Spiral motion & angular acceleration (Question)

Fjolvar

Hello, I'm in desperate need of some help with a problem regarding spiral motion. If a particle moves along a spiral path by degrees θ in time t with intial angular velocity ω_o, can the rotational kinematic equation (θ = ω_ot + (1/2)[itex]\alpha[/itex]t²) still be applied? I'm trying to solve for angular acceleration [itex]\alpha[/itex].

(See link for spiral picture) http://mathworld.wolfram.com/ArchimedesSpiral.html

So in other words, is spiral motion any different than standard rotational motion with regards to calculating angular velocity/acceleration? It seems I would need to account for arc length (s) and curvature (k) using the equations in the posted link.

Any help would be greatly appreciated.

clevermetal

When it completes a revolution it will have also gained 'height' as it will have moved 'up' the spiral. So you're angular velocity will have to account for that direction as well.

However with your angular acceleration, as long as it is moving 'up' the spiral at constant velocity the equations shouldn't change (if being referenced in the x-y plane that is).

just my 2c though

Fjolvar

In the link I posted above there are three equations: the polar equation, one for curvature (k) and one for arc length (s). In all equations there is a constant "a." What is this constant exactly and how is it measured(in a real application)?

r = aθ

k(θ) = [2+θ²]/[a(1+θ²)^3/2

s(θ) = (1/2)a[θ√(1+θ²) + ln(θ+√(1+θ²))

haruspex

Sounds like you're thinking of a helix, not a spiral. The question relates only to 2D.

haruspex

That formula is only valid for constant angular acceleration. Do you know that to be the case here?
If not, what determines the motion?

clevermetal

Woops sorry, I was thinking of something along the lines of:
z(t) = (acos(t), bsin(t), ct)
which I now realise is completely wrong.

kmarinas86

If a=1, then after one rotation (Δθ=2*pi) r increases by 2*pi.

If a=2, then after one rotation (Δθ=2*pi) r increases by 4*pi.

etc...

To measure a, you find how many rotations there have been total, and you measure r.

Once you find a, you can solve for k and s.

Darwin123

Only if the angular acceleration is constant.


By spiral motion, I suppose that you mean the distance from the particle to the
origin is increasing with time. The physical effect of this would be the moment of inertia
is increasing with time. If the external torque is constant and the moment of inertial is increasing in time, the angular acceleration must be decreasing in time. Thus, your formula is wrong if the torque is constant.
For planar motion,
Torque=(Moment of inertia)(Angular acceleration).
If the Torque is constant while the moment of inertia is increasing with time, then
the angular momentum has to be decreasing with time. Thus, you can't use a formula
that is valid for constant angular acceleration.

Fjolvar

The angular acceleration is decreasing as the particle approaches the center, but it stops before it reaches the center with a final angular velocity of zero (the particle hits a wall). So if I know the initial velocity, the distance or degrees travelled, and final velocity of zero.. what equation could I use to calculate the deceleration? I realize now that the moment of inertia is changing, but I'm not sure how to translate this into a calculation.

Also I need to calculate the deceleration of the particle as it approaches a point on one of the outer spirals (also hitting a wall and stopping) in the reverse direction, but that should be the same concept.

haruspex

Fjolvar, the conditions are still not clear to me. What is controlling the motion of the particle?
As far as I can tell from your posts, this is irrelevant: the particle is being made to move in some way, and all you want is an expression for the angular acceleration about the origin. That's easy: it's the second derivative of the angle wrt to time. That's true whether it's on a spiral, a circle or a straight line.
Or maybe you mean acceleration in the tangential direction? That's [itex]2\dot{r}\dot{\theta}+r\ddot{\theta}[/itex]. Given that [itex]r = a\theta[/itex], you can eliminate [itex]r[/itex] or [itex]\theta[/itex] from that.
Meanwhile, some of the posts bring up momentum and torque, which will be relevant if the particle is moving subject to known forces.
So which is it?

