Calculating Velocity and Distance for Ascending Double Cone on Rails

  • Thread starter franceboy
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In summary: But I think the answer is still the same]In summary, the double cone with a radius of R and an angle α (pike) and the mass m rolls on two rails with an opening angle β. The rails enclose the angle γ with the ground. The center of mass of the double cone is located vertically above the point A regarding the plane described by the two rails.
  • #36
franceboy said:
Okay that means if d>0, the lines do not cross each other orthogonally.

Yes, if you mean that line ##P_1P_2## is not orthogonal to the rail ##AB##. If this is not what you mean, can you specify the lines that you are talking about?

And d is zero if the double cone rolled a round, considered from the point A, right?

I'm not sure I understand this question. If you work out the "offset distance" ##d_0## as a function of ##d##, then I believe you will find that there must be some offset distance even at the starting point if you want the double-cone to rest on the rails at the starting point. So, the CM of the system would not actually be on a line through A and perpendicular to the plane of the rails at the starting point. Maybe this is what you are saying.
 

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  • #37
I am talking about the line AB and the line through the center of mass and slice point from AB with the double cone.
Maybe this can be solved by vectors in the form
AB=(1,sin(y),tan(b/2)) Ocm(x,R+h,0) but I don not know how to describe OS, the slice point by the coordinate x of the center of mass.
With OS we could build the product AB*(Ocm-OS). The lines are orthogonal if this product is zero, isn't it?
 
  • #38
To find the point where one of the rails touches the cone, find the equation for the conic section created by the intersection of the cone with a vertical plane containing the rail. This requires writing an equation for the surface of the cone and an equation for the vertical plane. Use these to find the equation for the intersection. You can express these equations relative to your coordinate system where the origin is at the center of mass of the cone, the x-axis passes through origin and is parallel to plane of rails, the y-axis runs along the axis of the double cone, and the z-axis is perpendicular to the plane of the rails. See diagram.
 

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  • #39
I failed at describing the double cone with vectors and finding an intersection equation. Could you help me?
 
  • #40
First find the equation for the cone in terms of the (x,y,z) coordinate system. Imagine the surface of the cone as made up of circular cross sections of varying radius.
 

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  • #41
Sorry for answering so late.
Okay we imagine that the CM is the point CM=(xCM, 0 , R+xCM*(tan(y)-tan(b/2)*tan(a/2)).
Then the varying is dependent from the value of y. We can say: r=R-tan(a/2)*ycircle
The centre of the circle is (xCM , ycircle , R+xCM*(tan(y)-tan(b/2)*tan(a/2)). The circle is in the y-z-plane, therefore we have the circle equation: (y-ycircle)2+(z-( R+xCM*(tan(y)-tan(b/2)*tan(a/2))2=r2
Moreover the rails can be described by: AB=t*(1 , tan(b/2) , sin(y))
What should I do next?
 
  • #42
When the cone is at a certain position, I found it easier to choose the axes relative to the cone as shown. The x-axis runs through the CM of the cone and parallel to the rails. The equation for the circular cross section is then just x2 + z2 = r2 where r is as given in your expression. You now have the equation for the surface of the cone in terms of x, y, and z.

Find an expression for the rail in the form y(x). This is also the equation for the plane that contains the rail and is parallel to the z axis. Use it in the equation for the cone so that you get a relation between x and z for the intersection of the the plane and the cone..
 

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  • #43
Okay y(x) is supposed to be y(x)=tan(b/2)*x
And y(x)=ycircle Then we would have the equation:
(R-tan(a/2)*tan(b/2)*x)2=x2+z2
And z= root((R-tan(a/2)*tan(b/2)*x)2-x2)
 
  • #44
franceboy said:
Okay y(x) is supposed to be y(x)=tan(b/2)*x

The rail has a non-zero y-intercept in the chosen coordinate system.
 
  • #45
Is y(x)=tan(b/2)*x+R ?
 
  • #46
franceboy said:
Is y(x)=tan(b/2)*x+R ?
Not quite. y = tan(##\small\beta /2##) x + b where b is the y-intercept as shown.
 

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  • #47
Thanks.
Are my next attempts right?
How do I solve this task?
 
  • #48
You have the right idea. Once you get the correct equation for intersection of the cone with the plane y = mx + b, you can then find the value of x that corresponds to the point where the cone touches the rail. Think about the value of dz/dx at this point.
 
