Question

Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exceed 0.02 rad/s. If the astronaut is moving parallel to the x axis and the position of her center of mass when she attaches is (- 1.8, - 0.9, 0) m, what is the maximum relative velocity at which she should approach the satelfae?

Answer 1

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

$\omega=\frac{v}{r}$

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

$|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}$

the maximum relative velocity is 0.04m/s

hope this helps!!

Answer 2

Answer:

The maximum relative velocity = 0.0296m/s

Explanation:

Data Given;

angular velocity = 0.02rad/s

r = -1.8m and -0.9m

To calculate the relative velocity, we use the formula;

v = wr-------------------------------1

where;

v = Linear velocity

w = angular velocity

r = radius

from the question, the resultant radius is calculated as;

r = √rx² + ry²

   =√ (-1.18² + -0.9²)

  =√(1.3924 +0.81)

   =√2.2024

r   = 1.48m

Substituting the values into equation 1, we have

v = wr

= 0..02 * 1.48

= 0.0296m/s