Question

Matlab

Answer 1

@jim_thompson5910

Answer 2

%% Problem 1
%{
Find: the value of w necessary to for the bar to remain constant angle
beta.
%}

% Clears command and workspace window
clc
clear

% Defined variables
syms g w L b B m;

% Omega rigid body
wRB = [-w*cos(B); w*sin(B); 0];

% Moment of inertia matrix
IG = [0 0 0; 0 ((1/12)*m*L^2) 0; 0 0 0];

% Angular momentum about the center mass of the bar
HG = IG*wRB;

% Time rate of change for angular momentum
dHG = cross(wRB, HG)

% Gravity vector
g = [(g*cos(B)) -(g*sin(B)) 0];

% Length vector from G to O
rGO = [(L/2) 0 0];

% Acceleration of center mass of bar
aG = [-w^2*((b*sin(B) + (L/2)*(sin(B))^2) -w^2*(b*cos(B) + (L/2)*sin(B)*cos(B)))];

% Reactions vector at pivot point O
FRO = m*(aG - g)

% Sum moments about center mass of bar
MG = cross(rGO, FRO)

Answer 3

Is there any way I can find w using a command in Matlab? I found it using hand calculations.

Answer 4

MG = dHG and you solve by hand.

Answer 5

cross refers to a matrix cross product?

Answer 6

Yes.

Answer 7

I don't see how
wRB = [-w*cos(B); w*sin(B); 0];
is a 3x3 matrix

Answer 8

cross products only work in R3

Answer 9

I made it a 3x1 because of the matrix multiplication between IG and wRB.

Answer 10

oh nvm, so wRB and HG are 3x1 matrices, ie vectors

Answer 11

Yes.

Answer 12

this command might work

https://www.mathworks.com/help/symbolic/solve.html

though not sure, I'm still trying to picture what the equation looks like

Answer 13

I have the solution written out by hand. Let me upload an image.

Answer 14

@jim_thompson5910

Answer 15

yikes, a lot more than what I was thinking

Answer 16

does the solve command work though?

Answer 17

You mean

eqn = solve(x) = 1;
solx = solve(eqn, x) ?

Answer 18

something like that, but idk if that format will work

Answer 19

It doesn't work.