A body rotates uniformly about a fixed vertical axis in an inertial...

Let the fixed vertical axis of rotation be the z-axis. The angular velocity of the body is $\vec{\omega} = \omega \hat{k}$.

Since the body rotates uniformly, the angular acceleration $\vec{\alpha} = 0$.

The position vector of a particle not on the axis can be written as $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.

The acceleration of the particle is given by the centripetal acceleration, as the tangential acceleration is zero:

$\vec{a} = \vec{\omega} \times (\vec{\omega} \times \vec{r})$

$\vec{a} = \omega \hat{k} \times (\omega \hat{k} \times (x \hat{i} + y \hat{j} + z \hat{k})) = \omega \hat{k} \times (\omega x \hat{j} - \omega y \hat{i}) = -\omega^2 x \hat{i} - \omega^2 y \hat{j}$

$\vec{a} = -\omega^2 (x \hat{i} + y \hat{j})$

According to Newton's second law, the resultant force on the particle is $\vec{F} = m\vec{a} = -m\omega^2 (x \hat{i} + y \hat{j})$.

This force has no z-component, meaning it is entirely horizontal. Furthermore, it is directed towards the origin of the xy-plane, which means its line of action intersects the axis of rotation.