# Normal Force of a hoop rolling down an inclined plane.

1. Apr 19, 2013

### mercrave

1. The problem statement, all variables and given/known data

I have a question from my homework. My homework is completed, but I've been running some thought experiments lately and I wish to conceptually discuss this.

The problem has to do with a hoop rolling down an inclined plane. I had to find the Normal Force using Lagrange's Equations.

Hoop has mass M, radius R, the hypotenuse of the incline is L and the angle from the bottom of the incline is α.

2. Relevant equations

I'm not sure what to place here, but I have set my origin to be at the tip of the inclined plane, and my x axis lies parallel to the bottom of the incline whereas the y axis is perpendicular to my x-axis. I have set θ to be the angle through which the hoop rotates.

3. The attempt at a solution

When all is said and done I get the following (the ' refers to a time derivative);

T = .5 m x'2 + .5 m y'2 + .5 m R2 θ'2
U = mgy
g1[x,y,θ] = y - x Tan[α] (= 0)<- hoop has to stay on the plane
g2[x,y,θ] = L - x Sec[α] - Rθ (= 0) <- hoop is rolling without slipping (L - x Sec[α] therefore refers to the portion of the plane that is traversed by the hoop)

I plug into Lagrange's equations and obtain the following;
λ1 = .5 mg(1+Cos2[α]) (normal force)
λ2 = -.5 mg Sin[α] (acceleration down the incline)

I did some checks in my head for the normal force; at α = .5$\pi$, we have -.5 mg for acceleration down the incline and normal force is at a minimum (.5 mg) whereas at α = 0, no acceleration and the normal force is effectively the mass of the hoop.

I am basically wondering if this is correct. My thought process right now is that hypothetically, it can roll down a .5$\pi$ incline, and to do that it requires a frictional force which then requires a normal force. It also means that it effectively doesn't roll when there is no incline (flat surface) so normal force is at a maximum and it is just supporting the hoop. Am I correct in thinking so?

Thank you very much, and I hope to hear from you. I am fairly confident in my answer, but my friends get conflicting results (notably, they get mgCos[α] for the normal force, which makes sense to me to some extent until we factor in the rolling) and we were supposedly supposed to recognize this other force. It could be possible that I need to find the x and y components of this force using the constraints on Lagrange's Equation's and probably find the magnitude to round it off, but I'm just curious to hear what everyone here thinks.

Great forum by the way, it has helped me through numerous homeworks in the past!

2. Apr 19, 2013

### haruspex

Why would the rolling make it otherwise? That adds a frictional force up the slope, so has no effect on the normal force. Maybe you could post your working.

3. Apr 19, 2013

### mercrave

I basically took double-time derivatives of each of the first two constraints, then used lagranges equations alongside them. My friends and I all double checked lagranges equations, so at this point I am wondering if the constraints are the issue or, better yet, the lagrangian?

It also helped that all algebra we did was performed on Mathematica.

Also, I thought rolling friction was normal force dependent even still.

4. Apr 19, 2013

### TSny

I get the same expressions for the lagrange multipliers λ1 and λ2.

But λ1 does not represent the normal force. Note that in setting up your constraint g2[x,y,θ] for rolling without slipping, you invoked the constraint g1[x,y,θ] to get the x Sec[α] part. So, both g1 and g2 include the constraint to stay on the plane.

You can find the normal component of each of the constraint forces. For example, the normal component of the constraint force associated with g1 is

$\lambda_1\frac{\partial{g_1}}{\partial n}=\lambda_1 (\frac{\partial{g_1}}{\partial x}\frac{\partial{x}}{\partial n}+\frac{\partial{g_1}}{\partial y}\frac{\partial{y}}{\partial n} ) = \lambda_1 (-\frac{\partial{g_1}}{\partial x}sin\alpha+\frac{\partial{g_1}}{\partial y}cos\alpha )$

where $n$ indicates normal direction. Similarly you can find the normal component of the force due to the second constraint. If you add the two normal components from the two constraints, you should get the total normal force.