A body rotates nonuniformly about a fixed vertical axis within an...

Let the fixed vertical axis of rotation be the z-axis. The angular velocity of the body is $\vec{\omega} = \omega \hat{k}$ and its angular acceleration is $\vec{\alpha} = \alpha \hat{k}$, where $\alpha \neq 0$ since the rotation is non-uniform.

Let the position vector of a particle of mass $m$ be $\vec{r} = r \hat{e}_r + z \hat{k}$, where $r > 0$ because the particle is not located on the axis.

The acceleration of the particle has two components:
Centripetal acceleration: $\vec{a}_c = -\omega^2 r \hat{e}_r$
Tangential acceleration: $\vec{a}_t = \alpha r \hat{e}_\theta$

The resultant force acting on the particle is given by Newton's second law:
$\vec{F} = m\vec{a} = -m\omega^2 r \hat{e}_r + m\alpha r \hat{e}_\theta$

Since the resultant force $\vec{F}$ has no component in the $\hat{k}$ (vertical) direction, it lies entirely in a horizontal plane.

The torque of this force about the z-axis is:
$\tau_z = r F_\theta = m\alpha r^2$

Since $\alpha \neq 0$ and $r \neq 0$, the torque $\tau_z$ is non-zero. If the line of action of the force intersected the z-axis, its torque about the axis would be zero. Because the torque is non-zero, the force does not intersect the axis of rotation.

A horizontal line that does not intersect the vertical axis is skew with the axis. Therefore, the resultant force is horizontal and skew with the axis.