I'm a high school student studying the motion of a particle inside a rigid
pipe. I'm considering straight pipes, toroidal and helical pipes. The
equation of motion for torus is relatively straightforward, however I'm
having trouble with the helix.

If a particle initially at rest inside a horizontal toroidal pipe with a
circular cross section of constant radius r, is given an initial velocity v
tangential to the centreline of the pipe, the particle will assume a stable
circular path in the bottom half of the torus. It follows that

(1) g*Tan[psi] = v^2/d

where d is the distance of the ball from the centre of the hole in the
torus, and psi is the inclination angle of the tangent plane to torus where
the ball touches the pipe. If the centreline of the torus is of radius R,
then

(2) d = R + r*Sin[psi]

Therefore

(3) v^2 = g*Tan[psi](R + r*Sin[psi])

In the limit as v -- oo, psi -- pi/2. This makes sense, since the ball
cannot slide more that half way up the tube.

A particle sliding in helical tube is more complicated. The tangent plane is
inclined by a fixed angle phi from one axis and a continously changing angle
psi(v) from a second orthogonal axis. For a helix with radius of curvature
rho,

(4) v^2 = g*Tan[psi](rho + r*Sin[psi])

A particle inside a helical tube is subject to the force of gravity mg
acting vertically down and the normal force N exerted by the pipe wall. The
normal force can be decomposed into three mutually orthogonal components; a
vertical component balancing the graviational force, a radial component
providing the centripetal force, and a component providing the tangential
acceleration along the path. The question is, how do I derive these forces
in terms of the inclination angles and the other forces? Any help would be
greatly appreciated.

Thanks in advance.

James

Well, considering that no-one appears to be sure about this, I'll ramble on
a little more.

As I understand it, the orientation of the tangent plane can be specified by
the
inclination angle of the helix together with the inclination angle of plane
from the principle normal vector, not to be confused with the surface normal
vector orthogonal to the tangent plane on the pipe surface. The angle
between the normal force N and the gravitational force mg is thus

(1) chi = (phi + psi)/2
(Does this follow?)

Therefore

(2) N * Cos[chi] = mg

(3) N^2 * Cos^2[chi] = m^2 * g^2

is the vertical component of N.

(4) N^2 * Sin^2[chi] = m^2 * g^2 * Tan^2[psi] + m^2 * a^2

where mg*Tan[psi]= mv^2/(R + r*Sin[psi]) is the radial force, ma is the
tangential force.

Dividing (4) by (3) gives

Tan^2[chi] = Tan^2[psi] + a^2/g^2

0 = a^2/g^2
a = 0

Clearly this is incorrect. Any ideas?

"James Stokes" wrote:
[snip]
A particle inside a helical tube is subject to the force of gravity mg
acting vertically down and the normal force N exerted by the pipe wall.
The
normal force can be decomposed into three mutually orthogonal components;
a
vertical component balancing the graviational force, a radial component
providing the centripetal force, and a component providing the tangential
acceleration along the path. The question is, how do I derive these
forces
in terms of the inclination angles and the other forces? Any help would
be
greatly appreciated.

Thanks in advance.

James

"James Stokes" wrote in message ...

A particle inside a helical tube is subject to the force of gravity mg
acting vertically down and the normal force N exerted by the pipe wall. The
normal force can be decomposed into three mutually orthogonal components; a
vertical component balancing the graviational force, a radial component
providing the centripetal force, and a component providing the tangential
acceleration along the path. The question is, how do I derive these forces
in terms of the inclination angles and the other forces?

You are having your first love afair with trigonometry. :-)

Didn't you ask about this before? IIRC, it was pointed out that you
must specify whether the particle is sliding or rolling. If it's
sliding, w/o friction, then its KE at each height is mgh, where h is
the height it has fallen ... whether falling straight or twisting
through a pipe, it doesn't matter -- pure energy conservation.

Given v(h) I suppose you can also find v(t), and what's more you can
find the accelerations necessary to keep it confined to your helix;
from these you get the normal forces.

If the particle is rolling, things are more complicated: now it may
store KE both in motion of the center of mass, and in rotation. Their
sum is now mgh, again without friction -- i.e., except for the static
rolling friction necessary to stop the thing from sliding. You have
some consistency conditions, like the angular velocity times the
radius matching the linear velocity. The normal forces as before are
given by the acceleration of the center of mass ...

Hmm... the helix is actually largely a distraction ... all you have
is a mass sliding or rolling down and inclined plane, with some
additional horizontal forces added in to to confine it to a circle (as
seen from above).