Consider a ball released from rest and allowed to roll down a frictionless
tube deformed into the shape of a circular spiral with a vertical axis. How
does the position vector of the ball vary as a function of time?

I've been looking on the net for a while and have found plenty of
information concerning helical motion with uniform angular speed but have
yet to find anyone who has considered the case of accelerated motion.

Thanks in advance.

James.

Correction, the tube is deformed into the shape of a circular helix, not a
spiral.

"James Stokes" wrote in message ...
Consider a ball released from rest and allowed to roll down a frictionless
tube deformed into the shape of a circular spiral with a vertical axis. How
does the position vector of the ball vary as a function of time?

I've been looking on the net for a while and have found plenty of
information concerning helical motion with uniform angular speed but have
yet to find anyone who has considered the case of accelerated motion.

Thanks in advance.

James.

The helix (not spiral) can be parametrized with the angle T (theta):
x = R*cos(T) R0 is radius of helix (constant)
y = R*sin(T)
z = -a*T a0 and z is pointing 'up'

Suppose the time derivatives (dots) are abbreviated as
T' = theta-dot (time derivative = dT/dt)
T" = theta-dot-dot (2nd time derivative = d^2T/dt^2)
Then the Lagrangian = kinetic energy - potential energy
= 1/2 m * (R^2+a^2) * (T')^2 - m*g*z
where g is gravitational acceleration
= 1/2 m * (R^2+a^2) * (T')^2 + m*g*a*T
The Lagrange equation
d/dt[@L/@T'] = @L/@T
gives
1/2 m*(R^2+a^2)*T" = m*g*a
equivalent with
T" = g*a/(R^2+a^2)
which can be integrated to
T(t) = T_0 + w_0 * t + g*a/(R^2+a^2)*t^2
Of course if the thing started from rest, then T_0 and
w_0 are both zero.
T(t) = g*a/(R^2+a^2)*t^2
and now you can calculate x(t), y(t) and z(t) if you like.

hth

Dirk Vdm

"Dirk Van de moortel" wrote in message
...

[snip]

T" = g*a/(R^2+a^2)

Ouch, typo: T" = 2*g*a/(R^2+a^2)
First review, *then* hit that send button ;-)

the remainder is okay.

Dirk Vdm

"Dirk Van de moortel" wrote in message
...

"James Stokes" wrote in message ...
Consider a ball released from rest and allowed to roll down a frictionless
tube deformed into the shape of a circular spiral with a vertical axis. How
does the position vector of the ball vary as a function of time?

I've been looking on the net for a while and have found plenty of
information concerning helical motion with uniform angular speed but have
yet to find anyone who has considered the case of accelerated motion.

Thanks in advance.

James.

The helix (not spiral) can be parametrized with the angle T (theta):
x = R*cos(T) R0 is radius of helix (constant)
y = R*sin(T)
z = -a*T a0 and z is pointing 'up'

Suppose the time derivatives (dots) are abbreviated as
T' = theta-dot (time derivative = dT/dt)
T" = theta-dot-dot (2nd time derivative = d^2T/dt^2)
Then the Lagrangian = kinetic energy - potential energy
= 1/2 m * (R^2+a^2) * (T')^2 - m*g*z
where g is gravitational acceleration
= 1/2 m * (R^2+a^2) * (T')^2 + m*g*a*T
The Lagrange equation
d/dt[@L/@T'] = @L/@T
gives
1/2 m*(R^2+a^2)*T" = m*g*a

Aarg ****, here was the typo:
that should be
m*(R^2+a^2)*T" = m*g*a
equivalent with
T" = g*a/(R^2+a^2)
which can be integrated to
T(t) = T_0 + w_0 * t + 1/2*g*a/(R^2+a^2)*t^2
Of course if the thing started from rest, then T_0 and
w_0 are both zero.
T(t) = 1/2*g*a/(R^2+a^2)*t^2
and now you can calculate x(t), y(t) and z(t) if you like.

sorry for the confusion ;-)
forget my previous correction

Dirk Vdm

"Dirk Van de moortel" wrote:
[snip]
Aarg ****, here was the typo:
that should be
m*(R^2+a^2)*T" = m*g*a
equivalent with
T" = g*a/(R^2+a^2)
which can be integrated to
T(t) = T_0 + w_0 * t + 1/2*g*a/(R^2+a^2)*t^2
Of course if the thing started from rest, then T_0 and
w_0 are both zero.
T(t) = 1/2*g*a/(R^2+a^2)*t^2
and now you can calculate x(t), y(t) and z(t) if you like.

