Question

To stimulate the acceleration of large rockets, astronauts are spun at the end of a long rotating beam of length $9.8m$. What is angular speed required to generate a centripetal acceleration $8$ times the acceleration due to gravity?

Answer 1

As we know the formula of centripetal acceleration. For finding the angular speed, we can substitute the values in the equation, ${a_c} = {\omega ^2}r$. Here $\omega$ is the angular speed and ${a_c}$is the centripetal acceleration.

Complete step by step answer:
According to the given statement,
${a_c} = 8g$
Here ${a_c}$ is the centripetal acceleration and $g$ is the gravity.
We need to find the angular speed required to generate a centripetal acceleration which is eight times of the gravity.
$\omega = ?$
As we know, formula for the centripetal acceleration,
${a_c} = {\omega ^2}r$
By using this formula, we can final angular speed, centripetal acceleration and the radius is given.
So,
${\omega ^2} = \dfrac{{{a_c}}}{r}$
Now, substitute value of ${a_c} = 8 \times 9.8,r = 9.8$
We get-${\omega ^2} = \dfrac{{8 \times 9.8}}{{9.8}} = 8$
So, ${\omega ^2} = 8$
$\omega = \sqrt 8$
$\therefore\omega = 2\sqrt 2 rad/\sec$

So, astronauts require, $\omega = 2\sqrt 2 rad/\sec$ (angular speed) to generate centripetal acceleration $8$ times the acceleration due to gravity.

Note: If any object is tracing a circular path that will have acceleration vector pointed towards the centre of that circle is called centripetal acceleration. The force that causes the acceleration is directed towards the centre of the circle is known as centripetal force. Both the centripetal force and centripetal acceleration will have the same direction. SI unit of centripetal acceleration is $m{s^{ - 2}}$ .