Maximum Sliding Distance in a Rotating Cone - Solving for r

[Alettix]

Homework Statement

I have gotten the following task: "A smal object is placed in a right circular cone turned "upside-down" with an apex angle equal to 90-2α degrees. The coefficient of friction is big enough to keep the object at rest when it's placed on the inne-side of the cone. After that, the cone is set in rotation with the period time T. If the object starts to slide or not, depends on its distance r to the vertikal axis the coen is spinning around. Show that the maximum value of r is given by:
$r = ( g T^2 / 4 π^2 ) (cot(α-β))$
where tanβ = μ "

Homework Equations

Relevant equtions are:
The equation for centripetal force: Fc = (mr4π^2/T^2)
The equation for friktion force : Ff = FN μ

The Attempt at a Solution

My thoughts are the following:

The object will feel that there's a force pushing it outwards, the centrifugal force (this is a ficitive force, but I have tried and the real centripetal force leeds to the same equation after a short while). The maximum value of r is when the valueforces pulling the object downward the slope equals those pulling it upwards. As r grows, so does the centrifugal (centripetal) force, causing the friction to change direction after a while. The force pulling the object downwards is one composant of the object's weight and the force pulling it upwards is one composant of the centrifugal force minus the friction. When the force pulling it upwards grows stronger than the force downwards, the object starts to slide upwards. r is therefore given by:
sinx Fc - Ff = F1
where x = 45 - α

The rest of my soloution can be seen in the attached picture (in the beginning of it I have renamed β to y).

The problem is that I end up with an euqation saying r = (g T^2 / 4 π^2) cos (x-y)
So all I have left to prove is that cos(x-y) = cot(α-β), but this is not really working out, partly becaus α-β = -45 degrees.

So it must be a mistake somewhere, eighter in my calculations or (and this is probably more possible) in my reasoning about the problem. I would be very happy if you could help me find it and show me how the problem should be solved. I do also appreciate pictures/figures very highly, I helps me understand things a lot better. :)
Thank you!

 

[BvU]

Hello Alettix, and welcome to PF.
Apparently a fairly difficult exercise, since there were no replies yet.
I have to ask a bit more about these angles: apex 90-2$\alpha$ and no sliding down when placed on the slope would mean a $\mu > 1$. Pretty hefty friction, therefore.

If I try to imagine a case where $\alpha = 0, \beta = \pi/4$. (Just) satisfies the no sliding down when not rotating. In that case the expression in the exercise would yield a negative r. This makes me wonder if something has gone wrong in rendering the exact exercise tekst ?

Then, a comment on your approach with the 'sensitive' object, 'feeling' a centifugal force. The bottom line physics is that the object executes a uniform circular motion, for which a net centripetal force is required. So the sum of ALL forces should have a horizontal component $-m\,\omega^2\, r$ and a vertical component 0. My experience is that with such an approach the chance of errors is much smaller than when working with a centrifugal force.

And now the bummer: I can't show what is are asked for in your post either...

 

[TSny]

Alettix,
In your hand written solution it looks like you let $\small \mu = \tan x$. Shouldn't that be $\small \mu = \tan \beta$?

I agree with BvU that something about the statement of the problem seems odd.

I can get the stated solution only if I assume that the apex angle is $\small 2 \alpha$, rather than $\small 90^o - 2 \alpha$. So, the wall of the cone makes an angle of $\small \alpha$ to the vertical.

I also need to assume $\small \pi/2 < (\alpha + \beta) < 2\alpha$. This is equivalent to saying $\small \alpha > \pi/4$ and $\frac{1}{\tan \alpha}$ $\small < \mu < \tan \alpha$ .

 

[Alettix]

Hello Alettix, and welcome to PF.
Apparently a fairly difficult exercise, since there were no replies yet.
I have to ask a bit more about these angles: apex 90-2$\alpha$ and no sliding down when placed on the slope would mean a $\mu > 1$. Pretty hefty friction, therefore.

If I try to imagine a case where $\alpha = 0, \beta = \pi/4$. (Just) satisfies the no sliding down when not rotating. In that case the expression in the exercise would yield a negative r. This makes me wonder if something has gone wrong in rendering the exact exercise tekst ?

Hello BvU! Thank you so much for your reply, I appreciate it. :) I have checked the exercise text once more, and it says that the apex angle is 90-2α. I have also noticed that this gives μ>1, but that's what the text says. I also imagined α<0, but working with an negative angle seems silly to me.
Hmm, I'm not used to counting with radians, how should I interpret " β = π/4 "?

Then, a comment on your approach with the 'sensitive' object, 'feeling' a centifugal force. The bottom line physics is that the object executes a uniform circular motion, for which a net centripetal force is required. So the sum of ALL forces should have a horizontal component $-m\,\omega^2\, r$ and a vertical component 0. My experience is that with such an approach the chance of errors is much smaller than when working with a centrifugal force.

And now the bummer: I can't show what is are asked for in your post either...

