Question

An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of

Answer 1

Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

$I_i\omega_i= I_f\omega_f$

$mr_i^2\omega_i=m r_f^2\omega_f$

$\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i$

$\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i$

we know,

centripetal acceleration

$a = \dfrac{v^2}{r}$

v = r ω

$a =\omega_f^2 r_f$

$a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f$

$a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}$

$r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}$

$r_f^3=20408.1632$

$r_f = 27.32\ m$

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m