What is the minimum time at a given angle on a cone with constant size?

In summary: No, I used only $$\frac{h}{sin(\alpha+\varphi/2)}=\frac{\frac{1}{2}gt^{2}cos\alpha }{sin(\varphi/2) }.$$Can you please post the equation for ##x(t)##?
  • #1
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Homework Statement
A hollow cone is placed on the ground like shown in example. Inside a cone there is string attached to base center. In the string there is a small bead. What must the angle between string and vertical be, so bead reaches cone surface fastest? (You can stretch the string). Cone base is parallel to the ground and cone has angle φ. String frictionless.
Relevant Equations
kinematic eqs
trig identites
I tried to come up with expression of time in terms of φ, α, and some constant value of cone size that does not depend on angle (I used R as cone radius). I though I could use that expression to see at what angle time is going to minimum, but I came up with expression from which I can't make a conclusion. I believe there is another solution method, or maybe there are some useful trig identities, but I was unable to find it.
Can someone help me out? Thanks

Angle φ is marked badly at my solution, but that doesn't change anything in particular.
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  • #2
Note that the angle you call ##\varphi## is half the angle given as ##\varphi## in the diagram.

My first approach would be to use the law of sines for the triangle that has angles ##\alpha##, ##\varphi/2## and ##\pi-(\alpha+\varphi/2).## Then minimize ##t^2## with respect to ##\alpha.##
 
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  • #3
kuruman said:
Note that the angle you call ##\varphi## is half the angle given as ##\varphi## in the diagram.

My first approach would be to use the law of sines for the triangle that has angles ##\alpha##, ##\varphi/2## and ##\pi-(\alpha+\varphi/2).## Then minimize ##t^2## with respect to ##\alpha.##
So I get t^2=(2 a sin(φ/2))/(g cosα sin(φ/2+α)), a is length OC. How can I minimize it to find value of α?
 
  • #4
bowes78 said:
So I get t^2=(2 a sin(φ/2))/(g cosα sin(φ/2+α)), a is length OC. How can I minimize it to find value of α?
How do you usually find a minimum of a function wrt its independent variable?
 
  • #5
haruspex said:
How do you usually find a minimum of a function wrt its independent variable?
So after differentiation with respect to α, setting t=0, I get α=90-(φ/2). Could this be the answer?
 
  • #6
bowes78 said:
So after differentiation with respect to α, setting t=0, I get α=90-(φ/2). Could this be the answer?
Consider what that looks like in your diagram. It would mean the string meets the cone at a right angle. It is easy to show that the answer must have the string meet the cone somewhat lower.
If you cannot find your error, please post your working.
 
  • #7
haruspex said:
Consider what that looks like in your diagram. It would mean the string meets the cone at a right angle. It is easy to show that the answer must have the string meet the cone somewhat lower.
If you cannot find your error, please post your working.
Yes, previous solution had a mistake. This time I tried to do something like this, but that seems to me that it doesn't make sense too. Because if we set (φ/2)=0, then α=32, which doesn't look right.
IMG_20221216_162355.jpg
 
  • #8
What about the work before what you posted? Specifically, what does ##a## stand for in the first equation giving ##t^2##?
 
  • #9
kuruman said:
What about the work before what you posted? Specifically, what does ##a## stand for in the first equation giving ##t^2##?
I just picked there wrong trigonometry formulas, so I got wrong answer. As I mentioned before, a is length OC in the triangle. I needed that for law of sines.
 
  • #10
The law of sines is always two equations. In this case is $$\frac{s}{\sin\alpha}
=\frac{\frac{1}{2}gt^2\cos\alpha}{\sin(\varphi/2)}
=\frac{h}{\sin(\alpha+\varphi/2)}.
$$
Did you use both? The first equation involving segment ##s## is needed to ensure that the point of intersection P is on the hypotenuse at time ##t##.

