Question

A spaceship is designed with a central hub and two large pods. The pods are attached to the hub via light weight cables, like this:

(pod)----(hub)----(pod)

At it's full extension, the distance from each pod to the hub is 50m. In order to simulate gravity the ship rotates once every 14.2s. During an emergency, the crew needs to retract the pods to a distance of 25m from the hub. Calculate the acceleration experienced by any unfortunate crew members remaining in the pods.

Note: you will need to recall last weeks discussion of centripetal acceleration.

Answer 1

Explanation

• To simulate the gravity the spaceship rotates at a speed such that it creates a centripetal acceleration equal to the value of gravity at the 2 pods in this case.
• Centripetal acceleration depends on the radial distance from the center of rotation & the angular velocity, so, when the 2 pods are retracted the radial distance reduces causing the acceleration to increase as the angular velocity increases due to angular momentum that is being conserved.
• Here the orbital period of a pod is 14.2s

Notations

• Refer to the sketch(spaceship & the path of rotation) attached.
• (1)= Before the retraction & (2)=after it

Calculations

• Moment if inertia of a single / both pods = mr² / 2mr²
• Initial angular momentum of a hub = final angular momentum of a hub

$\qquad\qquad \begin{aligned} \small \omega I &= \small \small \omega_1 I_1\\ \small \frac{2\pi}{14.2s}\times \cancel{m}(50m)^2 &= \small \omega_1 \times \cancel{m}(25m)^2\\ \small \omega_1 &= \small \frac{2\pi\times4}{14.2} s^{-1}\\ \end{aligned}$

• Now the centripetal acceleration is given by r$\omega^2$

$\qquad\qquad \begin{aligned} \small a_{new} &= \small 25m \times \omega_1^2\\ &= \small \bold{78.31ms^{-2}}\\ \small a_{initial} &= \small 50m \times (\frac{2\pi}{14.2s})^2\\ &= \small 9.79ms^{-2} \end{aligned}$

• So any unfortunate remains in the hub experience acceleration of 78.31ms^-2 which is about 8 times greater than the simulated gravity.

Good Luck