Chapter 8, Problem 33P

To determine

To Find: The required study force of each rocket in which the satellite is $32\text{rpm}$ in $5.0\min$ .

Answer to Problem 33P

  $20\text{N}$

Explanation of Solution

Given information:

The satellite has a mass of $3600\text{kg}$ and a radius of $4.0\text{m}$ , and that the rockets each add a mass of $250\text{kg}$ .

Formula used:

  $\tau =I\alpha$

Here,

  $\tau$ is the torque,

  $I$ is the moment of inertia,

  $\alpha$ is the angular acceleration.

Calculation:

The firing force of the rocket will create a net torque but no net force. Since each rocket fires tangentially, each force has a lever arm equal to the radius of the satellite and each force is perpendicular to the lever arm. So the net torque is product of force and lever arm. To determine the steady force of each rocket by knowing the net torque. To calculate the angular acceleration and then the torque by using the equation rotational Kinematics. On equating the both to find the force.

The mass of satellite $M=3600\text{}\text{kg}$ , the radius of circle $R=4\text{m}$ , the angular velocity $\omega =32\text{rpm}$ and time taken $t=5\text{min}$ .

When each rocket firing force perpendicular to the lever arm equal to the radius of the satellite, as there are four rockets the net torque is

  $\tau =4FR$

The Torque

  $\tau =I\alpha$

From above two equations

  $4FR=I\alpha$

But, moment of inertia of solid cylinder is $I=\frac{1}{2}M{R}^2$ and the angular acceleration is $\alpha =\frac{\Delta \omega }{t}$

  $4FR=\left(\frac{1}{2}M{R}^2\right)\frac{\Delta \omega }{t}$

Therefore steady force,

  $\begin{array}{l}F=\frac{1}{8}\frac{MR\Delta \omega }{t}\\ =\frac{1}{8}\frac{\left(3600\text{kg}\right)\left(4.0\text{m}\right)\left(32\text{rev/min}\right)\left(\frac{\text{2π}\text{rad/rev}}{\text{60}\text{s/min}}\right)}{8\left(\text{5}\text{.0}\text{min}\right)\left(\text{60}\text{s/min}\right)}\\ F=20\text{N}\end{array}$

Conclusion:

The value of force is $20\text{N}$ .