Cones and Particles: Newsletter Challenge (03/20/07)

By geometry, h=r.

mω²r = mg

r = g/ω²

= 9.81ms^-2/(2.2rads^-1)²

= 2.0m

Hi Davo, this was too fast! That's unless there is a catch that we haven't spotted yet.

Since I think you're perfectly right, how about "panel beating" the cone a bit to flare it out at the top, to (say) a cone that has the equation: h = 1-e^-r and then see at what height the particle must orbit for the same ω.

Maybe we can have even more fun by inviting proposals for different shapes of cone, provided that they are solvable, of course...

Regards, Jorrie

Dang tootin' that was fast! Kinda takes all the challenge out of the "challenge!"

As for flaring out the cone, I'm not quite sure such a shape is stable, and if it is, I'd bet a whole nickel that it can't stand any perturbations, such as centrepital accelerations of a not-perfectly-rigid cone.

I'm also not sure if I can waste enough time (when I should be doing something productive) to satisfy the requirements of the modified challenge.

Hi Bill, I think you're right about the instability of a particle in a flared cone. The curve I gave above does not seem to have a solution anyway, except at the origin!

For those with the time, what about a dish type cone? One that "looks like it must have" a solution is: h = e^r/2-1, but there should be many others...

Regards, Jorrie

hi jorrie......a thought here...assuming the particle is round, ...shouldn't we consider that the particle cannot be at the distance h, since it must be "inside the cone" and therefore must travel in a trace circle path that is at the center point of the particle, the contact point of the particle on the cone would be 2m, but the trace height would be at 2m + sine 45 * r of particle......depends on where we define the trace path or contact point....

Hi dberts, your concern about the radius of the particle is valid, but practically negligible. The challenge stated "very small particle" and compared to a height of 2 meters, you will get a few microns difference...

Regards, Jorrie

hi jorrie....true, a small particle negligible, but in another look, i believe the factor would be greater than stated; it would be twice as much....assuming contact point can not be the 2 m radius, rather the center of the particle trace (to meet period requirement), then the contact point would be 2*r* sine 45......changing the dimension h by that factor, so h cannot equal r......but h=2m + (2*r particle* sine45)....

Agreed, but the travelling object is not just small, it is a very small particle. Angular velocity has been given in with two significant figures. To push h up from 2.0m to 2.1m requires (2*r particle* sine45) > 0.05m, or r particle > 0.035m. That's a pretty big very small particle.

Would it be possible for the particle to follow an inclined elliptical path, similar to an orbiting body?

Hi Davo, you asked: "Would it be possible for the particle to follow an inclined elliptical path, similar to an orbiting body?"

I believe so, yes. The kinetic plus potential energy of the particle must remain constant like in orbits. Whether it will follow a precise Keplerian orbit I'm not sure about, but if it has a little too high an ω, it will tend to move higher, ending up with too small an ω and so on, possibly moving in a precessing, inclined, elliptical orbit.

Regards, Jorrie

If you design it so that the slope is inversely proportional to the distance from the center (horizontally), it would be a good simulation of a satellite moving in a gravitational field, and you should get your Keplerian orbits.

hi davo....i believe that would depend on if the statement of 2.2 radians per second is constant or average.

For this little idea I had ignored the constant ω. If it followed a true ellipse I think angular momentum would be conserved instead.

Hi Davo, am I right in stating that the height can be solved for any constant cone angle Φ, provided 0 < Φ < pi? I get r = tan(Φ/2) g/ω².

This has an important bearing on solutions for shaped cones...

Jorrie

I get:

Sin(Φ/2) = mg/F_n So F_n = mg/Sin(Φ/2)

Cos(Φ/2) = mω²r/F_n So F_n = mω²r/Cos(Φ/2)

mg/Sin(Φ/2) = mω²r/Cos(Φ/2)

r = g/(ω²Tan(Φ/2))

Hi Davo

I think we agree; I just used Φ to mean the angle that the cone subtends at the bottom, i.e., 'your Φ' = (90° - 'my Φ').

Thanks, Jorrie

Hi Jorrie,

How about this one: At what radius r will the very small particle sit if it is doing laps at of a hemispherical cup of radius R?

Or what about this: What would be the minimum escape velocity of the very small particle doing laps of a sine wave shaped bowl as shown below?

Hi Davo, the cases you gave are very interesting, but I haven't had the time to try them.

