Rotating Cone and instantaneous axis of rotation

[RicardoMP]

Homework Statement

Hi! I'm trying to solve a simple problem of mechanics, but I'm getting the wrong results and I suppose I don't yet grasp the concept of instantaneous axis of rotation very well.
So, a cone (see attached picture) is rolling without slipping on a plane. Vp is point P linear velocity on the laboratory frame with its origin on the cone's apex. I must choose "z" as the vertical axis and, instantaneously, "x" along the line of contact.
I have to find the instantaneous angular velocity vector [tex] \omega [/tex].

Homework Equations

[tex]\omega = \Omega + \omega _s[/tex]
\Omega = \frac{v_p}{hcos(\alpha)}
\omega _s = \frac{v_p}{R}

The Attempt at a Solution

My first thought was that there are 2 contributions for the angular velocity: [tex]\Omega[/tex] pointing upwards and relates to the rotation of the cone around the vertical axis going through the origin and [tex]\omega _s[/tex] which is the spin angular velocity of the cone. I simply decomposed the spin ang.vel in its x and z components and got
[tex]\omega = (\omega_s cos(\alpha))e_x+(\omega_s sin(\alpha) + \Omega)e_z[/tex]
And I thought that was it.
The solution states that the angular velocity only has component along x: \omega = (\omega,0,0) and I'm failing to see how. Is the line of contact an instantaneous axis of rotation? And what does that mean for this problem?

Thank you in advance

 

[SammyS]

Homework Statement

Hi! I'm trying to solve a simple problem of mechanics, but I'm getting the wrong results and I suppose I don't yet grasp the concept of instantaneous axis of rotation very well.
So, a cone (see attached picture) is rolling without slipping on a plane. Vp is point P linear velocity on the laboratory frame with its origin on the cone's apex. I must choose "z" as the vertical axis and, instantaneously, "x" along the line of contact.
I have to find the instantaneous angular velocity vector [tex] \omega [/tex].

Homework Equations

[tex]\omega = \Omega + \omega _s[/tex]
##\Omega = \frac{v_p}{hcos(\alpha)}##
##\omega _s = \frac{v_p}{R}##

The Attempt at a Solution

My first thought was that there are 2 contributions for the angular velocity: [tex]\Omega[/tex] pointing upwards and relates to the rotation of the cone around the vertical axis going through the origin and [tex]\omega _s[/tex] which is the spin angular velocity of the cone. I simply decomposed the spin ang.vel in its x and z components and got
[tex]\omega = (\omega_s cos(\alpha))e_x+(\omega_s sin(\alpha) + \Omega)e_z[/tex]
And I thought that was it.
The solution states that the angular velocity only has component along ##\ x: \omega = (\omega,0,0)\ ## and I'm failing to see how. Is the line of contact an instantaneous axis of rotation? And what does that mean for this problem?

Thank you in advance

I'm looking for the attached picture.

 

[RicardoMP]

I'm looking for the attached picture.

I'm so sorry! I completely forgot to attach it!

 

[haruspex]

The solution states that the angular velocity only has component along x: \omega = (\omega,0,0) and I'm failing to see how. Is the line of contact an instantaneous axis of rotation?

Yes. That line of the cone is in contact with the plane and is not slipping, therefore it is instantaneously stationary. That makes it the instantaneous axis of rotation.
Concentrate on the point P. You know its velocity, and you know it is rotating about that axis.