# Homework Help: Force balance in rod falling

1. Nov 7, 2009

### Ramar

1. The problem statement, all variables and given/known data
Hi all.
I've got a vertical rod on a surface with a fixed and frictionless pivot at bottom end. The rod of length L is released and begins to fall.
The forces acting on the CM of the rod should be:

- weight force: mg
- tangential force: m*L/2*alfa

2. Relevant equations
3. The attempt at a solution
I tryied to calculate the vertical and horizontal resulting force but without success. Assuming the positive sign in the sense of forces increasing teta angle, the balance of vertical components I think should be:

N = m*L/2*alfa*sin(teta) + mg + m*L/2*omega^2*cos(teta)

Expressing alfa and omega in terms of weigth force (respectively from tau = I*alfa and PE = 1/2*I*omega^2) I've got:

N = 3/4mg*sin^2(teta) + mg + 3/2mg*cos(teta)

Where first and third terms of right member are the vertical component of tangential and centripetal forces.

This result is clearly not correct because at 90° the normal force should be zero.
Could anybody check this result ?
Thank you.

2. Nov 8, 2009

### tiny-tim

Welcome to PF!

Hi Ramar! Welcome to PF!

(have a theta: θ and an alpha: α and an omega: ω and try using the X2 tag just above the Reply box )

The only force acting on the CM is the weight, mg.

The other force is the reaction force at the wall (which you can resolve along and perpendicular to the rod).

And your equations should be using the moment of inertia of the rod.

But anyway, what is the question asking for?

If it isn't specifically asking for the force, then wouldn't it be easier to use conservation of energy?

3. Nov 8, 2009

### Ramar

Re: Welcome to PF!

Yes, you are right, I've incorrectly added weight to its components.
What I'm trying to do is to get the balance for vertical and horizontal force of a falling rod. If mg is the only force acting on the rod, with rod in the vertical we should have:

N = mg;

When the rod begins to fall we should have:

N = mg minus something.

With moment of inertia I can get the expression for alfa and omega and I use it to express centripetal and tangential forces in terms of weight and teta. Intuitively I wrote:

N = mg - centripetal force*cos(teta);

Where centripetal force is m*L/2*omega^2. Applying Newton's second law and using I = 1/3*m*L^2 I've got:

omega^2 = 3*g/L

hence:

N = mg - 3/2*mg*cos(teta)

This result is clearly wrong because for teta = 90° I've got N = mg.
Where am I wrong ?