Question

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?

Short answer

a) Block’s angular velocity after 10 rev is 20.33 rad/sec or 194.1372 rpm

b) Tension in the rod after 10 rev is T=412.111 N

Step-by-step solution

Part(a) Step 1: Given Information

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

Part(a) Step 2: Explanation

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

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

Part(b) Step 1 : Given information

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.

Part(b) Step 2 : Explanation

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

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