I do absolutely agree with you on your point about the centrifugal force. Of course I prefer working with real forces, but after spending 4 hours trying to derive the formula from the centripetal force getting weirder and weirder results (one attached) I decided to try with the fictive one and it gave a result which at least resembled the wanted one a little (athough it was probably completely wrong).
However, back to the centripetal force. My thoughts are the followning: The centripetal force is the force resultant NEEDED to keep an object with mass m in a circular motion, in a circle with radius r, at the anglespeed of ω (Fc= mrω^2). If the sum of the forces acting on an object under these conditions do not equal the centripetal force needed, the object can not be keept in a circular motion and moves away (in this case slides upwards). Well this is at least how I have interpreted it. :)

 

[Alettix]

Alettix,
In your hand written solution it looks like you let $\small \mu = \tan x$. Shouldn't that be $\small \mu = \tan \beta$?

I agree with BvU that something about the statement of the problem seems odd.

I can get the stated solution only if I assume that the apex angle is $\small 2 \alpha$, rather than $\small 90^o - 2 \alpha$. So, the wall of the cone makes an angle of $\small \alpha$ to the vertical.

I also need to assume $\small \pi/2 < (\alpha + \beta) < 2\alpha$. This is equivalent to saying $\small \alpha > \pi/4$ and $\frac{1}{\tan \alpha}$ $\small < \mu < \tan \alpha$ .

Hello TSny and thank you for your response! :)

Firstly, sorry for my bad hand writing. What I think you read as μ = tanx, should be μ = cotx (= tanβ). And yes, I agree that the problem seems odd. Sadly I'm not really in a situation where I can ask if they have not forgotten to include some more information.

Hmm, sadly I have not learned so much about radians (will do it this term), so to be honest I don't fully understand why assuming $\small \pi/2 < (\alpha + \beta) < 2\alpha$ helps. But that's my own fault.

 

[TSny]

should be μ = cotx (= tanβ)

β is defined by the relation μ = tanβ. However, μ does not have to equal cotx. If μ = cotx, then that would mean that when you place the object on the cone with the cone not rotating, then the object would be on the verge of slipping down the cone. In general, all we can say is that μ > cotx.

Hmm, sadly I have not learned so much about radians (will do it this term), so to be honest I don't fully understand why assuming $\small \pi/2 < (\alpha + \beta) < 2\alpha$ helps. But that's my own fault.

$\small \pi / 2$ radians $\small = 90^0$. Sorry for switching to radians.

$\small \pi/2 < (\alpha + \beta)$ is equivalent to $\small \beta > 90^0 - \alpha$ which is equivalent to $\small \tan \beta > \tan(90^0 - \alpha)$. Using a trig identity, this is the same as $\small \tan \beta > \cot \alpha$, or $\small \mu > \cot \alpha$. This is just the condition that the object not slip on the cone when the cone is not rotating. [Here I am assuming that the symbol $\small \alpha$ represents the angle that the wall of the cone makes to the vertical; i.e., my $\small \alpha$ is the same as your $\small x$. I have to make this assumption in order to get the stated answer.]

The other half of the inequality is $\small (\alpha + \beta) < 2\alpha$. Or, in your notation, $\small (x + \beta) < 2x$. This says $\small \beta < x$. If $\small \beta \geq x$, then the object would not slip even for r going to infinity no matter how fast the cone rotates. So, there would be no maximum value for r in this case. At least that's what I get.

Sorry for the confusion with the notation. I think I should have just used your notation for $\small x$ as the angle between the wall and the vertical. Then, I would say that the answer for maximum r is the same as the answer stated in the problem except replace $\small \alpha$ by $\small x$. And the condition on $\small \mu$ is $\small \cot x < \mu < \tan x$. Note that this implies $\small \cot x < \tan x$ which requires $\small x > 45^0$.

EDIT: I noticed that there is a place in your notes where you take the normal force to be $\small F_N = mg \sin x - F_C \cos x$ .
I believe there is a sign error in this expression.

Once you fix that, see if you can proceed as you did in your notes. If you let $\small \mu = \tan \beta$, you should be able to use trig identities to get $\small F_C = mg \cot (x- \beta)$ .

 

[Alettix]

β is defined by the relation μ = tanβ. However, μ does not have to equal cotx. If μ = cotx, then that would mean that when you place the object on the cone with the cone not rotating, then the object would be on the verge of slipping down the cone. In general, all we can say is that μ > cotx.

μ = tanβ is given in the task. :) (se my original post)

$\small \pi / 2$ radians $\small = 90^0$. Sorry for switching to radians.

$\small \pi/2 < (\alpha + \beta)$ is equivalent to $\small \beta > 90^0 - \alpha$ which is equivalent to $\small \tan \beta > \tan(90^0 - \alpha)$. Using a trig identity, this is the same as $\small \tan \beta > \cot \alpha$, or $\small \mu > \cot \alpha$. This is just the condition that the object not slip on the cone when the cone is not rotating. [Here I am assuming that the symbol $\small \alpha$ represents the angle that the wall of the cone makes to the vertical; i.e., my $\small \alpha$ is the same as your $\small x$. I have to make this assumption in order to get the stated answer.]

