91Ó°ÊÓ

Mass of block = 500 g = 0.5 kg

Radius of rotation = 1.2-m-long

Thrust force of 4.0 N perpendicular to the tube.

The maximum tension the tube can withstand without breaking is 50 N.

As the thrust is perpendicular to the tube. so , the tangential component of the thrust force is F_t=4N

and radial component F_r=0

Lets find the angular acceleration of the block

$\mathrm{Î\pm }=\frac{F_t}{mr}$

Substitute the values

$\mathrm{Î\pm }=\frac{4N}{(0.5kg)×(1.2m)}=6.667rad/s^2$

We can find the angular velocity when it breaks by equating radial force with maximum withstand force

$$\begin{gathered} mr{\mathrm{Ó\neg }}^2=50\mathrm{N} \\ {\mathrm{Ó\neg }}^2=\frac{50\mathrm{N}}{mr}âˆ\pounds \\ {\mathrm{Ó\neg }}^2=\frac{50\mathrm{N}}{0.5\mathrm{kg}×1.2\mathrm{m}} \\ \mathrm{Ó\neg }=9.129\mathrm{rad}/\mathrm{s} \end{gathered}$$

Use this angular speed and acceleration to find the rotations made.

$$\begin{gathered} {\mathrm{Ó\neg }}^2={\mathrm{Ó\neg }}_0^2+2\mathrm{Î\pm }\mathrm{θ} \\ \mathrm{θ}=\frac{{\mathrm{Ó\neg }}^2-{\mathrm{Ó\neg }}_0^2}{2a} \\ \mathrm{θ}=\frac{{(9.129rad/s)}^2-(0rad/s)2}{2×(6.667rad/s^2)} \\ \mathrm{θ}=6.25\mathrm{rad}=0.995\text{rotation} \end{gathered}$$