While playing around, I found one very interesting parabolic shape, where the 2.2 rad/s particle can apparently circle at any height you choose! Or perhaps better stated, the particle will "choose" a height according to the energy you give it, while maintaining the angular velocity.

The curve is: h = r²/4, with a slope dh/dr = r/2.

Using r = dh/dr x g/ω², as determined before, this gives a solution of r = r (since g/ω²= 2). As a reality check, at r = 2, dh/dr = 1, just like in the straight cone case.

Does this make sense, or did I mess it up somewhere?

Regards, Jorrie

Jorrie,

If you spin a bucket of water, it assumes a parabolic shape. If it were to solidify and you put your marble anywhere, it should stay in place, because by definition, the water was "level" with respect to the forces on it. Makes sense to me.

RBW

Hi RBW

That's good elegant theory, but not if the marble has any height. If it is lossless, it will oscillate with the initial position as its highest point. If there is any loss it will always end up at the centre/bottom.

I remember trying this one as a kid, using objects that just floated (yes washing up was boring). The floating object always ends up at the lowest point in the centre. It's partly that the CofG of the floating object is raised (albeit ever so slightly), and partly air resistance, I think.

Fyz

Where can I buy frictionless cones and mini particals the size of a point?

So I can set up an experiment!

I can't spell hypothetical but I can see it when I see it.

(general truble maker)

Davo,

Your answer is too simple!

Abe

Abe, I thought it was only abundantly simple

Dave

I reckon you are assuming that the cone is fixed or mounted.

Rather the question says that the cone is simply standing on it's tip, and frictionless.

Therefore, assuming that the small particle is travelling North and has attainded it's 2.2 radian/sec velocity at it's initial contact with the cone; and assuming that the cone's tip is truly a point;

m>x will cause the cone to tumble somewhat West of North, or;

m<x will cause the cone to precess and assume an angle of greater than the subtended 90 degrees mentioned in the equation - thus causing the height to be greater than the 2 meters

x is of course a ratio of the mass of the particle to the mass of the cone.

So, the tough question is what is x?

tom

=-==

p.s. beats me, all I know is the devil pays me good money to give you eggheads headaches...

I didn't give you a cent ! Kindly refrain from reference to the dude on the left - I'll get the flack for it. For general info of all , I merely like to play 'devils advocate' and ID myself when doing so .

Thanks for your (hoped) understanding Guest.

Tom,

It sounds like you are referring to an initial interaction between a stationary cone and a moving particle. This really renders the question incoherent because we don't know the reference frame for describing the particle's motion in rad/sec. The question says that "a very small particle moves in a circle on the inside of the cone". To me this points towards a steady state rather than an inital interaction.

I agree with Davo's answer, but I think to be strictly correct the initial formula should read:

mg.sin(Theta)=m.r.ώ²cos(Theta)

where Theta is the angle of the incline

This is the equilibrium equation for the component of the forces parallel to the inclined plane. The answer of 2.0m is only correct because Theta is 45 degrees. Hence

sin(Theta)=cos(Theta)

and the equations simplifies to

mg= m.r.ώ²

But how fast do I have to turn the cone over to keep the particle in the same place?

Trouble maker

Frictionless? Then unless the particle is attached with a tether to the axis of the cone it would spiral out of the cone. The requirement to stay at a specific height requires not only centrifugal force to keep it in planar motion but also a resistance (friction) to keep it on the interior surface of the cone.

As shown so elegantly by Davo , the resultant force (and hence reaction F_n) is perpendicular to the surface . Hence no sliding tendency.

If one point on the cone is moving at 2.2 radian/sec, every point, at any height, is moving at the same speed, because it is an angular speed. If the cone is a rigid shape, every point will go on the same angle during the same time.

Interesting interpretation which renders the challenge trivial unless we ingore friction, in which case the original answer holds.

I don't intend to sound rude but this seems like a bit of mischievous sophistry. Expressed in Radians all point have the 'same' speed , but consider their velocity.

Not trusting Radians (or any other extra terrestrials !) I used velocity (as in centrifugal force=mv²/r and got the same result as Davo . However I find it hard to come up with a good and concise way of rebutting your point (other than to suggest that if the cone was elastic it would deform to a plate shape !).There are many ways in which the question might be re-interpreted , but I will return to this discussion later to see somebody more eloquent than myself answer your point.

