Path on Cone Fig.1: Angle ACP is 90°?

[Rikudo]

Homework Statement

A mountain is modeled as a cone with height H (see fig 1). On the cone, a path that goes around it and connects point A-B is made by wire with minimum length. If a particle move along the path from A to B, what is the minimum initial velocity it should have to reach point B? Assume that there is no friction.h = H/√15

Relevant Equations

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Fig.1

Fig 2 (the net of the cone)

Point C is the turning point. $\phi$= 90°.
I wonder why the angle ACP is 90°. Is this a coincidence, or the "wire of minimum length" has anything to do with this?
(Though, I thought the minimum length of the path can be acquired if ABP is a right angle)

 

[anuttarasammyak]

We see point C is higher than point B in Fig.2. It would matter the initial speed of the particle at A.

 

[jbriggs444]

Though, I thought the minimum length of the path can be acquired if ABP is a right angle.

For some choice of point B, ABP might be a right angle. For others it might be obtuse. In this case we are shown that it is acute.

Obviously, the point C (if any) where the path is at right angles to a line toward the apex will be the point of maximum height, the "turning point" as that term is used here.

 

[wrobel]

Looks like the path is an ideal constraint. Then the problem has one degree of freedom and the energy conservation is enough. Have I missed something?

 

[Orodruin]

I wonder why the angle ACP is 90°.

This is what defines point C, which is the point of maximal height. It may or may not exist on the path from A to B depending on the apex angle. That the path is the minimal length path in essence tells you how to find C.

Looks like the path is an ideal constraint. Then the problem has one degree of freedom and the energy conservation is enough. Have I missed something?

Energy conservation is enough, but not necessarily between A and B.

 

[wrobel]

Energy conservation is enough, but not necessarily between A and B.

I do not understand what you mean. The solution is
$$\frac{1}{2}mv^2=mgh\cos \theta$$

 

[jbriggs444]

I do not understand what you mean. The solution is
$$\frac{1}{2}mv^2=mgh\cos \theta$$

No. It is not. That gives you the energy requirement to be at point B with zero energy. But that is not what the question asks. You have to get there first.

 

[Orodruin]

I do not understand what you mean. The answer is
$$\frac{1}{2}mv^2=mgh\cos \theta$$

No. This is only true if there is no point on the trajectory that lies higher than B.

 

[wrobel]

No. It is not.

No. This is only true if there is no point on the trajectory that lies higher than B.

Oh! now I see. Thanks. Shame on me.

 

[bob012345]

The first drawing suggests the path goes all the way around the cone but ends up at a higher point. The second drawing suggests a different situation unless it is misleading the student on purpose. Is point $A'$ the same as point $A$?

 

[jbriggs444]

The first drawing suggests the path goes all the way around the cone but ends up at a higher point. The second drawing suggests a different situation unless it is misleading the student. Is point $A'$ the same as point $A$?

The drawings seem compatible to me. The one is an unrolling of the cone from the other. Clearly, the minimum path length on the regular cone corresponds to a straight line on the unrolled paper.

 

[bob012345]

The drawings seem compatible to me. The one is an unrolling of the cone from the other. Clearly, the minimum path length on the regular cone corresponds to a straight line on the unrolled paper.

In general yes, I made a paper model but if the path is one trip around but higher, I'm having trouble drawing a straight line for that.

 

[Orodruin]

In general yes, I made a paper model but if the path is one trip around but higher, I'm having trouble drawing a straight line for that.

just do what is done in figure 2 in the OP.

 

[bob012345]

just do what is done in figure 2 in the OP.

When I unrolled the cone to be flat the angle $\phi$ is 180°.

 

[Orodruin]

When I unrolled the cone to be flat the angle $\phi$ is 180°.

Make a cone with a smaller apex angle as the one in figure 2.

 

[jbriggs444]

When I unrolled the cone to be flat the angle $\phi$ is 180°.

That says that the cone is not pointy enough. The path of minimum length that makes a minimum of one lap around the apex intersects with the apex. (It's cheaper to go to the apex and back down than to go around the regular way).

If you were modeling the not-pointy-enough case on a corkboard with push pins and string, the optimal path would be one where the string went from the start point (A) bent around the apex (P) and then back down to the end point (B).

 

[Steve4Physics]

I wonder why the angle ACP is 90°

In addition to what has already been said, this may help.

In Post #1, Fig. 2, construct a circle centre P, passing through C (circle’s radius = PC).

The arc of the circle which overlaps the cone’s net corresponds to a line of constant-height on the cone.

B is shown outside the arc, so B is shown lower than C.
[Edit: - as already noted by @anuttarasammyak in Post #2. Sorry @anuttarasammyak, missed it!]

AB is a tangent to the arc at C, hence ∠ACP = 90º.

 

[bob012345]

That says that the cone is not pointy enough. The path of minimum length that makes a minimum of one lap around the apex intersects with the apex. (It's cheaper to go to the apex and back down than to go around the regular way).

If you were modeling the not-pointy-enough case on a corkboard with push pins and string, the optimal path would be one where the string went from the start point (A) bent around the apex (P) and then back down to the end point (B).

