Answer
a) $6.632\;\rm rad/s$
b) $42.8\;\rm N$
Work Step by Step
a) To find the block's velocity, we need to find its acceleration, and to find the acceleration, we need to find the net force exerted on it.
So we need to draw a force diagram, as shown below.
Now we can apply Newton's second law;
$$\sum F_x=T+F_T\cos\theta=ma_x $$
where $x$ here is the radial direction at any moment while $z$ here is the tangential direction.
Thus,
$$T+F_T\cos\theta=m\dfrac{v^2}{R}$$
where $R=L$ is the length of the massless rod and $v$ is given by $v=\omega R=\omega L$
$$T+F_T\cos\theta=m\dfrac{\omega^2 L^2}{L} $$
$$T+F_T\cos\theta =m\omega^2 L\tag 1$$
$$\sum F_z=F_T\sin\theta-f_k=ma_T$$
where $f_k=\mu_kF_n$, thus
$$F_T\sin\theta-\mu_kF_n=ma_T \tag 2$$
$$\sum F_y=F_n-mg=ma_y=m(0)=0$$
Thus,
$$F_n=mg $$
Plugging into (2);
$$F_T\sin\theta-\mu_kmg=ma_T $$
Solving for $a_T$;
$$a_T=\dfrac{F_T\sin\theta-\mu_kmg}{m}=\dfrac{3.5\sin70^\circ-0.6(0.5)(9.8)}{0.5}$$
$$a_T=0.698\;\rm m/s^2\approx \bf 0.7\;\rm m/s^2$$
Now we can find the angular speed after 10 revolutions; where $\rm 1\;rev=2\pi \;rad$
$$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$$
where $a_T=\alpha R$;
$$20\pi =0+(0)t+\dfrac{a_T}{2R} t^2$$
Solving for $t$;
$$t=\sqrt{\dfrac{40 \pi R}{a_T}}=\sqrt{\dfrac{40 \pi (2)}{0.7}}=\bf 18.95\;\rm s$$
Thus,
$$\omega=\omega_0+\alpha t=0+\dfrac{a_T}{R}t=\dfrac{0.7}{2}\times 18.95 $$
$$\omega=\color{red}{\bf 6.632}\;\rm rad/s$$
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b) We can find the tension in the rod by solving (1) for it;
$$T =m\omega^2 L -F_T\cos\theta $$
We know the angular speed after 10 revs which we found above in part [a].
So, we just need to plug the known;
$$T =0.5(6.632)^2 (2) -3.5\cos70^\circ $$
$$T=\color{red}{\bf42.8 }\;\rm N$$