sorry for the confusion ;-)
forget my previous correction

Very interesting, albeit way beyond my depth of understanding (yr 11 high
school). Now, a few questions. Firstly, how exactly did you derive

K = 1/2 m * (R^2+a^2) * (T')^2

Secondly, what is the significance of

d/dt[@L/@T'] = @L/@T

Also, once we have obtained

\theta(t) = \frac{gc}{2(R^2+a^2)t^2}

is it possible to simplify this equation at all?

Furthermore, it possible to derive an expression for T without using the
Lagrangian function at all?

Thanks again.

James

Dirk Vdm

Dirk Van de moortel wrote:

"Dirk Van de moortel" wrote in message
...

"James Stokes" wrote in message ...
Consider a ball released from rest and allowed to roll down a frictionless
tube deformed into the shape of a circular spiral with a vertical axis. How
does the position vector of the ball vary as a function of time?

I've been looking on the net for a while and have found plenty of
information concerning helical motion with uniform angular speed but have
yet to find anyone who has considered the case of accelerated motion.

Thanks in advance.

James.

The helix (not spiral) can be parametrized with the angle T (theta):
x = R*cos(T) R0 is radius of helix (constant)
y = R*sin(T)
z = -a*T a0 and z is pointing 'up'

Suppose the time derivatives (dots) are abbreviated as
T' = theta-dot (time derivative = dT/dt)
T" = theta-dot-dot (2nd time derivative = d^2T/dt^2)
Then the Lagrangian = kinetic energy - potential energy
= 1/2 m * (R^2+a^2) * (T')^2 - m*g*z
where g is gravitational acceleration
= 1/2 m * (R^2+a^2) * (T')^2 + m*g*a*T
The Lagrange equation
d/dt[@L/@T'] = @L/@T
gives
1/2 m*(R^2+a^2)*T" = m*g*a

Aarg ****, here was the typo:
that should be
m*(R^2+a^2)*T" = m*g*a
equivalent with
T" = g*a/(R^2+a^2)
which can be integrated to
T(t) = T_0 + w_0 * t + 1/2*g*a/(R^2+a^2)*t^2
Of course if the thing started from rest, then T_0 and
w_0 are both zero.
T(t) = 1/2*g*a/(R^2+a^2)*t^2
and now you can calculate x(t), y(t) and z(t) if you like.

sorry for the confusion ;-)
forget my previous correction

Powerful stuff that physics!

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

"James Stokes" wrote in message ...

"Dirk Van de moortel" wrote:
[snip]
Aarg ****, here was the typo:
that should be
m*(R^2+a^2)*T" = m*g*a
equivalent with
T" = g*a/(R^2+a^2)
which can be integrated to
T(t) = T_0 + w_0 * t + 1/2*g*a/(R^2+a^2)*t^2
Of course if the thing started from rest, then T_0 and
w_0 are both zero.
T(t) = 1/2*g*a/(R^2+a^2)*t^2
and now you can calculate x(t), y(t) and z(t) if you like.

sorry for the confusion ;-)
forget my previous correction

Very interesting, albeit way beyond my depth of understanding (yr 11 high
school). Now, a few questions. Firstly, how exactly did you derive

K = 1/2 m * (R^2+a^2) * (T')^2

Start with the parametrization of the helix:
x = R*cos(T)
y = R*sin(T)
z = -a*T
where T (theta) is our still unknown function of time t.
Taking the derivatives wrt time:
x' = dx/dt = -R*sin(T)*T' (you know the chain rule?)
y' = dy/dt = R*cos(T)*T'
z' = -a*T'
these are all functions of t.
Tangential speed v^2 = x'^2 + y'^2 + z'^2
= R^2*cos^2(T)*T'^2 + R^2*sin^2(T)*T'^2 + a^2*T'^2
= R^2*T'^2 + a^2*T'^2
= (R^2+a^2) * T'^2
Kinetic energy = 1/2 m * v^2
= 1/2 m * (R^2+a^2) * T'^2


