# Rotation about the axis of the cone.

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I am given the velocity $v$ of the point $C$ of the base of the cone which rolls without slipping on a horizontal plane. Now the question asks me to find the angular velocity of the cone.

Attempt:

I found the angular velocity of the cone about its own axis using the condition for pure rolling:

$v= R\omega_{OC} \implies \omega_{OC}= \dfrac v R$

Now, the answer is incomplete because we have to include the angular velocity due to rotation about the point O. I am facing difficulty in this part. How do I go about solving it further? The distance OC is $R\cot\alpha$

Edit:

I analysed it further and here are my workings:
Let $\omega$ be angular velocity about O

$\omega = \dfrac{d\theta}{dt}= \dfrac{ds}{R\cot \alpha dt}= \dfrac{v}{R\cot \alpha}$

Is my approach correct?

edited May 27, 2018
i don't get why the mathjax doesn't show there.
Perhaps because you omitted the underscore after omega.
No there's an underscore, when I go to edit it, it shows an underscore.  Also, everything's fine in the preview too...
Your question is similar (identical?) to http://physics.qandaexchange.com/?qa=2306/angular-acceleration-of-the-disc. The direction of the angular velocity vector of the cone is continuously changing, so there is angular acceleration. But the magnitude of this vector is constant and equal to $\omega = v/R$.
Yeah, I realise that there's angular acceleration but is my derivation for its value correct?
Yes, you are correct, there is also angular velocity about the vertical axis, equal to $v/OC$. The total angular velocity is the resultant of those two vectors. This resultant is continuously changing in direction, so there is angular acceleration. Now the question is : Where is the torque?!
How does torque matter? maybe some external agency would be providing it
If there is angular acceleration there must be torque. But it is not obvious to me which force(s) is providing it.

The cone has angular velocity $\omega_{OC}=\frac{v}{R}$ about its own axis OC. Meanwhile this axis is rotating about the vertical OZ with angular velocity $\omega_{OZ}=\frac{v}{R\cot\alpha}$.
These two angular velocities are at right angles so the magnitude of the total angular velocity is $$\omega=\sqrt{\omega_{OC}^2+\omega_{OZ}^2}=\sqrt{\frac{v^2}{R^2}+\frac{v^2}{R^2\cot^2\alpha}}=\frac{v}{R} \sqrt{1+\tan^2\alpha}=\frac{v}{R}\sec\alpha$$