4. a space station consists of two outer wings with a distance of 64 m and two inner wings with a distance of 48 m from the center as shown. the station rotates to generate artificial gravity and the astronauts at \a\ experience an acceleration of 8.0 m/s².
a. what is the period of the rotation of the space station? (3 marks)

Question

4. a space station consists of two outer wings with a distance of 64 m and two inner wings with a distance of 48 m from the center as shown. the station rotates to generate artificial gravity and the astronauts at \a\ experience an acceleration of 8.0 m/s².
 a. what is the period of the rotation of the space station? (3 marks)

Accepted Answer

# Explanation:
## Step1: Recall centripetal acceleration formula
The centripetal acceleration formula is $a = \frac{v^{2}}{r}$, and also $v=\frac{2\pi r}{T}$, where $a$ is centripetal acceleration, $v$ is linear speed, $r$ is the radius of the circular path, and $T$ is the period.

## Step2: Substitute $v=\frac{2\pi r}{T}$ into $a = \frac{v^{2}}{r}$
Substituting $v=\frac{2\pi r}{T}$ into $a = \frac{v^{2}}{r}$, we get $a=\frac{4\pi^{2}r}{T^{2}}$. Then, solve for $T$:
$$
\begin{align*}
T^{2}&=\frac{4\pi^{2}r}{a}\
T& = 2\pi\sqrt{\frac{r}{a}}
\end{align*}
$$

## Step3: Plug in the values
Given $r = 48m$ and $a=8.0m/s^{2}$,
$$
\begin{align*}
T&=2\pi\sqrt{\frac{48}{8}}\
&=2\pi\sqrt{6}\
&\approx 2\times3.14\times2.45\
&\approx15.4s
\end{align*}
$$

# Answer:
The period of the rotation of the space station is approximately $15.4s$.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall centripetal acceleration formula

Step2: Substitute \(v=\frac{2\pi r}{T}\) into \(a = \frac{v^{2}}{r}\)

Step3: Plug in the values

Answer: