A.) Consider a non-rotating space station in the shape of a long thin uniform rod of mass 4.96 x 10^6 kg and length 530 meters. Rocket motors on both ends of the rod are ignited, applying a constant force of F = 4.99 x 10^5 N to each end of the rod as shown in the diagram, causing the station to rotate about its center. If the motors are left running for 1 minutes and 28 seconds before shutting off, then how fast will the station be rotating when the engines stop?
Multiple choice
- 1.34 rpm
- 2.29 rpm
- 3.44 rpm
- 1.91 rpm
B.) This time we have a non-rotating space station in the shape of a long thin uniform rod of mass 4.89 x 10^6 kg and length 1182 meters. Small probes of mass 6526 kg are periodically launched in pairs from two points on the rod-shaped part of the station as shown, launching at a speed of 3389 m/s with respect to the launch points, which are each located 385 m from the center of the rod. After 13 pairs of probes have launched, how fast will the station be spinning?
Multiple choice
- 3.71 rpm
- 1.86 rpm
- 4.82 rpm
- 1.12 rpm
A) 1.91 rpm
moment of inertia of space station, I = M*L^2/12
angular acceleration, alfa = Torque/I
= 2*F*r/I
= 2*F*(L/2)/(M*L^2/12)
= 2*4.99*10^5*(530/2)/(4.96*10^6*530^2/12)
= 0.0022778 rad/s
angular velocity after t = 1 minute 28 s = 88 s
w = wo + alfa*t
= 0 + 0.0022778*88
= 0.2004 rad/s
= 0.2004*60/(2*pi) rev/min
= 1.91 rmp
B) 3.71 rpm
given
M = 4.89*10^6 kg
L = 1182 m
m = 6526 kg
v = 3389 m/s
r = 385 m
Apply conservation of momentum
angular momentum of rod = angular momentum of 13 pairs of probes
I*w = 2*13*m*v*r
(M*L^2/12)*w = 26*m*v*r
w = 26*m*v*r/(M*L^2/12)
= 26*6526*3389*385/(4.89*10^6*1182^2/12)
= 0.38886 rad/s
= 0.38886*60/(2*pi) rev/min
= 3.71 rpm
Get Answers For Free
Most questions answered within 1 hours.