It's probably easier and more straightforward to write equations for the position of the bead, ##x(t)## and ##y(t),## find the intersection with the hypotenuse and then minimize.
StringSHortestTime.png
 
  • #11
bowes78 said:
Yes, previous solution had a mistake. This time I tried to do something like this, but that seems to me that it doesn't make sense too. Because if we set (φ/2)=0, then α=32, which doesn't look right.
Not sure how you got the line following "=>". Please post the details of that step.
I assume you realise it is only necessary to maximise the denominator in the expression for ##t^2## since the numerator is constant.
 
  • #12
kuruman said:
The law of sines is always two equations. In this case is $$\frac{s}{\sin\alpha}
=\frac{\frac{1}{2}gt^2\cos\alpha}{\sin(\varphi/2)}
=\frac{h}{\sin(\alpha+\varphi/2)}.
$$
Did you use both? The first equation involving segment ##s## is needed to ensure that the point of intersection P is on the hypotenuse at time ##t##.

It's probably easier and more straightforward to write equations for the position of the bead, ##x(t)## and ##x(t)## find the intersection with the hypotenuse and then minimize.
View attachment 318945
No, I used only $$\frac{h}{sin(\alpha+\varphi/2)}=\frac{\frac{1}{2}gt^{2}cos\alpha }{sin(\varphi/2) }.$$

I think I'm getting your idea, but how I should write ##x(t)## equation? I suppose that $$y(t)=\frac{1}{2}gt^{2}.$$
Or it should be like $$y(t)=\frac{1}{2}gcos^{2}\alpha t^{2} $$
$$x(t)=\frac{1}{2}gsin\alpha cos\alpha t^{2}$$
Because string length is $${\frac{1}{2}gcos\alpha}t^{2}$$
 
  • #13
haruspex said:
Not sure how you got the line following "=>". Please post the details of that step.
I assume you realise it is only necessary to maximise the denominator in the expression for ##t^2## since the numerator is constant.
IMG_20221217_094227.jpg
 
  • #15
haruspex said:
a and φ are constants, so what is the derivative of ##2a\sin(\phi/2)##?
I see... It should be zero, but whole equation then disappears.
 
  • #16
bowes78 said:
I see... It should be zero, but whole equation then disappears.
Does it? What do you have left?
 
  • #17
haruspex said:
Does it? What do you have left?
Well, should I solve it like that?
$$-gcos(2\alpha+\varphi /2)(2asin(\varphi/2))=0$$
$$cos(2\alpha+\varphi /2)=0$$
$$\alpha =\frac{90^{\circ}-\varphi /2 }{2}$$
 
  • #18
bowes78 said:
Well, should I solve it like that?
$$-gcos(2\alpha+\varphi /2)(2asin(\varphi/2))=0$$
$$cos(2\alpha+\varphi /2)=0$$
$$\alpha =\frac{90^{\circ}-\varphi /2 }{2}$$
Right.
 
  • #19
haruspex said:
Right.
Thanks!
 

1. What is the "String in cone problem"?

The "String in cone problem" is a mathematical problem that involves finding the shortest length of string required to wrap around a cone that has a given height and base radius. It is also known as the "Cone wrapping problem".

2. What is the significance of the "String in cone problem"?

The "String in cone problem" has real-world applications in fields such as engineering, architecture, and packaging. It helps in determining the minimum amount of material needed to cover a cone-shaped object, which can reduce costs and optimize design processes.

3. What is the formula for solving the "String in cone problem"?

The formula for solving the "String in cone problem" is L = √(h^2 + r^2) + 2πr, where L is the length of string required, h is the height of the cone, and r is the base radius of the cone.

4. How do you approach solving the "String in cone problem"?

To solve the "String in cone problem", you can break it down into smaller, simpler components. First, calculate the length of the hypotenuse of the triangle formed by the height and the slant height of the cone. Then, add the circumference of the base of the cone to this length to get the final answer.

5. Are there any variations of the "String in cone problem"?

Yes, there are variations of the "String in cone problem" such as finding the shortest length of rope required to tie around a cone-shaped object, or finding the maximum length of string that can be wrapped around a cone without any overlaps or gaps. These variations may involve different formulas and approaches to solve them.

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