# Help with an equilibrium problem

1. Nov 11, 2004

### graphic7

There's a problem I've been working on for the last few days on:

(problem number 4)

http://www.maths.usyd.edu.au:8000/u/leonp/MATH2004-2904/LDtute11.pdf

I'm trying to find the center of mass on that rod, and I'm having quite a bit of trouble. I'm only worried about a special case, when the radius of the bowl equals 1. If I can find equations for the COM for the y coordinate, I can apply the potential energy equation, and differeniate - giving Newton's second law. Then I'll be able to set the forces equal to zero, and solve for that angle theta at which the rod meets with the horizontal.

Last edited by a moderator: Apr 21, 2017
2. Nov 12, 2004

### HallsofIvy

Staff Emeritus
I take it you mean the center of mass in some sort of coordinate system- but you don't specify the coordinates.

I'm going to take the (0,0) at O, x-axis along the horizontal diameter of the bowl, y-axis vertical. Since the bowl has radius a and the lower end of the rod is at angle 2θ below the horizontal (the external angle is equal to the sum of the two opposite angles), the coordinates of that lower end are x= -a cos(2θ) and
y= -a sin(2θ). The length of the rod is 2c and it is uniform so the center of mass is geometric middle: distance c from the lower end. Looking at the right triangle formed by the rod (as hypotenuse), the vertical dropped from the center of mass and the horizontal from the lower end, we have &Delta;x/c= cos(θ) and &Delta;y/c= sin(θ). The center of mass of the rod is at
x= -acos(2θ)+ c cos(&theta;), y= -asin(2θ)+ c sin(θ)

3. Nov 12, 2004

### graphic7

Much appreciated.