Secondly, what is the significance of

d/dt[@L/@T'] = @L/@T

It produces the differential equation(s) that describe the
motion, just like F=m*a does, but you don't have to know
or decompose the forces. It's very handy when the object
is constrained to a well known path, like here.
All you have to know, is the kinetic energy and the potential
energy.
You find an *excellent* introduction [one of the best I have
ever seen] to this in chapter 5 of
http://www.courses.fas.harvard.edu/~...outs/textbook/
Before you go for it, briefly review the first 4 chapters.
Also look at lectures 6 and 7 of
http://www.courses.fas.harvard.edu/~phys16/lectures/
It takes some knowledge of calculus though and I don't
know if you are ready for that... If not, I'm sure you will
be soon :-)


Also, once we have obtained

\theta(t) = \frac{gc}{2(R^2+a^2)t^2}

is it possible to simplify this equation at all?

I guess the 'c' in {gc} should be 'a'.
Unless some relations are given between g, a and R
it can't be simplified. Of course, if R a or a R
you can use a Taylor approximation.


Furthermore, it possible to derive an expression for T without using the
Lagrangian function at all?

Yes, use conservation of total Energy:
Total Energy = kinetic energy + potential energy
= 1/2 m * (R^2+a^2) * (T')^2 + m*g*z
where g is gravitational acceleration
= 1/2 m * (R^2+a^2) * (T')^2 - m*g*a*T
Since this value must be conserved, it must have the
same value at time t=0 as at an arbitrary time t.
But at time t=0, both T=0 and T'=0, since the object
starts at z=0 and with theta=0
so you see that with these initial conditions the total
Energy is calibrated to zero:
1/2 m * (R^2+a^2) * (T')^2 - m*g*a*T = 0
where T is the value at time t now. We should write T(t)
here.
Eliminating m from this equation, it can be rewritten as
T'^2 = 2k*T with k = g*a/(R^2+a^2)
This is again an easy differential equation that can be
solved like this:
T' = sqrt(2k*T)
dT/dt = sqrt(2k)*sqrt(T)
dT/sqrt(T) = sqrt(2k)*dt
Int{ dT/sqrt(T) } = Int{ sqrt(2k)*dt }
(integrating 0 to T(t) on the left and
0 to t on the right)
2*sqrt(T) = sqrt(2k)*t
T = 1/2*k*t^2
which is exactly what we had before :-)

I hope you were able to follow this ... I really don't know
what to imagine with yr 11 high school.

Dirk Vdm

James Stokes wrote:

Consider a ball released from rest and allowed to roll down a
frictionless tube

One or the other, please, but not rolling on a frictionless
surface.

deformed into the shape of a circular spiral with a
vertical axis. How does the position vector of the ball vary
as a function of time?

I've been looking on the net for a while and have found plenty
of information concerning helical motion with uniform angular
speed but have yet to find anyone who has considered the case
of accelerated motion.

In the frictionless case, you find the force along the path, and
then find the vertical component of that force. Then you ignore
the helical path, and calculate the height from the vertical
acceleration. In the rolling case, you get a good approximation
if you just use the second moment for a sphere to get the
acceleration, and then follow the previous procedure. In this
case, you get the same answer for a helix as you do for a
straight line path with the same slope. This is all high-school
physics.

The precise result for a rolling ball depends on finding the
precise line of contact of the ball with the walls of the tube.
This is almost certainly more trouble than it is worth, unless
you are thinking of introducing the spherical luge to the Winter
Olympics.

If the ball starts at rest, it accelerates up the wall of the
tube, and then down again toward the bottom. This oscillation
superimposed on the geeral downward progression results in a
differential equation without an analytic solution. You can
approximate the correct answer by assuming a simple periodic
motion, and substituting x for sin x in figuring the lateral
restoring force, just as in the pendulum equation. (First-year
college physics with calculus.)

For significant initial velocities, or for a steep helix, this
approximation does not hold. In some cases the ball can go up
and over the top of the tube as it descends, or go part way up
and fall, in which case you have to deal with elestic bouncing
or worse (vibration, plastic deformation, strength of
materials...). (Anywhere from second-year physics to tenth-year
aerospace engineering, depending on how far you want to push
it.)

Thanks in advance.

James.

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Encore Technologies (S) Pte. Ltd.
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