# Instantaneous axis of rotation and rolling cone motion

Suppose a cone is purely rolling (no slipping) around a fixed axis. I mean, it is revolving around a fixed axis perpendicular to the ground and passing through its vertex and also rotating, so the vertex is stationary. (sorry this might be a bit confusing but I hope you understand what I mean). Something like this, but rolling on plane surface instead of another cone: rolling cone http://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Rolling_cone.pdf/page1-1024px-Rolling_cone.pdf.jpg

Now the instantaneous axis of rotation (IAR) of the cone is the 'line' that is touching the ground right? So how do you find the velocity of any other point using that? I mean, in the rolling wheel, you multiply the angular velocity by the distance from the IAR to get the velocity. Is it the same here?

If it is, then consider the center of base of the cone. If the height of cone is $h$ then its distance from the IAR is clearly $h\sin x$ where $x$ is the cone's half apex angle. So its velocity should be $ah\sin x$, where $a$ is the angular velocity with which the cone is rotating. Is this right?

Now we can also analyse the cone's motion by considering it in two parts: rotation + revolution, right? So again considering the center of base of the cone, it has no velocity by virtue of rotation (since the cone is rotating about an axis through the center), right? And by virtue of it rotating in a circle (of radius $h\cos x$) around the axis passing through its vertex, it has velocity $bh\cos x$ , where $b$ is the angular velocity with which the cone is revolving.

Now these two must be same, so we get $b=a \tan x$.

But Wikipedia states here that the ratio is $\sin x$.

And at the same time, this video (which I found in the external links section of the Wikipedia page) states that $a=b\cot x$ which is same as what I got.

So I'm really confused. Is everything that I did right? If not please correct me. Thank you.

Edit: OK so as carl commented, I'm also confused about how the instantaneous linear velocity of the center of the cone's base is different from the velocity of the center of a rolling disk.

Edit 2: How to find velocity of any point on the cone? There ought to be two approaches , one using IAR and another by considering motion as rotation + revolution but I'm not able to do it.

• The wikipedia article describes a cone rolling on a cone. You apparently are trying to describe a cone rolling on a plane. – David Hammen Oct 6 '14 at 11:45
• @DavidHammen Yes but it also says " In the special case of a cone rolling on a flat surface, this ratio becomes sin alpha, where alpha is the cone's half apex angle" – A Googler Oct 6 '14 at 11:59
• Well, (thought experiment) how is the instantaneous linear velocity of the center of the cone's base different from the velocity of the center of a rolling disk? Solve that and convert into angular velocity about the vertex of the cone, maybe. – Carl Witthoft Oct 6 '14 at 13:15
• @CarlWitthoft Oh , so the cone's base is like a rolling disk but it is touching the ground at an angle. If that angle is theta then I'd guess linear velocity of center to be um $w*r*sin(theta)$, is that right? edit: Now I'm thinking it's just $w*r$, hmm. – A Googler Oct 6 '14 at 18:10
• @AGoogler See my revised answer. I believe I've found my mistake: I divided $\cos$ by $\sin$ and somehow got $\tan$, whereas it should of course have been $\cot$! Now the answer makes a great deal more sense. The computation is exactly as you say: the instantaneous axis is along the contact plane and you multiply $h\sin\alpha$ by the instantaneous angular speed. – Selene Routley Oct 12 '14 at 4:39

Let the cone lie on the $\hat{X}\wedge \hat{Y}$ plane (z=0) and let the $z$ axis pierce this plane at the cone's apex. If the cone's half angle is $\alpha$, then its axis of symmetry as a function of time is defined by the vector

$$A(t)=\cos\alpha \left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)+\sin\alpha \hat{Z}$$

where $\omega_0 = 2\pi/\tau$ and $\tau$ is time it takes the cone to make exactly one circuit on the $\hat{X}\wedge \hat{Y}$ plane. Thus the cone's axis of symmetry rotates with angular velocity $\omega_0\,\hat{Z}$. I define my directions and symbols below: If the cone doesn't slip, this means that the rotation about the axis $A(t)$ has an angular velocity $- \omega_0 A(t)/\sin\alpha$. Sketch the cone near the apex to see this: at a distance $R$ along the edge (defined by the vector $\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}$) in the $\hat{X}\wedge \hat{Y}$ plane where the cone meets the plane, the tip of this edge moves at speed $v_e=\omega_0\,R$. The circular cone cross section (orthogonal to the cone's axis of rotational symmetry) through this point is like a wheel of radius $r=R\,\sin\alpha$ cambered inwards at angle $\alpha$. This "wheel" must spin at angular velocity $-\omega_0\,R\,A(t)/r$ so that its rim's velocity is $-v_e=-\omega_0\,R$ to offset the velocity $v_e=\omega_0\,R$ of the edge at this point and keep the point of the wheel in contact with the ground stationary.

We add these two angular velocities to get:

$$\Omega(t) = \omega_0\left(\hat{Z} - \frac{A(t)}{\sin\alpha}\right)=-\omega_0 \,\cot\alpha\,\left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)$$

which, as you correctly guessed, is always along the line where the cone meets the plane.

The instantaneous speed of a point on the cone's axis of symmetry a distance $h$ from the base is $|\Omega|\,h\,\sin\alpha = \omega_0\,h\,\cot\alpha\,\sin\alpha = \omega_0 \,h\,\cos\alpha$ ($|\Omega|$ times as you say, the orthogonal distance $h\,\sin\alpha$ of the point from the instantaneous axis of rotation.

Note that we get the same answer by simply working this speed out for an angular velocity $\omega_0\,\hat{Z}$, which is valid because the axis of the cone has no velocity owing to the rotation in the direction $A(t)$. The point on the cone's axis of symmetry is a distance $h\,\cos\alpha$ from the $\hat{Z}$ axis. Thus the speed is, as before, $\omega_0\,h\,\cos\alpha$.

• Hi , thanks for the answer. Your notation and way of solving the problem was a bit new for me but I'm trying to understand. A few questions: – A Googler Oct 10 '14 at 9:49
• 1. the final angular velocity is wtanx along the contact line. So if i wanted to find velocity of any point on the cone , would i just multiply wtanx by the perpendicular distance of that point from the line? 2. If so then velocity of centre of base of cone would be hsinxwtanx where h is height. But it should be hcosxw since hcosx is distance from z axis and centre has no velocity due to rotation of cone. Please help,I'm confused. – A Googler Oct 10 '14 at 9:57
• @AGoogler Let me come back to this: am a little busy at the moment: can see your problem and there must be a difference between our notations somewhere. – Selene Routley Oct 10 '14 at 10:34
• @AGoogler Still haven't forgotten about you. I'll be mulling over your question over the weekend - sometimes I do a great deal of slow thinking whilst watching my children play, and I'll take your problem with me. – Selene Routley Oct 11 '14 at 0:47
• Thanks , your help is really appreciated. Meanwhile can you suggest a good physics book for me ? (mostly classical mechanics , rotational mechanics with an emphasis on problem solving) – A Googler Oct 11 '14 at 7:15