Answer
See the detailed answer below.
Work Step by Step
a) From the figure below, we can apply Newton's second law to find the ball's angular velocity.
$$\sum F_x=F_n\cos\theta=ma_r=m\dfrac{v^2}{r}$$
where $\cos\theta=\dfrac{r}{R}$; thus
$$ F_n\dfrac{r}{R}=m\dfrac{v^2}{r}$$
where $v=\omega r$
$$ F_n\dfrac{\color{blue}{\bf\not} r}{R}=m\dfrac{\omega^2 \color{blue}{\bf\not} r^{\color{red}{\bf\not} 2}}{\color{red}{\bf\not} r}$$
Hence,
$$\omega=\sqrt{\dfrac{F_n}{mR}}\tag 1$$
Now we need to find the normal force;
$$\sum F_y=F_n\sin\theta-mg=ma_y=m(0)=0$$
Thus,
$$F_n=\dfrac{mg}{\sin\theta}$$
where $\sin\theta=\dfrac{R-y}{R}$
So that
$$F_n=\dfrac{mgR}{R-y}$$
Plugging into (1);
$$\omega=\sqrt{\dfrac{ \color{red}{\bf\not} m \color{red}{\bf\not} R g}{ \color{red}{\bf\not} m \color{red}{\bf\not} R(R-y)}}$$
$$\boxed{\omega=\sqrt{\dfrac{ g}{ R-y }}}$$
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b) The minimum value of $\omega$ to move on a circle is approach as $R-y$ approaches $R$.
In other words, we have the minimum value of $\omega$ when $R-y\approx R$ and hence,
$$\boxed{\omega_{mini}=\sqrt{\dfrac{g}{R}}}$$
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c) $y=\frac{1}{2}R$ since the ball is halfway up.
Plugging the given into the boxed formula in part (a).
$$ \omega=\sqrt{\dfrac{ 9.8}{ R-\frac{1}{2}R }}=\sqrt{\dfrac{ 9.8}{ \frac{1}{2}\times0.2 }} =\bf 9.9\;\rm rad/s$$
$$ \omega=\rm \dfrac{9.9\;rad}{1\;s}\times\dfrac{60\;s}{1\;min}\times \dfrac{1\;rev}{2\pi \;rad}$$
$$ \omega=\color{red}{\bf 94.5}\rm rpm $$