Not sure I understand your point here Guest:

If the particle is constrained to move at 2.2 radians per second (say the cone was spinning and the particle was in a groove) then Davo's original answer would still work as unstable equilibrium. But, if the particle moved up slightly then the cone would force more energy (scalar velocity) into the particle, and, it would spiral out of the cone. Conversely if it moved down slightly then the cone would remove energy (scalar velocity) and it would spiral down.

However with the particle sliding on the inside surface of the cone then if it varies upward a bit then it would lose angular velocity for two reasons: conservation of angular momentum, and, potential height energy. If it was perturbed down the opposite would happen. So in this case it would be in stable equilibrium.

The 'frictionless' cone and 'very small' particle will feel invalidated by use of such name calling . Inapt use of words 'required' and 'should' will lead to egocentric behaviour and denial . This will in turn causing friction . End result - family in crisis .

Therapist required . Apply below.

nick name-

The centrifugal force r*ω^2 should be equal to gravity = g thus r= g/ω^2

This is only valid in an homogenous gravitational field and because of the 90° total cone angle. It is also assumed that the central line of the cone is parallel to the gravity field (for earth conditions vertical) and that the particle mouves on a circular path in a plane perpendicular to gravity field direction. For earth conditions g=9.81 m/s², on an other planet the radius will be different.

In metric units the radius is 2.0269 m or 79.798 ".

Dear nick,

"This is only valid ... and because of the 90° total cone angle." The 90° angle was specified in the question. Of course the answer is only valid if we believe the question.

You write "in an homogenous gravitational field" but the gravity field only requires uniformity in the plane of the particle's motion. Is that a lot to ask?

"It is also assumed that the central line of the cone is parallel to the gravity field (for earth conditions vertical) and that the particle mouves on a circular path in a plane perpendicular to gravity field direction." We can deduce perpendicularity of the cone's axis to the plane of motion since the particle traces out a circle. We can deduce that the cone's axis is parrallel to the gravity field from the constancy of the particle's motion.

"For earth conditions g=9.81 m/s², on an other planet the radius will be different." The question states "You have a frictionless cone...". What planet other than Earth have you had a cone?

1 The problem is not so narrow as you take it and this what I wanted to stress. The problem could have been stated with a total cone angle 2θ.

2 It is nowhere mentionned that the cone is circular or am I wrong?

3 There are a lot of things which are assumed to be right but this is some time the wrong way to touch the problem

4 Gravity is variable, ok it is constant for small dimensions, but as you very well know (I assume) it is not constant all over the earth. And since it is related to mass the vector can be under circumstances not right "vertical".

5 How do you know that the problem cannot be considered for moon conditions? It is as far as I know a planet and how do you know that there are not cones on other planets or do you believe that the only planet in the univers is earth or that other civilaisations if they exist do not know what a cone is? Can you demonstarte that there are no cones some where else as on earth?

I would like to have this demonstration.

6. The problem of the particle mouvement can be even more interesting if it is stated that it is thrown at the start with a speed of V [m/s] on the cone and with an angle α with respect to the plane perpendicular to the cone axis. How will be the trajectory? Does it reach a stable position or not? What do you think?

From the other side I appreciate your sense of investigation and humor.

The problem states that the frequency must remain at 2.2 radians/second, not the angular speed.

For circular motion: frequency=ω/2π

Thus ω=2.2(rad/s)*2π = 13.82 rad/sec

Using this value for ω in Davo's equations gives:

r = g/ω²

r = 9.81(m/s²)/((13.82 (rad/sec))2

r = 0.051 m = 51 mm

"Using this value for ω in Davo's equations gives:"

How dare you mistreat my poor defenseless equation! Your equation frequency=ω/2π uses frequency measured in Hertz (s^-1) but frequency has been given in radian/s. You can't convert radian/s to radian/s by multiplying it by 2π.

Davo , Guest #29 has (cunningly) introduced 'must ' instead of 'should ' . Your poor equation may be experiencing anxiety separation . Where is Oprah when you need her ?

Dear Oprah,

You don't need to be shy since we all know it's you anyway. The equation would love to appear on your show, but it's appearance fee has just gone up from 2.2 rad/s to 13.82 rad/s.

Lynn Truss will also be appearing on the show, to counsel the apostrophe you have so cruelly abused by forcing it into a position it should not rightly be...in fact a place where it must not be. Its misery has been multiplied by being forced to share the post with a correctly placed apostophe, the shame and humiliation of this situation is almost unbearable and the poor little mite has only survived on the thought of all the compensation it will be able to extract from you.

Should you require any further information, please contact Swindell Solicitors

Are they your local branch of Swindell, Chargit and Cheetham?