Thank you for switching back to degrees! Hm, I understand your first inequality. But I still can't understand how cotα is not equal to tanβ if they are two acute angles in a right triangle?

The other half of the inequality is $\small (\alpha + \beta) < 2\alpha$. Or, in your notation, $\small (x + \beta) < 2x$. This says $\small \beta < x$. If $\small \beta \geq x$, then the object would not slip even for r going to infinity no matter how fast the cone rotates. So, there would be no maximum value for r in this case. At least that's what I get.

I think I understand the thing with r going to infinity. But in the task, the angle from the side of the cone to the vertical = 45-α (not your alfa). And because μ = tanβ (this is given), β should be the angle from the side of the cone to the "ground". Therefore: 90 = β +(45-α) → β = 45 + α and so β>α unless α<0. Or am I wrong?

Sorry for the confusion with the notation. I think I should have just used your notation for $\small x$ as the angle between the wall and the vertical. Then, I would say that the answer for maximum r is the same as the answer stated in the problem except replace $\small \alpha$ by $\small x$. And the condition on $\small \mu$ is $\small \cot x < \mu < \tan x$. Note that this implies $\small \cot x < \tan x$ which requires $\small x > 45^0$.

Thank you for explaining your notation. :) TheHow did you get: $\small \cot x < \mu < \tan x$ ? $\small x > 45^0$ once again brings up the idea of α<0 (x=45-α in the original notation)

EDIT: I noticed that there is a place in your notes where you take the normal force to be $\small F_N = mg \sin x - F_C \cos x$ .
I believe there is a sign error in this expression.

Once you fix that, see if you can proceed as you did in your notes. If you let $\small \mu = \tan \beta$, you should be able to use trig identities to get $\small F_C = mg \cot (x- \beta)$ .

If I change to $\small F_N = mg \sin x + F_C \cos x$ I end up with Fc = mg 2 cosx. How can I proceed? Or have I done something wrong?

Once again; thank you a lot for your help, I appreciate it highly. PF seems to have a really nice community. :)

 

[TSny]

...the angle from the side of the cone to the vertical = 45-α (not your alfa). And because μ = tanβ (this is given), β should be the angle from the side of the cone to the "ground".

I don't understand why you are saying that β should be the angle from the side of the cone to the ground. I think it will help a lot if we can get this cleared up. β is determined by the coefficient of friction through the equation tanβ = μ. How do you go from this relation to the conclusion that β is the angle from the side of the cone to the ground?

Thank you for explaining your notation. :) TheHow did you get: $\small \cot x < \mu < \tan x$ ? $\small x > 45^0$ once again brings up the idea of α<0 (x=45-α in the original notation)

I agree that if you define the apex angle to be 90^o - 2α, then you end up with the contradiction that α < 0. That's one of the reasons I think there's something wrong with the way the problem is stated. In order to get the answer that is given in your first post, I have to assume that the apex angle is 2α rather than 90^o - 2α.

If I change to $\small F_N = mg \sin x + F_C \cos x$ I end up with Fc = mg 2 cosx. How can I proceed? Or have I done something wrong?

I believe $\small F_N = mg \sin x + F_C \cos x$ is correct. But I don't get Fc = mg 2 cosx. F_c should depend on both x and β.

 

[Alettix]

I don't understand why you are saying that β should be the angle from the side of the cone to the ground. I think it will help a lot if we can get this cleared up. β is determined by the coefficient of friction through the equation tanβ = μ. How do you go from this relation to the conclusion that β is the angle from the side of the cone to the ground?



I agree that if you define the apex angle to be 90^o - 2α, then you end up with the contradiction that α < 0. That's one of the reasons I think there's something wrong with the way the problem is stated. In order to get the answer that is given in your first post, I have to assume that the apex angle is 2α rather than 90^o - 2α.



I believe $\small F_N = mg \sin x + F_C \cos x$ is correct. But I don't get Fc = mg 2 cosx. F_c should depend on both x and β.

Okay, so I think I know why we disagreed. I thought that β was the angle between the side of the cone and the ground because if an object is at rest on an inclined plane, the coefficient of friction equals tanγ , where γ is the angle of inclination. My very stupid mistake was to assume that tanγ is euqal to the maximal coefficient of friction, which I just realized probably is wrong. I'm very storry for stealing your time. Hopefully I will get it right now!
Thank you. :)

 

[TSny]

Okay, so I think I know why we disagreed. I thought that β was the angle between the side of the cone and the ground because if an object is at rest on an inclined plane, the coefficient of friction equals tanγ , where γ is the angle of inclination. My very stupid mistake was to assume that tanγ is euqal to the maximal coefficient of friction, which I just realized probably is wrong.

Right. All you can say is that the coefficient of friction must be greater than (or equal to) tan γ. That means that β ≥ γ

I'm very storry for stealing your time. Hopefully I will get it right now!

Don't worry about that. I hope it works out now. (I still think there's something fishy about the 90-2α apex angle.)