  • #49
I was not able to find the equation for the intersection. Could you name this equation?
And what should I do after solving this equation?
 
  • #50
The equation for the cone is of the form ##x^2 + z^2 = r(y)^2##.

You already found the function ##r(y)## for the right hand side back in post #41.

The equation of the plane that passes through the rail and is parallel to the z -axis is of the from ##y = mx + b##.

You need to find the values of ##m## and ##b## for the right hand side. I believe you have already found the correct value for ##m## in post #45.

##b## is the y-intercept of the plane, where the y-axis is the axis of the cone that passes through the two apexes. You should be able to express ##b## in terms of ##\beta## and ##d##. See the shaded triangle in the figure.

The intersection of the plane with the cone is found by substituting the plane equation into the cone equation.
 

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  • #51
Thank you.
The function for the left rail is y=tan(b/2)*x+tan(b/2)*d . Then we have the equations x2+z2=r(y)2
The conclusion would be x2+z2=(R-tan(a/2)*(tan(b/2)*x+tan(b/2)*d))2
I think z can be expressed by z(x)=tan(y)*x+tan(y)*d.
Back to the exercise: How I find out where the rail touches the double cone? And what will happen at A?
 
  • #52
franceboy said:
The conclusion would be x2+z2=(R-tan(a/2)*(tan(b/2)*x+tan(b/2)*d))2

I think this is correct. This equation gives the projection into the x-z plane of the intersection of the cone and the plane (containing the rail and parallel to the z-axis). This intersection is the purple curve in the figure. There are a couple of ways you can try to find the point of contact of the rail and the cone:

1.) Note that the point of contact is the point where z takes on a maximum value on the purple curve. So, what is ##dz/dx## at the point of contact?

or

2.) Rearrange your equation for the intersection into the form $$\frac{(x+x_0)^2}{A^2} + \frac{z^2}{B^2} = 1$$ for certain constants ##A##, ##B##, and ##x_0##. Interpret.
 

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  • #53
Since z takes on it`s maximal value, z`(x)=0.
z(x)=root((R-tan(a/2)*(tan(b/2)*x+tan(b/2)*d))2 - x2)
I can solve the first equation with my calculator. Then I got the real x-value which differs from the x value of the center of mass.
What is the next step?
 
<h2>1. What is the formula for calculating velocity of an ascending double cone on rails?</h2><p>The formula for calculating velocity is v = √(2gh), where v is velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the double cone on the rails.</p><h2>2. How do you calculate the distance traveled by the ascending double cone on rails?</h2><p>To calculate the distance traveled, you can use the formula d = v²/2g, where d is the distance traveled, v is the velocity, and g is the acceleration due to gravity.</p><h2>3. Can the velocity and distance be calculated if the height and acceleration due to gravity are not known?</h2><p>No, the velocity and distance cannot be calculated without knowing the height and acceleration due to gravity. These two variables are essential in the formulas for calculating velocity and distance.</p><h2>4. How does the velocity and distance of the double cone change if the height is increased?</h2><p>If the height is increased, the velocity and distance will also increase. This is because the higher the double cone is on the rails, the more potential energy it has, which is converted into kinetic energy as it descends.</p><h2>5. Can the formulas for calculating velocity and distance be used for any type of object on rails?</h2><p>Yes, the formulas for calculating velocity and distance can be used for any object on rails, as long as the object is only affected by the force of gravity and there is no other external force acting on it.</p>

1. What is the formula for calculating velocity of an ascending double cone on rails?

The formula for calculating velocity is v = √(2gh), where v is velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the double cone on the rails.

2. How do you calculate the distance traveled by the ascending double cone on rails?

To calculate the distance traveled, you can use the formula d = v²/2g, where d is the distance traveled, v is the velocity, and g is the acceleration due to gravity.

3. Can the velocity and distance be calculated if the height and acceleration due to gravity are not known?

No, the velocity and distance cannot be calculated without knowing the height and acceleration due to gravity. These two variables are essential in the formulas for calculating velocity and distance.

4. How does the velocity and distance of the double cone change if the height is increased?

If the height is increased, the velocity and distance will also increase. This is because the higher the double cone is on the rails, the more potential energy it has, which is converted into kinetic energy as it descends.

5. Can the formulas for calculating velocity and distance be used for any type of object on rails?

Yes, the formulas for calculating velocity and distance can be used for any object on rails, as long as the object is only affected by the force of gravity and there is no other external force acting on it.

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