Answer
a) $ T=\dfrac{mgL}{\sqrt{L^2-r^2 }}$
b) $ \omega=\sqrt{\dfrac{ g }{\sqrt{L^2-r^2}} } $
c) $5.0\;\rm N,\;30.2\;\rm rpm$
Work Step by Step
a) First of all, we need to draw the force diagram of the ball, as we see in the figures below.
$$\sum F_y=T\cos\theta-mg=ma_y=m(0)=0$$
Thus,
$$T=\dfrac{mg}{\cos\theta}\tag 1$$
Where $\cos\theta$, as we see below, is given by
$$\cos\theta=\dfrac{h}{L}$$
where $h$ is the height of the cone and is given by the right-triangle
$$h^2=L^2-r^2\\
\rightarrow h=\sqrt{L^2-r^2}$$
So that,
$$\cos\theta=\dfrac{\sqrt{L^2-r^2}}{L}$$
Plugging into (1);
$$\boxed{T=\dfrac{mgL}{\sqrt{L^2-r^2}}}$$
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b) The ball's speed is given by
$$\sum F_r=T\sin\theta=ma_r=m\dfrac{v^2}{r}$$
Thus,
$$T\sin\theta =m\dfrac{v^2}{r}$$
where $v=r\omega$
$$T\sin\theta =m\dfrac{\omega^2r^2}{r}=m\omega^2r$$
Plugging $T$ from the boxed formula above;
$$\dfrac{ \color{red}{\bf\not}mgL}{\sqrt{L^2-r^2}}\sin\theta = \color{red}{\bf\not}m\omega^2r$$
$$\dfrac{ gL}{\sqrt{L^2-r^2}}\sin\theta = \omega^2r$$
where $\sin\theta=r/L$;
$$\dfrac{ g \color{red}{\bf\not}L}{\sqrt{L^2-r^2}}\dfrac{ \color{red}{\bf\not}r}{ \color{red}{\bf\not}L} = \omega^2 \color{red}{\bf\not}r$$
$$\dfrac{ g }{\sqrt{L^2-r^2}} = \omega^2 $$
Therefore,
$$\boxed{\omega=\sqrt{\dfrac{ g }{\sqrt{L^2-r^2}} }}$$
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c) To find the tension, we need to plug the known into the first boxed formula above.
$$ T=\dfrac{mgL}{\sqrt{L^2-r^2}} =\dfrac{0.5\times 9.8\times 1}{\sqrt{1^2-0.2^2}} $$
$$T=\color{red}{\bf 5.0}\;\rm N$$
..
To find the angular speed, we need to plug the known into the second boxed formula above.
$$ \omega=\sqrt{\dfrac{ 9.8}{\sqrt{1^2-0.2^2}} } =3.16\;\rm rad/s$$
Finding it in rpm;
$$ \omega= 3.16\;\rm rad/s\left(\dfrac{1\;rev}{2\pi\;rad}\right)\left(\dfrac{60\;s}{1 \;min}\right)$$
$$\omega=\color{red}{\bf 30.2}\;\rm rpm$$
