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All of the questions on this problem concern a space station, consisting of a lo

ID: 1623492 • Letter: A

Question

All of the questions on this problem concern a space station, consisting of a long thin uniform rod of mass B. 1.2 x 10^6 kg and length C. 417 meters, with two identical uniform hollow spheres, each of mass D. 2.4 x 10^6 kg and radius E. 147 meters, attached at the ends of the rod, as shown below. Note that none of the diagrams shown is drawn to scale! (a) Suppose that the station starts out at rest (not rotating). What we want is to get it spinning about an axis passing through its center of mass, at an angular velocity of F.0.14 rad/s, which is just what’s needed to produce 1-g of artificial gravity at the end points. To achieve this, we use rocket motors to apply a constant force F = G. 2.0 x 10^6 N to each sphere as shown, directed toward the centers of the spheres. How long must the motors fire in order to bring the station from rest up to an angular velocity of F. 0.14 rad/s? Answer: _________.___ minutes

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Explanation / Answer

Initial angular momentum of the station about the axis is zero, as it is initially at rest.

Finally the station rotates around its axis with and angular velocity = 0.14 rad/s.

Let the mass of rod be M, length of the rod be L, mass of the sphere be m and radius of the sphere be R.

The moment of inertia of the station about its axis is,

I = ML2/12 + 2[(2/3)mR2 + m(R + L/2)2]

or, I = (1.2X106 kg)(417 m)2/12 + 2(2.4X106 kg)[(2/3)(147 m)2 + (147 m + 417 m/2)2]

or, I = 693162.9X106 kg-m2  

So final angular momentum is,

L = I = (693162.9X106 kg-m2 )(0.14 rad/s)

or, L = 97042.806X106 kg-m2/s

Net torque provided by the forces is,

T = 2F(L/2 + R) = 2(2X106 N)(417 m/2 + 147) = 1422X106N-m

Let the time taken be t, then, angular impulse is,

J = Tt

Also, J = L2 - L1 = L = 97042.806X106 kg-m2/s

or, Tt = 97042.806X106 kg-m2/s

or, (1422X106)t = 97042.806X106 kg-m2/s

or, t = 68.24 s

or, t = 1.14 minutes

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