A dancing marble

First, consider the marble at equilibrium and use spherical coordinate. By Newton's second law we obtain

$m\dot{\phi_0^2}R\cos \theta = N \cos \theta_0$ (1)...

$mg=N\sin \theta_0$ (2)...

Divide (1) and (2) we get

$\dot{\phi_0^2}=\frac {g}{R\sin \theta_0}$ (3)...

Now let's look at the motion of the marble after given a perturbation and again use spherical coordinate for convenience

The angular momentum of the system is conserved so we can write

$mR^2\cos^2 \theta_0 \dot{\phi_0} = mR^2 \cos^2 \theta \dot{\phi}$

$\dot{\phi} = \dot{\phi_0} \frac {\cos^2 \theta_0}{\cos^2 \theta}$ (4)...

and the energy of the system is also conserved, given as

$E = \frac {1}{2}m(R^2 \cos^2 \theta \dot{\phi^2}+R^2 \dot{\theta^2})-mgR \sin \theta$ (5)...

Substitute eq (4) to (5) and then set the derivative of energy with respect to time equals zero due to energy conservation and using eq (3) we obtain

$0 = \frac {gR \cos^4 \theta_0 \sin \theta}{\sin \theta_0 \cos^3 \theta}+R^2 \ddot{\theta}-gR\cos \theta$

The particle is given a small kick in z-direction hence the small change in theta. Let's this change named as $\epsilon$ so according to Taylor expansion up to first order

$\sin \theta \approx \sin \theta_0 + \epsilon \cos \theta_0$

and

$\cos \theta \approx \cos \theta_0 - \epsilon \sin \theta_0$

This simplifies equation to (maintaining the first order of $\epsilon$ only)

$0=\ddot{\epsilon}+\frac {g}{R} (4 \sin \theta_0 + \cot \theta_0 \cos \theta_0)\epsilon$

So, the angular frequency is given by

$\omega = \sqrt{\frac {g}{R}} \sqrt{4 \sin \theta_0 + \cot \theta_0 \cos \theta_0}$

Plug in $r = 0.5R$ we get

$F(r=0.5R) = \sqrt{2\sqrt{3}+\frac {\sqrt{3}}{6}} = 1.94$