Ok, so do you agree in this case as Fig 2 is drawn that point B is one exactly loop around above point A?

 

[jbriggs444]

Ok, so do you agree in this case as Fig 2 is drawn that point B is one exactly loop around above point A?

Yes.

 

[Rikudo]

In fig 2, the cone is divided into 8 identic parts by dashed lines. Why the point C is located on the 7th line, not the 5th?

EDIT: the line where the point A is located is assumed as the 1st line

 

[bob012345]

In fig 2, the cone is divided into 8 identic parts by dashed lines. Why the point C is located on the 7th line, not the 5th?

EDIT: the line where the point A is located is assumed as the 1st line

They defined the problem that way. I think it is a big clue to solve the problem.

 

[Rikudo]

They defined the problem that way. I think it is a big clue to solve the problem.

The only figure that is given in the problem is fig 1. Fig 2 is given in the solution paper.

If what you say is true, I think solving the problem is utterly impossible.

 

[bob012345]

The only figure that is given in the problem is fig 1. Fig 2 is given in the solution paper.

If what you say is true, I think solving the problem is utterly impossible.

Obviously I thought fig 2 was provided. Sorry. The problem is not impossible. Try using the equation of the cone to figure point C.

 

[jbriggs444]

The only figure that is given in the problem is fig 1. Fig 2 is given in the solution paper.

If what you say is true, I think solving the problem is utterly impossible.

The answer will involve H and $\theta$.

The discussion so far has identified three cases.

1. $\theta$ is large (the cone is not very pointy). The shortest path from A to B that makes a lap around the apex actually goes to the apex and back down. The high point in the path is at the apex and the required answer in this case is easy to find.

Edit to add: One can make a strong argument that this case invalidates the assumption of conservation of energy that holds for gently curved friction-free paths. The laws of physics are based on differential equations. You cannot directly apply them to a non-differentiable path. For technical correctness, one would need some weasel words around this case.

2. $\theta$ is medium (the cone is somewhat pointy). The angle ABP is acute and the highest point on the path is somewhere in the middle. Finding that high point will require some trigonometry but is possible. The required answer can then be found.

3. $\theta$ is small (the cone is very pointy). The angle ABP is either right or obtuse and the highest point on the path is at its far end. The height of the end point is straightforward to calculate and the required answer is then easy to find.

 

[bob012345]

It is straightforward and instructive to derive a relationship between $\theta$ and $\phi$. Then one can see the boundary for case #1 above. The boundary between the second and third cases is when the line between points $A$ and $B$ becomes tangent to the circle defined by the radius $\frac{H}{cos(\theta)} - h$.

 

[Steve4Physics]

... The boundary between the second and third cases is when the line between points $A$ and $B$ becomes tangent to the circle defined by the radius $H - h$.

The side-length of the cone is $\frac H {cos\theta}$. I think the required radius is $\frac H {cos\theta} - h$.

 

[bob012345]

The side-length of the cone is $\frac H {cos\theta}$. I think the required radius is $\frac H {cos\theta} - h$.

Sorry, I misunderstood and thought $H$ was defined as the edge length. You are correct but it gives an secondary condition between $\theta$ and $\phi$ so one needs to solve two equations together for that case.

 

[haruspex]

The answer will involve H and $\theta$.

The discussion so far has identified three cases.

1. $\theta$ is large (the cone is not very pointy). The shortest path from A to B that makes a lap around the apex actually goes to the apex and back down. The high point in the path is at the apex and the required answer in this case is easy to find.

Edit to add: One can make a strong argument that this case invalidates the assumption of conservation of energy that holds for gently curved friction-free paths. The laws of physics are based on differential equations. You cannot directly apply them to a non-differentiable path. For technical correctness, one would need some weasel words around this case.

2. $\theta$ is medium (the cone is somewhat pointy). The angle ABP is acute and the highest point on the path is somewhere in the middle. Finding that high point will require some trigonometry but is possible. The required answer can then be found.

3. $\theta$ is small (the cone is very pointy). The angle ABP is either right or obtuse and the highest point on the path is at its far end. The height of the end point is straightforward to calculate and the required answer is then easy to find.

For case 1 ($\phi\geq\pi, \theta\geq\pi/6$) the number of laps becomes indeterminate, so arguably cannot be specified as "exactly one lap". More generally, n laps implies $\phi<\pi/n, \sin(\theta)<1/(2n)$.

I did wonder whether the given h/H ratio might rule out a case, but it seems not. The case 2:3 boundary is $h=(1-\cos(2\pi\sin(\theta)))\sec(\theta)H$. For the given ratio that gives $\theta\approx 6.7^o$.

 

[anuttarasammyak]

We see point C is higher than point B in Fig.2. It would matter the initial speed of the particle at A.

From Figs 1 and 2 the height of C from the ground is
$$(PA-PC) \cos \theta$$ where
$$PC=\frac{PA*PB \sin\phi}{AB}$$
where
$$\phi = 2\pi \sin \theta$$
$$AB=\sqrt{(AA')^2+(BA')^2-2AA'*BA' \cos \frac{\pi-\phi}{2}}$$
where
$$AA'=2PA\sin \frac{\phi}{2}$$
$$BA'=PA-PB$$