Questions concern a space station, consisting of a long thin uniform rod of mass

Question

Questions concern a space station, consisting of a long thin uniform rod of mass 3.7x 10^6 kg and length 621 meters, with two identical uniform hollow spheres, each of mass 2.8 x 10^6 kg and radius 188 meters, attached at the ends of the rod, as shown below. Note that none of the diagrams shown is drawn to scale!

(c) 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 F rad/s, If the length of the rod is reduced to K. meters, what will be the new angular velocity of the space station? Answer: rad/s

Explanation / Answer

lnitialy,

M = mass of rod = 1.2*106 kg

L = length of rod = 417 m

Irod = moment of inertia of rod = ML2 / 12 = (1.2*106) * (417)2 /12 = 1.73*1010 kgm2

m = mass of each hollow sphere = 2.4*106 kg

r = radius of sphere = 147 m

Isphere = moment of inertia of sphere about the axis = (2/3) mr2 + m (L/2)2

Isphere = (2/3)*2.4*106 * (147)2 + (2.4*106) (417 / 2)2 = 13.89*1010 kgm2

wi = initial angular velocity = 0.14 rad/s

After changing rod length,

Irod = moment of inertia of rod = ML'2/12 = (1.2*106) * (208)2 / 12 = 0.432*1010 kgm2

Isphere = (2/3) mr2 + m (L/2)2

Isphere = (2/3)*2.4*106 * (147)2 + (2.4*106) (208 / 2)2 = 6.053*1010 kgm^2

Let, Final angular velocity = wf

using conservation of angular momentum

Ii wi = If wf

(Irod + 2 Isphere) wi = (Irod + 2 Isphere) wf

(1.73*1010 + 2*13.89*1010)*0.14 = (0.432*1010 + 2*6.053*1010) * wf

wf = 0.33 rad/s

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