# Interaction forces

1. Jan 4, 2006

### Benny

Q. A bead (particle) of mass m travels along a path described by $$\mathop r\limits^ \to = a\cos \left( {\omega t} \right)\mathop I\limits^ \to + a\sin \left( {\omega t} \right)\mathop J\limits^ \to + bt\mathop K\limits^ \to$$ with respect to the inertial system of coordinates XYZ.

Determine the expression for the interaction force between the spiral slide (the path) and the bead.

$$\mathop R\limits^ \to = \left( { - ma\omega ^2 \cos \left( {\omega t} \right) + \frac{{mga\omega b}}{{a^2 \omega ^2 + b^2 }}\sin \left( {\omega t} \right)} \right)\mathop I\limits^ \to$$
$$+ \left( { - ma\omega ^2 \sin \left( {\omega t} \right) - \frac{{mga\omega b}}{{a^2 \omega ^2 + b^2 }}\cos \left( {\omega t} \right)} \right)\mathop J\limits^ \to$$
$$+ \left( {\frac{{mga^2 \omega ^2 }}{{a^2 \omega ^2 + b^2 }}\sin \left( {\omega t} \right)} \right)\mathop K\limits^ \to$$

Note: I think that it is implicit in the question and answer that there is no friction between the slide and particle.

The main thing I am having trouble with is determining the forces acting and the basic (Newton's 2nd law) equations to set up. Looking at this problem I can see that while there are quite a few computations to be made, the level of difficulty is fairly low.

My initial thought was that the 'natural' coordinate system to work with is the intrinsic (tangential, normal and binormal 'axes') coordinate system. I've found expressions for the unit tangential, normal and binormal vectors of this system in terms of the 'usual' I, J and K vectors so conversions between the two coordinate systems are simple.

Then, to work out the interaction forces between the particle I will use Newton two to resolve forces along the tangential, normal and binormal directions. This is where I run into problems. Lets just say that I start with resolving forces in the normal direction.

F = ma right? F is the net force acting on the particle in the normal direction. The weight of the particle is mg in the K direction. When I use F = ma do I just consider the normal component of the weight force (can be found using a scalar product)? If do then the net force F, that I get is zero in the normal direction. However if I use the RHS (ie. ma) in the normal direction then I get: $$F_n = \left( { - ma\omega ^2 \cos \left( {\omega t} \right)} \right)\mathop I\limits^ \to + \left( { - ma\omega ^2 \sin \left( {\omega t} \right)} \right)\mathop J\limits^ \to$$ which is clearly non-zero.

In short I'm having trouble resolving forces. I've chosen intrinsic coordinates to simplify the working but I'm open to suggestions. Any help is appreciated, thanks.

2. Jan 4, 2006

### Chi Meson

Edit.

I said something useless, never mind.

3. Jan 5, 2006

### Benny

I've thought a little more about the question. Along the binormal and normal directions, there will be reaction forces. However, I'm inclined to believe that there is no reaction force along the tangential direction since there is no friction. In that case the expression that I obtain for the interaction force between the slide and the bead is:

$$\mathop R\limits^ \to = \left( { - ma\omega ^2 \cos \left( {\omega t} \right) + \frac{{mga\omega b}}{{a^2 \omega ^2 + b^2 }}\sin \left( {\omega t} \right)} \right)\mathop I\limits^ \to + \left( { - ma\omega ^2 \sin \left( {\omega t} \right) - \frac{{mga\omega b}}{{a^2 \omega ^2 + b^2 }}\cos \left( {\omega t} \right)} \right)\mathop J\limits^ \to + \left( {\frac{{mga^2 \omega ^2 }}{{a^2 \omega ^2 + b^2 }}} \right)\mathop K\limits^ \to$$

That is the same as the given answer except that the answer has in extra sin(wt) factor in the K component. I wonder what's wrong, I've checked my working a few times. I resolved forces along directions (normal and binormal to the path) which I believe are relevant in terms of "interaction" forces between the slide and the bead.