acceleration vs. time. This will help, but only if you actually draw them. Not think about drawing them, not deciding that you know what they look like so you don't have to draw them, but actually draw them.
 
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  • #26
Dale
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Here's something that will help. Draw three graphs - one of the coin's position vs. time, one of the coin's velocity vs. time and one with the coin's acceleration vs. time. This will help, but only if you actually draw them. Not think about drawing them, not deciding that you know what they look like so you don't have to draw them, but actually draw them.
+1 on this. A very important suggestion
 
  • #27
vela
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Well, I would have assumed it was zero. But, I assumed that this zero was the sum of two forces. Since now I'm being told there's only one force in operation while the coin is in motion, I'm left wondering how does that become zero?,

I used a = f/m. Since the net forces in the numerator come to zero, this yields zero acceleration.
Your assumption that a=0 at the top is wrong. That's the idea you need to let go of. You know the only force on the coin is mg, and you know Newton's second law tells you a=F/m. It follows then that a=g.

The question you want to consider is, why do you think a=0 at the top of the trajectory? That's where your misconception lies.

If you're using the 12th edition of Conceptual Physics, check out figure 3.8 on page 48.
 
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  • #28
Beanyboy
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Your assumption that a=0 at the top is wrong. That's the idea you need to let go of. You know the only force on the coin is mg, and you know Newton's second law tells you a=F/m. It follows then that a=g.

The question you want to consider is, why do you think a=0 at the top of the trajectory? That's where your misconception lies.

If you're using the 12th edition of Conceptual Physics, check out figure 3.8 on page 48.
Thanks for the help. I'm beginning to see where my problem lies - conflating velocity with acceleration. I assumed, wrongly, that since v = 0, then a must be 0 too.

However, since a = F/m and since F is mg, then , a = g. Thus, the acceleration is a constant throughout the trajectory.
I do have the 12th edition, but Fig. 3.8 is on p.74 in mine. Is it the one with the stone being launched from the cliff-top?
Many thanks again for taking time out to explain this.
 
  • #29
Beanyboy
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Here's something that will help. Draw three graphs - one of the coin's position vs. time, one of the coin's velocity vs. time and one with the coin's acceleration vs. time. This will help, but only if you actually draw them. Not think about drawing them, not deciding that you know what they look like so you don't have to draw them, but actually draw them.

Very helpful. Yes, I can see that the slope of the position vs time graph gives me the velocity; slope of velocity vs time graph gives me the acceleration. Most notably, the acceleration remains constant throughout.

Interestingly, your suggestion struck me initially as being "a lot of bloody work for one coin toss" and so I hesitated. Then, a voice inside me said: it's the details you idiot, the details that matter! Thanks again.
 
  • #30
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Then, a voice inside me said: it's the details you idiot, the details that matter!
Awesome! Listen to that voice. That is the one that will drive your understanding if you are willing to put in the work too.
 
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  • #31
Beanyboy
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I am sincerely convinced you are not stupid. It's just that you have difficulty distinguishing position ##x##, velocity ##v = {dx\over dt} ## and acceleration ##a = {dv\over dt} = {d^2 x\over dt^2}##
Awesome! Listen to that voice. That is the one that will drive your understanding if you are willing to put in the work too.
 
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  • #32
Beanyboy
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Awesome! Listen to that voice. That is the one that will drive your understanding if you are willing to put in the work too.

My stupidity never ceases to amaze me. I try to compensate with stubbornness. Thanks for your support.
 
  • #33
Stephen Tashi
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"Of all the intellectual hurdles which the human mind has confronted and has overcome in the past fifteen hundred years, the one which seems to me to have been the most amazing in character and the most stupendous in the scope of its consequences is the one relating to the problem of motion.... " (Butterfield)

Learning the scientific approach to motion requires ignoring everyday experience! Teachers of physics have less sympathy with students' intuitions about motion that historians of science, like Butterfield. Physicists are used to "saving the phenomena" (a phrase with long history in philosophy and science). After one knows a scientific theory, it becomes natural to interpret experience in a way that it conforms to the theory. Students are expected to "save the phenomena" rather than save the scientific theory when it appears to be contradicted by everyday experience.

It is our everyday experience that to cause motion and keep an object in motion, we must continue to exert a force on the object. (e.g. pushing a wheelbarrow). The scientific (Newtonian) view is that it requires zero force to keep an object moving at a constant velocity. Newton's approach to motion was an extraordinary achievement. People accustomed to saving the phenomena of ordinary experience by interpreting it in terms of Newton's approach may think that Newton's approach is intuitively obvious. The history of science shows otherwise.

As a student, your main job is to understand the theory completely (e.g. F = MA, so when velocity is constant F = 0. F is the net force, not the force exerted only by yourself, etc.) If the theory is an outstanding intellectual achievement in the history of science, don't expect it to be an obvious consequence of everyday experience.
 
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  • #34
vela
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I do have the 12th edition, but Fig. 3.8 is on p.74 in mine. Is it the one with the stone being launched from the cliff-top?
Yes, that's the figure. I was looking at the electronic version, which might be formatted differently than the print version.
 
  • #35
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Thanks for the help. I'm beginning to see where my problem lies - conflating velocity with acceleration. I assumed, wrongly, that since v = 0, then a must be 0 too.
I hope you understand now that they are very different concepts.

Another of your misunderstandings is thinking there are two forces on the coin. At the very beginning of the experiment, when the person tosses the coin into the air, there are two forces: an upward push and the force due to gravity acting downward. However, as soon as the coin leaves the person's hand, the only force on the coin is that due to gravity. All we're interested in for this problem is the trajectory of the coin once it has left the person's hand.
 

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