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In-s [12.5K]
3 years ago
12

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

Physics
1 answer:
nataly862011 [7]3 years ago
3 0

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

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