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Galileo, inclined plane and geometry.

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Galileo in seeking to discover the laws governing the motion of bodies under the action of their weight, conducted a series of experiments on inclined planes. Choosing as unit of length the distance traveled by the ball the first unit of time, measuring at subsequent time points the distance traveled and repeating the experimental procedure with different mass ball finds the same numerical result: the distance [itex]s[/itex] traveled by any ball in the inclined plane under the influence of the weight, is square proportional to the time elapsed: [itex]s(t)=kt^2[/itex].
    Gradually increasing the slope of the plane, finds that the value of [itex]k[/itex] increases to a maximum value which takes at the vertical gradient, i.e. freefall. Can you determine the relation between [itex]k[/itex] and the slope of the inclined plane?

    2. Relevant equations

    [itex]s(t)=kt^2[/itex]

    3. The attempt at a solution

    I tried this one: let the inclined plane have angle [itex]\theta[/itex], then [itex] sin \theta= \frac{AB}{AO}[/itex], we use now the fact that the distance traveled by the ball is [itex]s(t)=kt^2[/itex] so we can find the distances [itex]AO[/itex] and [itex]AB[/itex], that is [itex] sin \theta= \frac{AB}{AO} = \frac{gt^2}{kt^2} = \frac{g}{k}[/itex], but while [itex]\theta[/itex] increases, [itex]k[/itex] decreases, where is the wrong???

    I tried also this: the first unit of time the ball travels distance: [itex]s(1)=k[/itex],
    the second unit of time distance: [itex]s(2)=4k[/itex]
    and the third unit of time distance: [itex]s(3)=9k[/itex].
    I tried to use the Thales' theorem (Intercept theorem), but I don't know how to move on.
     

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  3. Nov 12, 2012 #2
    k should be related to the component of the weight along the incline.
     
  4. Nov 12, 2012 #3
    That's wrong. [itex]k[/itex] is related to the slope of the inclined plane and is independent from the weight. If we use Newton's law we have [itex]mk=mgsin \theta \Rightarrow k=gsin \theta[/itex] I must prove this (without using Newton's law).
     
  5. Nov 12, 2012 #4

    haruspex

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    Basic_Physics means that it's related to the fraction of the weight that acts along the slope. You have k sin(θ) = g, which makes k > g. Clearly that's wrong. What should the equation say?
     
  6. Nov 13, 2012 #5
    The equation should say k=sin(θ)g, im writing above what i have tried. Can someone help me to prove it?
     
  7. Nov 13, 2012 #6
    For Galileo's choice of units of distance k will always be one:
    " Choosing as unit of length the distance traveled by the ball the first unit of time..."
    A graph of s vs t2 will be directly proportional with gradient k, but the graph goes through the point 1,1. This means its gradient will be k=1.
     
  8. Nov 13, 2012 #7
    sin(θ) ≤ 1.0 so you are fine. With a little bit of reasoning one can came to the right conclusion then. When the plane is flat k should be 0 (no acceleration). When the plane is vertical k should be equal to g.
     
  9. Nov 13, 2012 #8
    This ball is rolling not slipping
     
  10. Nov 13, 2012 #9
    I must determine the relation between k and the slope of the inclined plane. Im writing at the top of the topic what I have tried, I found g=ksin(θ) this is wrong. Im trying to use the trigonometry of the problem (no sees that???????). Can someome really help me to find the relation between k and the slope of the inclined plane? (without using Newton's law, but only the problem's data)
     
  11. Nov 13, 2012 #10
    One meaning that can be given to k is that it is the distance travelled along the incline when t = 1. Galileo did these "diluted acceleration" experiments to determine the relation beween distance covered and time for uniform accelerated motion. The component of gravity along the incline is
    g[itex]\;[/itex]sin([itex]\theta[/itex])
    This means that the speed obtained after 1 time unit will be
    v[itex]_{1}=g\;sin(\theta)[/itex]
    The average speed is then
    v[itex]_{avg}=\frac{g\;sin(\theta)}{2}[/itex]
    if it starts out of rest. The distance covered along the incline for 1 time unit will then be such that
    [itex]\frac{g\;sin(\theta)}{2}\;t=k\;t^{2}[/itex]
    so that
    [itex]\frac{g\;sin(\theta)}{2}\;=\;k[/itex]
     
  12. Nov 13, 2012 #11
    If the vertical acceleration is 'a' then:

    [itex] sin \theta= \frac{AB}{AO} = \frac{0.5at^2}{kt^2} = \frac{a}{2k}[/itex]

    When the plane is vertical (ie freefall), [itex]sin \theta = 1[/itex] and [itex]k = \frac{a}{2}[/itex].

    From Newton's Law, at freefall, [itex]s(t) = \frac{1}{2}gt^2[/itex]. From Galileo, [itex]s(t) = kt^2 = \frac{1}{2}at^2[/itex] so that the vertical acceleration a = g.

    Is this the answer being sought? The mistake was using [itex]gt^2[/itex] as the vertical distance AB since the vertical acceleration is only equal to g in freefall.
     
  13. Nov 13, 2012 #12
    You haven't understand the problem at all, anyway I found the solution.
     
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