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B Toss of a coin

  1. Jun 10, 2018 #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?
     
  2. jcsd
  3. Jun 10, 2018 #2

    ZapperZ

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    Initial upward velocity.

    Zz.
     
  4. Jun 10, 2018 #3

    BvU

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    The fact that it already had been given an upward velocity
     
  5. Jun 10, 2018 #4

    BvU

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    Inertia makes that this velocity is reduced (by g m/s per second) only 'gradually'
     
  6. Jun 10, 2018 #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?
     
  7. Jun 10, 2018 #6

    BvU

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    Who says we don't ? The scientific term is 'momentum' : mass x velocity
     
  8. Jun 10, 2018 #7

    ZapperZ

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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.
     
  9. Jun 10, 2018 #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.
     
  10. Jun 10, 2018 #9

    ZapperZ

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    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.
     
  11. Jun 10, 2018 #10

    Beanyboy

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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.
     
  12. Jun 10, 2018 #11

    ZapperZ

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    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.
     
  13. Jun 10, 2018 #12

    jtbell

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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.
     
  14. Jun 10, 2018 #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
     
    Last edited: Jun 11, 2018
  15. Jun 10, 2018 #14

    Beanyboy

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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. 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.
     
  16. Jun 10, 2018 #15

    sophiecentaur

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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.
     
  17. Jun 10, 2018 #16

    ZapperZ

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    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.
     
  18. Jun 10, 2018 #17

    Beanyboy

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    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?
     
  19. Jun 10, 2018 #18

    russ_watters

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    You tell us. I'll give you a hint: you've already said it, here in this thread!
     
  20. Jun 10, 2018 #19

    Beanyboy

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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.
     
  21. Jun 10, 2018 #20

    russ_watters

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    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!
     
  22. Jun 10, 2018 #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.
     
  23. Jun 10, 2018 #22

    BvU

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    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}##
     
  24. Jun 10, 2018 #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.
     
  25. Jun 10, 2018 #24

    Dale

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    The initial upwards velocity is not a force. The units are different.
     
  26. Jun 10, 2018 #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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