Rotation of a sphere on two different axes

Question

You have a small globe, which is mounted so that it can spin on the polar axis and can be spun about a horizontal axis (so that the south pole can be on top). Give the globe a quick spin about the polar axis, and then, before it stops, give it a spin about the horizontal axis. Are there any points on the globe that are at rest?

(A) There are two points, fixed on the globe, that are at rest.

(B) There are two points that are instantaneously at rest, but these two points move around the globe in an apparently random fashion.

(C) At some times two points are instantaneously at rest.

(D) There are no points at rest until the globe stops spinning.

My approach was to add the angular velocity vectors $w=w_1+w_2$ to obtain net angular velocity vector with the points on the globe which are parallel and antiparallel to it being at rest with the answer coming A.

But the answer provided was B. Kindly explain why the instantaneous axis of rotation and consequently the points at rest are changing seemingly randomly.

Answer 1

The only point hat remains at reset is the north pol point.

starting with the rotation matrix

$$\mathbf S= \left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\ \sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right]\, \left[ \begin {array}{ccc} 1&0&0\\ 0&\cos \left( \varphi \right) &-\sin \left( \varphi \right) \\ 0 &\sin \left( \varphi \right) &\cos \left( \varphi \right) \end {array} \right] $$

and the position vector of a point on the sphere surface

$$\mathbf R=r\,\left[ \begin {array}{c} \cos \left( \lambda \right) \cos \left( \vartheta \right) \\ \cos \left( \lambda \right) \sin \left( \vartheta \right) \\ \sin \left( \lambda \right) \end {array} \right] $$

where $~\lambda~$ is the longitudinal angle and $~\vartheta~$ the azimuth angle.

from here , the transformed vector coordinates are

$$\left[ \begin {array}{c} x\\ y\\ z\end {array} \right] =\mathbf S^T\,\mathbf R\tag 1$$ solving $~x=0~,y=0~$ you obtain that $$\varphi=\pm \arctan\left(\frac{1}{\tan(\lambda)}\right)\quad ,\psi=\vartheta \pm\frac\pi2$$ substitute those results in equation (1) you obtain:

$$ \left[ \begin {array}{c} x\\ y\\ z\end {array} \right]= \left[ \begin {array}{c} 0\\ 0\\ \pm r\end {array} \right]$$

thus , if you rotate the globe first about the z axes with $~\psi=+\psi(\vartheta)~$ and then rotate about the $~x'~$ axes with $~\varphi=+\varphi(\lambda)~$ the only point that remains at reset is the north pol point.

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the rotation matrix $~\mathbf S=\mathbf S(\,\psi(\vartheta)_\pm~,\varphi(\lambda)_\pm\,)~$ is then

$$\mathbf S_+= \left[ \begin {array}{ccc} -\sin \left( \vartheta \right) &-\cos \left( \vartheta \right) \sin \left( \lambda \right) &\cos \left( \lambda \right) \cos \left( \vartheta \right) \\ \cos \left( \vartheta \right) &-\sin \left( \vartheta \right) \sin \left( \lambda \right) &\cos \left( \lambda \right) \sin \left( \vartheta \right) \\ 0&\cos \left( \lambda \right) &\sin \left( \lambda \right) \end {array} \right]\\ \mathbf S_-=\left[ \begin {array}{ccc} -\sin \left( \vartheta \right) &-\cos \left( \vartheta \right) \sin \left( \lambda \right) &-\cos \left( \lambda \right) \cos \left( \vartheta \right) \\ \cos \left( \vartheta \right) &-\sin \left( \vartheta \right) \sin \left( \lambda \right) &-\cos \left( \lambda \right) \sin \left( \vartheta \right) \\ 0&\cos \left( \lambda \right) &-\sin \left( \lambda \right) \end {array} \right] $$

$\Rightarrow$

the angular velocity obtained in inertial system

$$\mathbf\omega= \left[ \begin {array}{c} \sin \left( \vartheta \right) \dot\lambda \\ -\cos \left( \vartheta \right) \dot\lambda \\ \dot\vartheta \end {array} \right] $$