All of the questions on this concern a space station, consisting of a long thin
ID: 2077662 • Letter: A
Question
All of the questions on this concern a space station, consisting of a long thin uniform rod of mass 2.3 x 106kg and length 561 meters, with two identical uniform hollow spheres, each of mass 1.7 x 106 kg and radius 258 meters, attached at the ends of the rod, as shown below. Note that none of the diagrams shown is drawn to scale!
(a) Now, another feature of this station is that the rod-shaped section can change its length (kind of like an old-fashioned telescope), without changing its overall mass and remaining uniform in its density. Suppose that, however it was accomplished, the station is now rotating at a constant angular velocity of 0.11 rad/s. If the length of the rod is reduced to 279 meters, what will be the new angular velocity of the space station? Answer: ____rad/s
(b) Let’s start again with the station not rotating, and back to its original size, with the rod again at 561 meters long. This time, we would like to get the station rotating at a rate of 0.11 rad/s, but now about an axis that passes straight down the length of the rod. We will accomplish this by placing a pair of rocket engines as shown, each again with a thrust of 1.4 x 106 N, on one of the spherical end modules. How long must these engines fire in order to get the station’s angular velocity up to 0.11 rad/s?
Answer: _______ minutes
Explanation / Answer
for the moment of inertia of the space station
I1 = 2 * 2/3 * 1.7 * 10^6 * 258^2 + 2 * 1.7 *10^6 * (258 + (561/2)^2) + 1/12 * 2.3 *10^6 * 561^2
I1 = 479589975000 Kg.m^2
a)
after decreasing the length
new moment of inertia , I2 = 2 * 2/3 * 1.7 * 10^6 * 258^2 + 2 * 1.7 *10^6 * (258 + (279/2)^2) + 1/12 * 2.3 *10^6 *279^2
I2 = 232839975000 Kg.m^2
Using conservation of angular momentum
I1 * w1 = I2 * w2
479589975000 * 0.11 = 232839975000 * w2
w2 = 0.226 rad/s
the final angular speed is 0.226 rad/s
b)
moment of inertia , I = 2 * 2/3 * 1.7 * 10^6 * 258^2
I = 150878400000 Kg.m^2
0.11 * 150878400000 = 2 * 1.4 *10^6 * 258 * t
solving for t
t = 23 s
the time taken is 23 s