Figure 6 shows a rotating spacecraft that is proposed to carry astronauts to Mars - AQA - A-Level Physics - Question 4 - 2021 - Paper 1

To derive the relationship for $r_A$, we can equate the centripetal force acting on both A and B. The centripetal force for mass A is given by $F_A = m_a \frac{v_A^2}{r_A}$ and for mass B, $F_B = m_b \frac{v_B^2}{r_B}$. As both parts are connected, the angular velocities ($\omega$) are the same, and we have:

$v_A = \omega r_A$
$v_B = \omega r_B$

By substituting $v_A$ and $v_B$ into the centripetal force equations, equating forces, and considering that the distances relate by $r_A + r_B = L$, we arrive at the equation for $r_A$ as follows:

$r_A = \frac{m_b L}{m_a + m_b}.$

Since $m_a < m_b$, it follows that $r_A < r_B$. Given that linear speed ($v$) relates to the radius of rotation, the linear speed of A (which is closer to the center of mass) will be lower than that of B. Thus, the centre of mass of B rotates with a greater linear speed.

To suggest a suitable diameter for the rod, we must consider the maximum stress the rod can endure. The safety factor should be taken into account, with the maximum stress for the material being less than 300 GPa from the stress-strain curve. Given the loaded conditions, we can calculate the stress exerted on the rod using:

$\text{Stress} = \frac{F}{A}$

Where $F$ is the force and $A = \frac{\pi d^2}{4}$. By rearranging and substituting the appropriate values, we can estimate an appropriate diameter ensuring it does not exceed the material's maximum stress capacity. A calculated diameter in the range of 0.01m (10mm) may be suitable, ensuring ample safety and structural integrity under load.