Jose Guzman(*) and Ignacio Delgado
version 1.0.1
December 09, 2009
(*) This file is under construction. Please send any suggestions or comments to nin@neurohost.org
1. Introduction:
We will consider the equation to describe the vertical motion of a ball, and analyze and solve with Sage. Code is extensively commented through the notebook and should be self-explanatory. Comments and suggestions are greatly appreciated.
2. Ball motion equation:
From Newton's second law of motion we can write a mathematical formulation of a ball motion. With this equation we can calculate the ball's vertical position (y) according to the time as follows:
$y(t) = V_ot -\frac{1}{2}gt^2$
where $y$ is the vertical distance, $V_o$ is the initial velocity of the ball, $g$ is the acceleration of gravity, and $t$ is time.
︡6661011c-5169-4b96-9f7c-8fdf42b2ac83︡{"html": "Jose Guzman(*) and Ignacio Delgado
\r\nversion 1.0.1
\r\nDecember 09, 2009
\r\n(*) This file is under construction. Please send any suggestions or comments to nin@neurohost.org
\r\n\r\n
1. Introduction:
\r\nWe will consider the equation to describe the vertical motion of a ball, and analyze and solve with Sage. Code is extensively commented through the notebook and should be self-explanatory. Comments and suggestions are greatly appreciated.
\r\n\r\n
2. Ball motion equation:
\r\nFrom Newton's second law of motion we can write a mathematical formulation of a ball motion. With this equation we can calculate the ball's vertical position (y) according to the time as follows:
\r\n$y(t) = V_ot -\\frac{1}{2}gt^2$
\r\nwhere $y$ is the vertical distance, $V_o$ is the initial velocity of the ball, $g$ is the acceleration of gravity, and $t$ is time.
"}︡ ︠43e770bf-22cc-47c3-bf04-f512024b58eb︠ # define equation var('t,v0,g') # symbolic variables y(t)=v0*t-(g*t**2)/2 # ball motion equation #result = y.subs(v0=5,g=9.81,t=1) #print result.pyobject() # we need pyobject() to get the value ︡08771877-8109-41aa-a63d-90910f59d962︡︡ ︠068557f7-cc50-43c5-8233-4df903fdb7f3i︠ %html3. Solving the equation with Sage:
We can look for for solutions to the equation a y=0, to know the time it takes for the ball to move upwards and returns to y again.
$V_ot -\frac{1}{2}gt^2 =0 ; t(V_o -\frac{1}{2}gt) = 0$
We have two possible solutions, corresponding to
$t=0$ or $t=2V_o/g$
We can easily solve this with Sage.
︡0fae2c8d-f379-4b9a-84e3-a0dee576e837︡{"html": "3. Solving the equation with Sage:
\r\nWe can look for for solutions to the equation a y=0, to know the time it takes for the ball to move upwards and returns to y again.
\r\n$V_ot -\\frac{1}{2}gt^2 =0 ; t(V_o -\\frac{1}{2}gt) = 0$
\r\nWe have two possible solutions, corresponding to
\r\n$t=0$ or $t=2V_o/g$
\r\nWe can easily solve this with Sage.
"}︡ ︠abf2755d-edad-42fa-bf54-285d7ae947e1︠ solve(y==0,t) # solve equation y=0 with respect to t ︡73c7817f-a836-4b4d-975a-0159141aa12d︡{"stdout": "[t == 2*v0/g, t == 0]"}︡ ︠c77262c7-93c6-4b91-89e7-d1a9edd4ededi︠ %htmlWe can calculate the maximal vertical distance reached by simply taking the derivative of the function and solve it against time.
${\displaystyle\frac{dy}{dt}=V_o -gt}$
Once Again, Sage allows us to perform this simple calculation.
︡5cc0fdf7-107b-436b-996e-02d70e23d37b︡{"html": "We can calculate the maximal vertical distance reached by simply taking the derivative of the function and solve it against time.
\r\n${\\displaystyle\\frac{dy}{dt}=V_o -gt}$
\r\nOnce Again, Sage allows us to perform this simple calculation.
"}︡ ︠67676ecd-9e14-4575-8e42-f5e930a07860︠ diff(y,t) # calculate dy/dt ︡dbe979d5-7e86-4cc4-b3b1-f7e317647485︡{"stdout": "t |--> -g*t + v0"}︡ ︠14fc8ffb-a448-4a42-8e81-269c178e63edi︠ %htmlIf we equal the equation to zero.
${\displaystyle\frac{dy}{dt}=V_o -gt=0 :}$
we can calculate the time at which the maximum vertical distance is reached
$t=V_o/g$
In Sage, we simply solve the differential equation with respect to t.
︡5dee036f-a622-4ba4-8a1e-31cc971cfc3c︡{"html": "If we equal the equation to zero.
\r\n${\\displaystyle\\frac{dy}{dt}=V_o -gt=0 :}$
\r\nwe can calculate the time at which the maximum vertical distance is reached
\r\n$t=V_o/g$
\r\nIn Sage, we simply solve the differential equation with respect to t.
"}︡ ︠27f79e4f-b73c-49ca-b399-04f1fa58a412︠ solve(diff(y,t)==0,t) ︡2e543a66-35a9-4b54-8532-560da2a46fbe︡{"stdout": "[t == v0/g]"}︡ ︠b5a4f761-a620-445d-9289-4d9caf328980i︠ %html4. Graphic representation:
We can plot it now all together.
︡e3de5140-2761-4e27-b082-8e5133c192f1︡{"html": "4. Graphic representation:
\r\nWe can plot it now all together.
"}︡ ︠ce45c5fc-1135-40f0-a50d-488e00a829fb︠ import numpy as np class Hyperbola(): def __init__(self, velocity, color): """ creates a hyperbolic curve of the ball motion at a given initial velocity. Arguments: velocity -- initial velocity color -- color to plot the curve """ # public variables #f(v) = v*t - (g*t**2)/2 v,t=var('v,t') # symbolic variables g=9.81 self._v = velocity self._color = color self.f = velocity*t - (g*t**2)/2 # this only depends on t now self._tend = 2*velocity/g self.__g=9.81 self._tmax = self._v/self.__g def maxcoor(self): """ returns x,y coordinates of the point at the maximum """ ymax = self.f(self._tmax) return (self._tmax, ymax) def plot(self): """ creates a plot object with the curve, label, maximun point and the coordinates of the max point """ curve = plot(self.f,0,self._tend,rgbcolor=self._color) circle = point(self.maxcoor(), rgbcolor='white', pointsize=30,faceted=True) location = (self._tmax,self.f(self._tmax)*1.2) label = text("V=%s m/s"%self._v, location,rgbcolor=self._color) return curve + circle + label # plot functions # create some objects myfunc = Hyperbola(velocity=6,color='orange') #print myfunc.plot() #print myfunc.maxcoor() figure = myfunc.plot() v=[2,3,4,5] # list of velocities colors = ['red','green','black','blue'] # list of colors for i,j in zip(v,colors): myfun = Hyperbola(velocity=i, color=j) figure += myfun.plot() figure.fontsize(14) figure.show(xmax=1.50) ︡1edd5168-65ff-4d3a-9eb2-488dac68987d︡{"html": "
"}︡