The distance does not matter. The small particle will need its speed adjusted at each different depth in the cone to maintain the 2.2 radians per second. The higher in the cone the faster the particle will travel and vice versu.

It's true that the angular velocity, ω, can be maintained by changing velocity to suit the radius, but this will cause an imbalance of forces.

Two forces are acting on the particle. Gravity is pulling the particle down with magnitude mg. The surface of the cone provides a perpendicular reaction force, since it is "frictionless". For the particle to be in equilibrium the sum of these forces must be equal to mω²r and be directed towards the axis of the cone, since this is the force required to keep a body in circluar motion. The equations and diagram in #1 show how to sum the two forces.

What if the particle was made of iron and the cone was full of ice...

Add a splash of Vodka for a

Rusty Mary!!!

tom

=-==

An earlier poster opined that a more generalized formula should be constructed to account for different included angles in the hollow cone. Let's add to that the radius (r) of a spherical object rolling on the inside surface of the cone.

Now the equation for equalibrium becomes:

mω²Rtan(Θ/2) = mg

R = g/ω²tan(Θ/2)

And the height becomes:

H = (R/tan(Θ/2)) + (r/cos(Θ/2))

or

H = (g/ω²tan²(Θ/2)) + (r/cos(Θ/2)

Now you can plug in the derivative of the formula for a curved surface to generate values for Θ/2 and H and another layer of complexity may be accommodated.

Lots of fun.

I think that this is the equation for a sliding sphere. A rolling sphere is a much more difficult problem than that, and not one that I'm inclined (!) to tackle for fun. However, we can say that a sphere will need to be lower down, because the moment of inertia will cause it to keep going. (I haven't checked, but I fear that modelling gyroscopic effects would make it tricky)

Fyz

*Apostrophes placed especially for Lynne Truss.

What!! This answer is crazy!! The equation h = g/(ω².tan²θ) must use radians per second as units for angular velocity, giving 2.0m as the correct height. What a disappointing answer, specially after the ice block fiasco. More care is needed CR4.

I agree Davo, the question setter has illustrated the dynamics of their own cranium.

Two neurons and they would be dangerous .

After the furore over the ice blocks I can't believe that we have reached Wednesday without this being noticed. Its a good thing we aren't designing bridges here.

Davo,

The important outcome in this question is that the solution equation (which is the only thing important in this matter) is correct. You got it the first day.

Abe

The application of the equation is equally important - many engineering mistakes that find their way through to cause real problems are still caused by incorrect units (cycles/sec instead of rads/sec here, or mil for mm and degF for degC for NASA) or simple numerical errors.

Hi Guest,

I totally agree. If I designed something for a client, spent $400,000 of the client's money making it and then had it failed due to a mistake in units I don't think the client would be very understanding.

An equation that is correct "in principle" and incorrect in units is just incorrect.

Davo

No need to convert radian/s to cycle/sec. the correct answer for h = 2.03m approx.

>Manas, India

In the supposed correct answer, the algebra is correct but the arithmetic not. Since the tangent of 45 degree is 1, we merely need g/(omega squared), and that is 2.03 m approximately, not 80!

Ken Gundry

None of the quibbles I have seen so far have considered the rotation of the earth, which I assume this cone to be supported on. Unless it was at one of the poles, the tangential velocity of the surface of the cone is a function of both its height ( distance from the center of the earth) and the latitude of its location---which differs between its northward and southward surfaces (the 'Coriolis' effect). This effect is far from negligible; the rotation of water falling a few inches in a draining tub reverses when the tub is relocated a few feet each side of the equator! When you are exactly on the equator the water does not rotate. This is a demonstration I was shown at a small museum in Ecuador. I believe that this would make it impossible for the small particle to remain in a planar path inside the 'stationary' cone.

hi octogenarian....who said anything about earth in the problem....it is best not to add complication to these challenges and deal only with information given. Davo has eloquently given the equation correctly....and i agree with not converting radians to cps.

Hi Octogenarian,

I agree that the coriolis effect would prevent a true circular path, but as for being far from negligible, have a look at http://en.wikipedia.org/wiki/Coriolis_effect

Regards, Davo

As Davo says , check out the web . There is a fantastic experiment in construction by Forum members - I can't remember the link , but use search (on the right ) and 'Coriolis'

Kris

It can't be anywhere on earth ... its frictionless... we aint got no crap like that. Or do we?

Trouble maker