Question

PLEASE HELP SOLVE WITH MATLAB LANGUGE.

2 Coriolis Force In a rotating frame-of-reference,the equations of motion of a particle, written in co- ordinates fixed to the frame, have additional terms due to the rotation of the frame itself Consider such a rotating frame, with its origin at the center of rotation.In these coor- dinates, the equations of motion for a point-mass subjected to forces F, and F S m, are F(0 #(t)-20y(t)-r(t) (t)+2i (t)- y(t) This is shown in the figure below ball with an initial speed Now, consider a person located at r (relative to the rotating frame) of vo, with a direction as shown in the figure below (defined by the angle 0). R,y 0, launches

Below are hints to the problem. THANKS A LOT!!

Answer 1

THE FUNCTION odefun IS NESTED SO THE ENTIRE CODE BELOW SHOULD BE SAVED AS ONE FILE coriolis.m:

%

function coriolis

clc,clear,close all

%% coriolis.m
% let dxdt = p, dydt = q

%% part a)
%
R = 1;
v0 = 1.1854; theta = -1.0039; Tfinal = pi;
x0 = R; y0 = 0; p0 = -v0*cos(theta); q0 = v0*sin(theta);
cond = [x0 y0 p0 q0];
% solving with ode45
t = linspace(0,Tfinal,100);
[T,SOL] = ode45(@odefun,t,cond);
X = SOL(:,1)
Y = SOL(:,2)
% plotting the solution
figure
plot(X,Y,'- b')
xlabel('x(t)'),ylabel('y(t)')
title(['(v0,theta,Tfinal) = (' num2str(v0) ',' num2str(theta) ',' num2str(Tfinal) ')']),grid on

%% part b)
%
R = 1;
v0 = 1.0223; theta = -1.3617; Tfinal = 3*pi;
x0 = R; y0 = 0; p0 = -v0*cos(theta); q0 = v0*sin(theta);
cond = [x0 y0 p0 q0];
% solving with ode45
t = linspace(0,Tfinal,100);
[T,SOL] = ode45(@odefun,t,cond);
X = SOL(:,1);
Y = SOL(:,2);
% plotting the solution
figure
plot(X,Y,'- b')
xlabel('x(t)'),ylabel('y(t)')
title(['(v0,theta,Tfinal) = (' num2str(v0) ',' num2str(theta) ',' num2str(Tfinal) ')']),grid on

%% part c)
%
R = 1;
v0 = 1.0081; theta = -1.4442; Tfinal = 5*pi;
x0 = R; y0 = 0; p0 = -v0*cos(theta); q0 = v0*sin(theta);
cond = [x0 y0 p0 q0];
% solving with ode45
t = linspace(0,Tfinal,100);
[T,SOL] = ode45(@odefun,t,cond);
X = SOL(:,1);
Y = SOL(:,2);
% plotting the solution
figure
plot(X,Y,'- b')
xlabel('x(t)'),ylabel('y(t)')
title(['(v0,theta,Tfinal) = (' num2str(v0) ',' num2str(theta) ',' num2str(Tfinal) ')']),grid on

%% function to express the ode as a system of equations
%
function [ ode ] = odefun(t,in)
% dxdt = p, dydt = q
%
omega = 1;
%
x = in(1); y = in(2); p = in(3); q = in(4);
%

dxtd = p; % x'
dydt = q; % y'
dpdt = 2*omega*q + omega^2*x; % x''
dqdt = -2*omega*p + omega^2*y; % y''
%
ode = [
dxtd;
dydt;
dpdt;
dqdt;
];

end

end

(v0, theta, Tfinal) = (1.1854,-1.0039,3.1416) y(t) 0 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0. 9 1 0.5 X(t)

(v0,theta, Tfinal) = (1.0223,-1.3617,9.4248) y(t) -0.4 -0. 2 0 0.2 0.4 0.6 0.8 1 12 X(t)

(v0, theta, Tfinal) = (1.0081,-1.4442,15.708) y(t) -0.8 -0.6 -0.4 -0.2 0 0.4 0.6 0.8 1 0.2 X(t)

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