A 500 g steel block rotates on a steel table while attached to a 2.0-m-long massless rod. Compressed air fed through the rod is ejected from a nozzle on the back of the block, exerting a thrust force of 3.5 N. The nozzle is 70^o­ from the radial line, as shown in FIGURE P8.62. The block starts from rest.
a. What is the block’s angular velocity after 10 rev?
b. What is the tension in the rod after 10 rev?

Mass of block = 500 g =0.5 kg
Length of rod = 2 m
Thrust force = 3.5 N.
The nozzle is at an angle of 70^o from the radial line as shown.
The block starts from rest

The centripetal force is calculated by

mv²/r ……………………..(1)

The angular velocity of the mass revolving around a circle is calculated by

${\omega }^2={\omega }_0^2+2\alpha \theta ...............................\left(2\right)$

Now break the thrust in tangential and radial components

$$\begin{gathered} F_t=3.5sin(70°)N \\ {F}_r=3.5cos(70°)N \end{gathered}$$

Now find angular acceleration of the block

$\alpha =\frac{F_t}{mr}$

Substitute values

$\alpha =\frac{(3.5N)\sin (70°)}{(0.5kg\times 2m)}=3.289rad/s^2$

Substitute in equation (2) to get angular velocity after 10 rev

$$\begin{gathered} {\omega }^2=0^2+2(3.289rad/s^2)\left(20\pi rad\right) \\ \omega =20.33rad/s \end{gathered}$$

Mass of block = 500 g =0.5 kg
Length of rod = 2 m
Thrust force = 3.5 N.
The nozzle is at an angle of 70^o from the radial line as shown.
The block starts from rest.

The centripetal force acting on the block is calculated by substituting values in equation (1) $\left(m{v}^2\right)/r$

substitute v= ωr

$\frac{m{v}^2}{r}=\frac{\left(m\right(r\omega {)}^2)}{r}=mr{\omega }^2$

Tension in the rod after 10 rev is

$$\begin{gathered} T=mr{\omega }^2-F_r \\ Substitutevalues \\ T=(0.5kg)\left(2m\right)\left(2\right(3.289rad/s^2)\left(20\pi rad\right)-3.5\cos (70°) \\ T=412.11N \end{gathered}$$