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

To analyze the forces on A and B, we use the formula for centripetal force, which can be expressed as:

$F_c = \frac{m v^2}{r}$

For mass A rotating at radius $r_A$, the centripetal force is given by:

$F_{cA} = m_a \frac{v_A^2}{r_A}$

For mass B, which is further from the center at radius $r_B = L - r_A$, the centripetal force is:

$F_{cB} = m_b \frac{v_B^2}{r_B}$

Since the forces exerted by the rod on A and B are equal, we equate the two expressions and solve for $r_A$. After substituting $v_A = r_A \omega$ and $v_B = r_B \omega$, we can find that these relate back to write $r_A$ in terms of $m_a$, $m_b$, and L.

Given that $m_a < m_b$, we can analyze the distances. The linear speed is directly proportional to the radius; hence, as $r_A$ is smaller than $r_B$, the linear speed of the center of mass of B will be greater than that of A. Therefore, the centre of mass of B rotates with a greater linear speed.

To suggest a suitable diameter, we need to consider the maximum stress the rod will undergo. Assuming a safety factor of 3 (as it can withstand the operational stress), we calculate the maximum allowable stress. Given that the maximum stress from the curve in Figure 7 should not exceed 300 MPa, we can use:

$\sigma_{max} = \frac{F}{A} = \frac{m_a g + m_b g}{\frac{\pi d^2}{4}}$

From this, we can rearrange to solve for the diameter $d$. After plugging in the values for masses and the force due to gravity, we can derive a suitable diameter that ensures the rod can withstand the applied forces without failing.