Fjolvar

So to clarify some, my ultimate goal is to calculate the torque/force of the particle's collision with the wall, which is in place near the end of the spiral. The particle starts from the outer ring of the spiral, has constant velocity ω_o, and moves inwards until about half way when it begins to decelerate and then eventually collides into a wall near the origin bringing it to a complete stop. The driving force moving the particle is irrelevant in this case. I'm trying to satisfy the equation: Torque = Moment of Inertia * Angular Acceleration (Deceleration in this case), but I'm not sure if this equation can be used now since the moment of inertia varies with time.

Since the moment of inertia is changing, I'm stuck trying to derive an equation to make this calculation. Hopefully I've explained this situation clearly enough. Feel free to ask any more questions, as I still have yet to derive the proper equation =/

haruspex

You cannot calculate those. How a force varies over the duration of an impact depends on subtle considerations of the elasticity, plasticity and density of the materials. What you usually can can calculate is momentum and/or angular momentum of the impact.
Is the deceleration constant in magnitude? Constant in terms of angular acceleration? If not, what determines the speed of the particle at impact? Why would there necessarily be any remaining speed?
[/QUOTE]

Fjolvar

Read below for my engineering project, which I tried to describe above in terms of a particle hitting a wall, since the same principles can be applied.. I thought it would've been easier to explain that way.

There is a spinning wheel about an axle with a spiral groove on the side (where the rim of the wheel would be). The wheel is spinning at a constant velocity, controlled by an electric current. Then the current is cut off and the wheel begins to decelerate spinning freely losing velocity. A locking bolt is then engaged perpendicular into the spiral until it hits a hard stop/wall located within the spiral groove, near the center of the spiral. I am trying to calculate the torque/force exerted on this hard stop. I only know the initial velocity before the current is cut off, the distance travelled to the hard stop after, and the time it takes to reach the hard stop. Maybe it would have been easier to have explained it this way to begin with. Also the answer doesn't have to be 100% exact, just a close estimate, so we can ignore friction for now.

Fjolvar

So I'm looking for an equation that takes the following into consideration:

1. Intial angular velocity
2. Angular deceleration
3. Changing moment of inertia with respect to time
4. Torque

This is very important to me, so any help is greatly appreciated, even a hint at how to derive such an equation.

haruspex

OK, if I've understood that correctly:
- the wheel decelerates steadily for reasons unconnected with the bolt;
- the bolt's progress is not considered to affect the wheel until it hits the stop; that means we're ignoring both friction and the inertial mass of the bolt;
- the bolt slides both in the groove on the wheel and in a stationary groove set along a radius of the wheel
Since the spiral is arithmetic (r = Aθ), and the wheel's rotation rate ω = ω₀ - Bt, the bolt's radius r = r₀ - Aω₀t + ABt²/2.
You can solve using the final value of r to find the travel time, and hence the final value of ω. The impulse, moment of inertia of wheel × ω, comes from the wheel being brought abruptly to a halt. As I said, you cannot easily calculate a torque or force. If there is very little give in the system you will get a very high torque and would be in danger of shearing the bolt or damaging the groove. A better solution might be to attach a spring to the bolt so that it helps to decelerate the wheel while as it slides in.

Fjolvar

All is correct in your understanding haruspex. So with these givens, you are saying the torque cannot be calculated then, correct?

haruspex

It's a more elaborate version of a question that comes up every week or three on this forum: if a mass m travelling at speed v hits the ground, what is the force of impact? Answer: there is not a single value of the force. If you watched events microsecond by microsecond you'd see the force increase, more-or-less linearly at first, up to some maximum. It might stay at the max a little while before dying away. The integral of the force over the duration is mv, the change in momentum, but the peak value etc. depend very much on how long the process lasts and the force versus time profile. That depends on the elasticities, plasticities, and densities of the impacting bodies. Consider dropping an egg a metre onto concrete versus onto hay.