The position and velocity of the object as a a series of calculations as follows

Question

The position and velocity of the object as a a series of calculations as follows: as followsbject as a finction of time can then be approximated by Acceleration Step Time F(C,) FO FC,) Position F(t,) You may work in groups of up to three students if you desire. Turn in one assignment per group. 2 m (5 pts) Consider a mass sliding down a frictionless curve in the shape of a quarter circle of radius 2.00 m as in the diagram. Assuming it starts from rest, use Euler's method to approximate both the time it takes to reach the bottom of the curve and its speed at the bottom. You may either use a spreadsheet like MS Excel or you may write and execute a computer program in the language of your choice. Do three trials: 1-02s, -0.02s, and -0.002 s. Compare the predicted speed at the bottom for each case to the accepted value of 6.261 m/s. (Calculate a percent error.) Does the approximation improve as At becomes smaller? a) b) (5 pts) Repeat (a), but this time assume a constant kinetic friction coefficient of .- 0.200. Again determine the time to the bottom and the speed at the bottom. .You need only run one trial: compare, do not calculate a percent error At = 0.002 s. As you do not have a "correct" value to

Explanation / Answer

I will answer this question with a Fortran Code. Please install gfortran in your computer along with gedit. The program can be reproduced in the gedit as a write it here. After writing the code in a file by opening the gedit as from terminal (if you use Ubuntu/linux): gedit file_name.f

file_name is the name of the file, you can write whatever name you like.

Now some theoretical work.

The total force at any point in the motion will be the sum of: 1) A component of Gravitational Force along the Path and, 2) A centrifugal force

Now as per the diagram, The gravitational force component will be Fg= mcos(theta)

The centrifugal force will be Fc = mv2/r , this force balances the normal reaction.

here theta and v are time dependent quantities and will be calculated in each time step and r=2m, and will be used as such in the code.

The theta at a given time after n steps tn is given by = xn/r (in radian)

acceleration at a given time tn is given by = gcos(xn/r) + v2/r - R

                                                           a = gcos(xn/r)

now, let us begin the fortran code:

        program euler

        real*8 x(1000),t(1000),a(1000),v(1000),theta(1000),r

        x(1) = 0

        t(1) = 0

        v(1) = 0

        r = 2

        dt = 0.2

do i = 2,1000

        theta(i-1) = x(i-1)/r

        a(i-1) = 9.81*cos(theta(i-1))

        v(i) = v(i-1) + a(i-1)*dt

        x(i) = x(i-1) + v(i)*dt

        write(10,*)x(i),v(i),t(i),a(i),theta(i)

        if(theta(i) .gt. 3.141529/2)stop

        end do

        end program

END OF CODE; now save the file, and compile it by "gfortran file_name.f" and execute by "./a.out" and required data will be stored in fort.10 file.

For adding the Friction, you just have to change the form of "a" in this code to

a = g*cos(theta(i)) - 0.200*v(i)*v(i)/r

and that will include the friction. Change value of dt to 0.02 and 0.002 for different cases. If required increase iterations from 1000 to 2000 at all places where 1000 is written.

       

      

      

       

       

     

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