A cone rolls without slipping on a table. The half-angle at the vertex is $\alpha$, and the axis has length $h$. Let the speed of the center of the base, point $\text{P}$ in the figure, be $v$.
Question: What is the angular velocity of the cone with respect to the lab frame at the instant shown?

Attempted solution

Let us consider a frame rigidly attached to the cone with basis vectors $\hspace{.1cm}\hat{x_1}$,$\hspace{.2cm}\hat{x_2}$ and $\hspace{.1cm}\hat{x_3}$ as shown below, where $\hat{x_2}$ points out of the page, and let the coordinate axes fixed to lab be $\hat{z}\hspace{.2cm}$and $\hat{y}$. Let the cone rotate with $\omega_z$ around $\hat{z}$ and with $\omega_3$ around $\hat{x_3}$.

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Since the point A is at rest with respect to the lab frame, $$\vec{v_a} = \vec{0}$$

This therefore implies,$$\omega_3\cdot(h\cdot\tan\alpha) = v$$ Also, since it is given that $\vec{v_a} = v\hat{x_2}$,$$\omega_z\cdot(h\cdot\cos\alpha) = v$$

Therefore, we have the angular velocity in lab frame at the current instant, $\vec{\omega}$ such that, $$\vec{\omega} = -\omega_z\hat{z} + \omega_3\hat{x_3}$$ And breaking $\hat{x_3}$ into its components along $\hat{z}\hspace{0.2cm}and\hspace{0.2cm}\hat{y}$ and substituting the values of $\omega_3\hspace{0.2cm}and\hspace{0.2cm}\omega_z$, $$ \vec{\omega} = \frac{-v}{h \cos\alpha}\hat{z} + \frac{v}{h \tan\alpha} \left[ \cos{\left(\alpha \right)} \, \hat{y}+ \sin{\left( \alpha \right)} \, \hat{z} \right]$$

And thus the angular velocity is, $$\vec{\omega} = \frac{v}{h}\left[ \left( \cos{\alpha} - \frac{1}{\cos{\alpha}} \right) \hat{z} + \frac{ \left(\cos{\alpha}\right)^2}{ \sin{\alpha}}\hat{y}\right]$$

Why is this solution wrong?

  • $\begingroup$ Looking from above, the cone rolls around its vertex at some angular velocity. Looking at the "base" of the cone, it rotates around the point P at some angular velocity. Which does the question ask about? $\endgroup$
    – DJohnM
    Jul 24, 2017 at 4:14
  • $\begingroup$ The total angular velocity of the cone at the current instant. In fact, even the direction is not asked for, just the magnitude is asked. $\endgroup$
    – BabaYaga
    Jul 24, 2017 at 4:57
  • 1
    $\begingroup$ Have you looked here, on Page 429? books.google.ca/… $\endgroup$
    – DJohnM
    Jul 24, 2017 at 5:22

2 Answers 2


You are right that angular velocities can be added as vectors, so total angular velocity can be expressed as sum

$$ \boldsymbol{\omega} = \omega_z \hat{\mathbf z} + \omega_3\mathbf{x}_3 \tag{1} \,.$$

The point where you go wrong is

Since the point A is at rest with respect to the lab frame, $$\vec{v_a} = \vec{0}$$

This therefore implies,$$\omega_3\cdot(h\cdot\tan\alpha) = v$$

It is true that point A has zero velocity, but that does not imply the given expression for $\omega_3$.

Actually, it is much easier to find the total angular velocity and then find $\omega_3$. Here is how.

Every solid body with one or more points at rest (zero velocity) is either at rest as a whole or is rotating around some axis. The cone is said to roll without slipping, which means that whole line of points of the cone that is in contact with the table is at rest. But since the cone is rotating around some axis, this contact line has to be the axis of rotation.

Then, the velocity of the point $P$ can be expressed as $$ v = \omega\, h \sin \alpha $$ and from this we obtain the magnitude of the sought angular velocity $$ \omega = \frac{v}{h\sin\alpha}, $$ the direction being that of the contact line.

Now, we could calculate $\omega_3$ from the equation (1). The result is

$$ \omega_3 = \frac{v}{h\sin\alpha \cos\alpha}. $$


Point P travels along a circle of radius $R = h\cos(\alpha)$, so its speed is $v = R\omega = h\cos(\alpha)\omega$, which finally gives

$$ \omega = \frac{v}{h\cos(\alpha)}$$

  • $\begingroup$ What are you referring to? The answer by the way, is v(h*sin(alpha)) $\endgroup$
    – BabaYaga
    Jul 23, 2017 at 22:20
  • $\begingroup$ I was referring to the angular speed of the axis $OP$ where $O$ is the vertex of the cone. This answer is not homogeneous so is wrong... $\endgroup$
    – Spirine
    Jul 23, 2017 at 22:26
  • $\begingroup$ I'm sorry i meant v/h(sin(alpha)) $\endgroup$
    – BabaYaga
    Jul 23, 2017 at 22:51
  • $\begingroup$ This supplied answer is the rotational velocity seen by looking at the "bottom" of the cone $\endgroup$
    – DJohnM
    Jul 24, 2017 at 5:15

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