Calculating Velocity and Distance for Ascending Double Cone on Rails

[franceboy]

Homework Statement

We consinder a doble cone with a radius R and an angle α (pike) and the mass m. It is located on two rails with an opening angle β. The rails enclose the angle γ with the ground. A is the lowest point of the rails.
First, the center of mass of the double cone is locatd vertically above the point A regarding the plane described by the two rails. The double cone ascends the rails. You can assume that base is in the center between the rails and that the line through the center of mass and the contact points with the rails is orthogonal to the rails.

a) Why does the double cone ascends the rails? State the conditions for the angles α,β,γ so that this happen.
b) Proof that the moment of inertia regarding the connecting line through through the pikes is I=3/10*m*R²
c) Determine the function v(d). v is the velocity of the center of mass and d is the rolled distance on the rails.
d) Calculate for the values α=50°; β=40°; γ=5°; R=10cm and m=100g the distance L which the double cone rolls on the rails and the maximal reached velocity.

Homework Equations

The Attempt at a Solution

a) The reason is that the height of the center of mass decreases while the double cone is ascending the rails. Condition: tan(γ)<tan(α/2)*tan(β/2)
b) Easy proof with the integral-formula of the moment of inertia.

However, I do not have ideas how to solve c) and d). Can you help me?

 

[gneill]

Consider conservation of energy. If you can determine the loss of gravitational PE with distance then it must go somewhere...

 

[BiGyElLoWhAt]

Well d is related to c, you can't do d without c. This actually looks like a good time for the application of some conservation laws. Or you could do some calc, but I think that can be avoided. I'm not sure, I still need to work it out.

I'm picturing something like so:

 

[franceboy]

Hi gneill,
Thank you for your help again.
Of course we have the equation: m*g*Δh_cm=1/2*m*v²+1/2*I*ω² with ω=v/r.
But i do not have a relation between h and d :(

 

[BiGyElLoWhAt]

Is there any relationship between delta h and delta L? A trigonometric relationship perhaps?

 

[franceboy]

Yeah there must be one relation but i do not how to formulate with the angles...

 

[BiGyElLoWhAt]

Look at the side view of my picture above. Draw yourself a right triangle, label the angles appropriately and it should be a simple relationship. SOHCAHTOA

 

[franceboy]

i am thinking of:
Δh=x*(tan(γ)-tan(α/2)*tan(β/2))
With x=cos(γ)*d

 

[TSny]

i am thinking of:
Δh=x*(tan(γ)-tan(α/2)*tan(β/2))
With x=cos(γ)*d

That looks good to me.

 

[franceboy]

Okay, now I have the function v(d). What about d), how can I determine L, the maximal reached distance? The maximal reached velocity is v(L), isn't it?

 

[nasu]

Well d is related to c, you can't do d without c. This actually looks like a good time for the application of some conservation laws. Or you could do some calc, but I think that can be avoided. I'm not sure, I still need to work it out.

I'm picturing something like so:

I think they mean something like this:
https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcQc_PS2Csv62bBTZJ6KLXM1xuXSog7RyBogHwnUwz7MA5AUljZuiQ

 

[franceboy]

Nasu, I think you are right. However, does that change any results?

 

[nasu]

No, my post was for bigyellowhat. I thought he may be confused about the geometry.
I don't think you used that picture for your calculation, did you?

 

[franceboy]

No I thought of the double cone you have presented :)

 

[TSny]

The maximum distance traveled is determined by where the double cone will no longer extend across the rails.

Note that as the cone approaches this point the radius of rolling, r, approaches zero. So, what do you expect the speed V to do as the cone approaches this point?

 

[franceboy]

If the radius of rolling approaches zero, the velocity must be zero!
That means I have to solve the equation v(L)=0 and than integrate the function v(d) from zero until L. Is that right? But the function v(d) does not become zero except for d=0

 

[TSny]

You should be able to determine L from the geometry.

You then find that v(d) goes to zero at d = L.

 

[franceboy]

My idea is to look when Δh=R.
Another idea is to look when the double cone would drop between the rails: R/tan(α/2)<tan(β/2)*d
Is one of the ideas right? Why is the v(d)=0 at this point?

 

[TSny]

My idea is to look when Δh=R.
Another idea is to look when the double cone would drop between the rails: R/tan(α/2)<tan(β/2)*d
Is one of the ideas right? Why is the v(d)=0 at this point?

You are right that v(d) is not zero at the point where the cone drops between the rails.

[EDIT: I now think v(d) = 0 at that point. Ugh!]

 

[BiGyElLoWhAt]

Hmmm... Is double cone a standard geometric shape? I guess I'm just not familiar with it, I actually googled it before I posted that picture (doh!). Stupid google, leading me astray.

And as far as the recent posts go, I think it makes physical sense that v is non zero at d=L. Why would the mass stop moving if it still has GPE w.r.t. the zero of your system? (if the center of mass is free to move down, neglecting friction, it will)

 

[franceboy]

So, is this equation: R/tan(α/2)=tan(β/2)*L right? Is v maximal the velocity after the distance of L: v(L)?

 

[TSny]

So, is this equation: R/tan(α/2)=tan(β/2)*L right?

This would be correct if L is measured from point A where the two rails meet. The question seems to define the distance d as the distance rolled. At the starting position, the point of contact of the cone with the rails is already some distance, D say, from A. So, if L is meant to be the maximum distance rolled from the starting point, then you would need to subtract D from your expression for L. This has caused some confusion for me.

[EDIT: Sorry, I had the initial position drawn incorrectly. The initial point of contact is at A and the CM is vertically above A, so there is no offset distance D. Your expression for L looks correct to me.]

Is v maximal the velocity after the distance of L: v(L)?

