Answer
$3.76\;\rm rev$
Work Step by Step
We need to find the number of revolutions until the tension force reaches 50 N and to do so we need to find its angular velocity at this time.
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;
$$\sum F_x=T=ma_x $$
where $x$ here is the radial direction at any moment while $z$ here is the tangential direction.
Thus,
$$T =m\dfrac{v^2}{R}$$
where $R=L$ which is the length of the massless rod and $v$ is given by $v=\omega R=\omega L$
$$T =m\dfrac{\omega^2 L^2}{L} $$
$$T =m\omega^2 L $$
Solving for $\omega$ when $T=50$ N;
$$\omega=\sqrt{\dfrac{T}{mL}}=\sqrt{\dfrac{50}{0.5\times1.2}}=\bf 9.13\;\rm rad/s\tag 1$$
$$\sum F_z=F_T -f_k=ma_T$$
where $f_k=\mu_kF_n$, thus
$$F_T -\mu_kF_n=ma_T \tag 2$$
$$\sum F_y=F_n-mg=ma_y=m(0)=0$$
Thus,
$$F_n=mg $$
Plugging into (2);
$$F_T -\mu_kmg=ma_T $$
Solving for $a_T$;
$$a_T=\dfrac{F_T -\mu_kmg}{m}=\dfrac{4-0.6(0.5)(9.8)}{0.5}$$
$$a_T= \bf 2.12\;\rm m/s^2\tag 3$$
Now we need to find the time it takes to accelerate from 0 rad/s to 9.13 rad/s and then we can find the number of revolutions. So we can apply the kinematic formula of
$$\omega=\omega_0+\alpha t$$
where $\omega_0=0$ and $\alpha=\dfrac{a_T}{R}=\dfrac{a_T}{L}$
Plugging from (1) and (3);
$$9.13 =0+\dfrac{2.12}{1.2} t$$
Hence, the time iterval is given by
$$t=\bf 5.17\;\rm s$$
Now we can use;
$$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$$
$$\Delta \theta =(0)t+\dfrac{a_T}{2L} t^2$$
Plugging the known;
$$\Delta \theta = \dfrac{2.12}{2\times 1.2} \times5.17^2=\bf 23.6\;rad$$
where $\rm 1\;rev=2\pi \;rad$
Therefroe,
$$\Delta \theta=23.6\;\rm rad\times \dfrac{1\;rev}{2\pi\;rad}$$
$$\Delta \theta=\color{red}{\bf 3.76}\;\rm rev$$
