Spiral motion & angular acceleration (Question)

In summary: If my understanding is correct, the problem consists of a particle moving under the influence of a central force. The solution is r=(L^2)/(mk) for some constants L and k. You are given that theta=k*t for some constant k. You are also given theta(0)=0. You are asked for alpha. Is this correct?In summary, the problem at hand involves a particle moving along a spiral path with a variable angular velocity and a constant external torque. The question is whether the standard rotational kinematic equation can still be applied in this case, and if not, how to calculate the angular acceleration. The solution involves considering the increasing moment of inertia and using the formula
  • #1
Fjolvar
156
0
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 ωo, can the rotational kinematic equation (θ = ωot + (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.
 
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  • #2
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
 
  • #3
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))
 
  • #4
clevermetal said:
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).
Sounds like you're thinking of a helix, not a spiral. The question relates only to 2D.
 
  • #5
Fjolvar said:
If a particle moves along a spiral path by degrees θ in time t with intial angular velocity ωo, can the rotational kinematic equation (θ = ωot + (1/2)[itex]\alpha[/itex]t2) still be applied? I'm trying to solve for angular acceleration [itex]\alpha[/itex].
That formula is only valid for constant angular acceleration. Do you know that to be the case here?
If not, what determines the motion?
 
  • #6
Woops sorry, I was thinking of something along the lines of:
z(t) = (acos(t), bsin(t), ct)
which I now realize is completely wrong.
 
  • #7
Fjolvar said:
In the link I posted [Archimedes' Spiral] 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))

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.
 
  • #8
Fjolvar said:
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 ωo, can the rotational kinematic equation (θ = ωot + (1/2)[itex]\alpha[/itex]t2) still be applied? I'm trying to solve for angular acceleration [itex]\alpha[/itex].
Only if the angular acceleration is constant.


Fjolvar said:
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.
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.
 
  • #9
Darwin123 said:
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.

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.
 
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  • #10
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?
 
  • #11
haruspex said:
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?

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 ωo, 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 =/
 
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  • #12
Fjolvar said:
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.
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.
The particle starts from the outer ring of the spiral, has constant velocity ωo, 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.
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]
 
  • #13
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 traveled 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.
 
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  • #14
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.
 
  • #15
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.
 
  • #16
haruspex said:
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.

All is correct in your understanding haruspex. So with these givens, you are saying the torque cannot be calculated then, correct?
 
  • #17
It's a more elaborate version of a question that comes up every week or three on this forum: if a mass m traveling 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.
 
  • #18
If the elasticity, plasticity, and density is known.. Does anyone know what motion/force equations I would have to reference?
 
  • #19
In your case, I'm sure you don't want to enter the plastic phase as that implies damage. In the elastic phase it behaves like a spring: force is proportional to deformation.
The issues will be the shear force on the bolt and the pressure of the bolt on the edges of the slots. I'm not a trained mechanical engineer, so don't really know how to go about analysing all of that.
Why not go with my suggestion of making the bolt work against a spring, so you can arrange that almost all of the energy has been absorbed before the bolt hits the stop?
 
  • #20
Unfortunately I cannot implement any design changes, since the product is already in production. I'm only assigned to determine the force of the locking bolt against the plastic hard stop of the gear. This requires using equations that will take into account change of inertia, since the spiral is directed inwards towards the center of the wheel or axle. Do you happen to know any equations to recommend?
 
  • #21
As I said, I'm not a qualified engineer, but my judgment is that a calculation sufficiently accurate will be a much more time consuming and difficult task than simply measuring it. You would need to model the detailed shape of the bolt and the grooves it contacts, plug in the properties of the materials, think about the different failure modes, the uncertainties in the variables...
If you have to go down that path:
- the bolt is a cylinder which sits in a groove at each end (yes?)
- there is some slack around the bolt and between the vertical plates carrying the grooves
- at impact, the longitudinal section through the bolt will look like a rectangle ABCD at some small angle to the horizontal; A is in space in the 1st groove, B contacts the upper side of the 1st groove, C is in space, D contacts the lower side of the 2nd groove; point E on BC contacts the upper lip of the 2nd groove; point F on DA contacts the lower lip of the 1st groove; there is a horizontal gap between the two plates
- there is a shear force on the bolt section between the plates
- there is a bending moment about E and about F
- there are compressive forces on the bolt and grooves at all contact points
Each of these should be analysed wrt elastic limit.
In practice, the bolt corners will be or will become a little rounded. This increases its ability to deal with compressive forces at B and D, but increases the slack, allowing the skew angle to increase; not sure whether that makes the shear worse, but if the angle gets steep enough it will create a significant force prising the plates apart. One possible failure mode is that the bolt escapes a groove entirely.
 

1. What is spiral motion?

Spiral motion is a type of motion in which an object moves in a curved path while also rotating around a central axis. This results in the object moving both in a circular path and along a spiral path.

2. What causes spiral motion?

Spiral motion is caused by the combination of a linear force and a rotational force acting on an object. The linear force causes the object to move along a curved path, while the rotational force causes the object to rotate around a central axis.

3. What is angular acceleration?

Angular acceleration is a measure of the change in an object's rotational velocity over time. It is represented by the symbol alpha (α) and is measured in radians per second squared (rad/s²).

4. How is angular acceleration related to spiral motion?

Angular acceleration is directly related to spiral motion, as it determines how quickly an object will rotate around its central axis. In spiral motion, the object is both rotating and moving along a curved path, so a change in the object's rotational velocity will also affect the path it takes.

5. Can an object have spiral motion without angular acceleration?

No, an object cannot have spiral motion without angular acceleration. As mentioned before, angular acceleration is directly related to spiral motion, so in order for an object to have spiral motion, it must also have angular acceleration.

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