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Approximations and forces

  1. Sep 28, 2005 #1
    Approximations.... and forces

    Just out of curiosity, why is it super important to round everything in physics, even stuff such as 2.666(place recurring sign above 6) rather than being able to use recurring signs and such? Wouldn't it be better to use recurring signs and radical notation and such? (My teacher usually tells us off if we do so... but aren't they better for purely precise answers?)
    Also, is it common in physics, and perhaps general mathematics, to use taylor series approximations and such for tedious-to-evaluate functions involving differentials and integrals?
    Two perhaps unrelated questions, but another out-of-the-blue question:
    Suppose I'm in this magical realm and on a frictionless surface. Let's say there's a rock... I don't know, 2 meters away from me? And I am "infinity" away from any other objects, borders or anything. (i.e. they are so far away, I'm going to consider their gravitational forces negligible) Now, according to Newton's law of Universal Gravitation, the rock is going to be exerting a gravitational force on me, and I will be exerting a force on it.....
    So we'll be moving towards each other until we collide (Please correct me if I'm pulling this all out of my "poop chute") and then I'd be moving in the direction the rock is moving (assuming it's more massive than I am) because of my lack of inertia, correct?
    If I wanted to calculate the distance the rock travels in, oh, say 5 seconds, I'd first have to use conservation of momentum for JUST before it hit and just after the collision and F = ma to find the acceleration of the two "stuck-together" masses, and then use the common physics formulas, right?
    Or did my understanding totally go down the drain at some point or another?
    I hope someone can help clarify my understanding of all of these interrelated subjects.
    Thanks alot for any responses.
  2. jcsd
  3. Sep 28, 2005 #2


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    This looks wrong to me. If you think of your collisiong with the rock as an elastic one, you'll both end up going in the opposite direction as you were before colliding. If you think of your collision with the rock as inelastic (more realistic imo), then after collision, you'll end up stuck together, both with velocity 0.

    Both the scenario respect conservation of momentum: total momentum is 0 at all time before and after collision.

    And it is not super important to round anything in theoretical physics. However, in experimental physics, it important to respect the number of significant digit. For exemple, if you know a value to be precise at 3 decimal places, say 1.345 and the final answer you get after computations is [itex]1.345*\sqrt{2}[/tex], then you know nothing about the value of the 4rth decimal, and giving [itex]1.345*\sqrt{2} = 1.902117241...[/tex] as the answer would implie that you supposed the 4th, 5th, 6th, etc decimal of 1.345 to be 0. But you don't know what they are, so it's false to claim that the answer is [itex]1.345*\sqrt{2} = 1.902117241...[/tex]. All you can tell is that the correct answer to 3 decimal places is 1.902.

    Finally, yes Taylor series are very useful in physics for approximations. An example of the use of Taylor serie that you might have met but not recognized is the seemingly harmless problem of the simple pendulum. This problem is very hard to solve analytically (i.e. exactly) because the equation to solve is [itex]d^2\theta / dt^2 = -gsin(\theta)/l[/itex] and that's not easy. However, an expansion of [itex]sin(\theta)[/itex] in Taylor serie tells us that as long as [itex]\theta[/itex] is small, [itex]sin(\theta)[/itex] is approximately equal to [itex]\theta[/itex]. So we can at least solve the simple pendulum problem for small angles by re-wrinting it [itex]d^2\theta / dt^2 = -g\theta/l[/itex]. But this is just the equation caracteristic of Simple Harmonic Motion for which the solution is well know to be [itex]\theta(t) = \theta_0 cos(\sqrt{g/l}t +\phi)[/itex]!
  4. Sep 28, 2005 #3

    For the first part, I can only point you to the concept of significant figures. Basically, a result can't be more precise than the input data.

    Taylor series expansions are commonly used in physics and math to reduce the complexity of an equation. It is done to simplify an equation either to make it easier to solve (for whatever reason). You have what kind of accuracy the approximate equation is going to have as well as the domain of convergence.

    For the you and the rock, the total momentum of the system is 0 at the start since both objects are at relative rest. When you collide you will both have the same magnitude of momentum but in opposite directions (since you each had the same magnitude of force acting on you but in opposite directions for equal time). If you and the rock collide and stick together then you'll once again be at rest as momentum is conserved and the total momentum is still 0.

    I'm not too clear on the rest of what you're saying. You want to know when the rock and you collide or what the final velocity is? I'm just not clear on what your final goal is.
  5. Sep 28, 2005 #4
    Nevermind my final goal.... I just wanted to see if my thinking was right-- but I forgot about the fact that the forces are equal and thus once they "hit" each other they are at rest. (That's it, right?)
    Thanks for the replies guys, appreciate it...
    quasar987: Yeah, I saw something about the small angle approximation in one of those MIT video lectures on physics.
  6. Sep 28, 2005 #5

    Doc Al

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    Not exactly. When two things collide the forces they exert on each other are always equal (Newton's 3rd law), so that's not it. As Geoff explained, the reason why the final speed is zero (assuming that rock and person stick together after colliding) is that momentum is conserved: Assuming it starts out with zero momentum, it ends up with zero momentum.
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