According to what I'm getting, the maximum velocity is not at the distance L. I believe the maximum velocity occurs well before L. After rechecking my work, I am again getting that v approaches zero as d approaches L.

I have been trying to really convince myself that it is OK to use v = ωr where r is the radius of the instantaneous cross section of the cone at the point of contact. If you painted a stripe on the cone corresponding to the points of contact of the cone with one of the rails, you would get a "helical" stripe of decreasing radius. The cone is rolling on this stripe. It is not rolling on a circular cross section. I need to think some more about the relation between v and ω.

 

[franceboy]

Well, I think I did not express myself very well. d is meant as the translation distance of the center of mass. So, my equation is right, isn´t it?.
I did not get, why v(L)=0. What happened to the potential energy?
I worked in c) with ω=v/R and I got:
v(d)=√(20/13*g*tan(γ)*d) Is that wrong?
How did you determine the maximal distance L?

 

[TSny]

Well, I think I did not express myself very well. d is meant as the translation distance of the center of mass. So, my equation is right, isn´t it?.

Yes, your equation is correct.

I did not get, why v(L)=0. What happened to the potential energy?

L is the position where r = 0. So, V = ωr = 0 at that point.

When r = 0, The potential energy has been entirely converted to rotational KE.

I worked in c) with ω=v/R and I got:
v(d)=√(20/13*g*tan(γ)*d) Is that wrong?
How did you determine the maximal distance L?

I assumed ω=v/r where the radius of rotation, r, decreases as the cone moves. r depends on d, R, and the angles α and β. I end up with a more complicated expression for v(d) that includes the angles α and β and the radius R. Did you derive an expression for r in terms of d, R, α and β?

I agree with your expression for L. L = R/[tan(α/2)*tan(β/2)]

 

[franceboy]

I always thought that the radius R is the same as r. How can a body rotate with another radius r?
No I do not have a expressions for r. In school we did not learn that r is not R. Do you understand what I mean?

 

[TSny]

As shown in the picture, r decreases as the cone moves along the rails.

 

[franceboy]

As shown in the picture, r decreases as the cone moves along the rails.

Thank you for the sketch.
r must be something like r=R-h-sin(y)*d with h as the difference of the height of the center of mass. But the conversation of energy is still right: m*g*h=1/2*m*v^2+1/2*I*w^2 and w=v/r and h=cos(y)*d*(tan(y)-tan(a/2)*tan(b/2))
Sorry, that I did not write the greek letters. I can not work with the "new" forum, yet.

 

[franceboy]

So if my consideration is right, I have to to look for a local maximum of the function v(d) with 0<d<L.
Is that right TSny?

 

[TSny]

So if my consideration is right, I have to to look for a local maximum of the function v(d) with 0<d<L.
Is that right TSny?

Yes, you'll have to find where v(d) is maximum. I get a fairly messy expression for v(d).
Solving v'(d) = 0 algebraically appears to lead to solving a cubic equation in d.

 

[franceboy]

Are the equations in post 27 right? Then the function v(d) and v'(d) appear very complicated...

 

[TSny]

Are the equations in post 27 right? Then the function v(d) and v'(d) appear very complicated...

Yes, your expression for h looks good except for the overall sign. You want h to be a positive number.

Your energy equation also looks good.

Your expression for r agrees with mine if you fix the overall sign of h. The expression for r will simplify somewhat.

 

[franceboy]

Thank you for your help.
What are your soulutions?
I got L=58.92cm and the maximal velocity is v(0.287m)=0.47m/s

There is also a part e) I have forgotten to post:
e) The assumption that the connecting line through the center of mass and the two pikes is always orthogonal to the rails is wrong. Analyse, where the rails really touch the double cone in the case of motion and find out what really happen in the point A.
I do not know how to deal with such kind of exercises... Could you give me some hints?

 

[TSny]

Yes, that's what I got except for a little round off difference. For d at max velocity I got 0.286 m and for the max velocity I got 0.464 m/s. I got exactly your answer for L.

I'll have to think about part (e).

 

[TSny]

Imagine a vertical plane that contains one of the rails. The plane will slice through the cone. The rail will contact the cone at the lowest point of the cross section. See the attached figure where you are looking at the cone along the axis connecting the apexes. For convenience, I drew the picture for the case where the plane of the rails is horizontal. The blue curve represents the cross section. The red dot shows the point of contact of the cone with the rail. You can see that the point of contact is offset from the apex by some distance $d_0$.

 

[franceboy]

Okay that means if d>0, the lines do not cross each other orthogonally. And d is zero if the double cone rolled a round, considered from the point A, right?

 

[TSny]

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.

 

[franceboy]

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?

 

[TSny]

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.

 

[franceboy]

I failed at describing the double cone with vectors and finding an intersection equation. Could you help me?

 

[TSny]

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.

 

[franceboy]

Sorry for answering so late.
Okay we imagine that the CM is the point CM=(x_CM, 0 , R+x_CM*(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)*y_circle
The centre of the circle is (x_CM , y_circle , R+x_CM*(tan(y)-tan(b/2)*tan(a/2)). The circle is in the y-z-plane, therefore we have the circle equation: (y-y_circle)²+(z-( R+x_CM*(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?

 

[TSny]

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..

 

[franceboy]

Okay y(x) is supposed to be y(x)=tan(b/2)*x
And y(x)=y_circle 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)

 

[TSny]

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.

 

[franceboy]

Is y(x)=tan(b/2)*x+R ?

 

[TSny]

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.

 

[franceboy]

Thanks.
Are my next attempts right?
How do I solve this task?

 

[TSny]

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.

 

[franceboy]

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?

 

[TSny]

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.