Dr. Doom on Planet IPhOO

The evil Dr. Doom seeks to destroy his enemy, the Intergalactic Federation, and has devised a plan to despin the Federation's space station. The hoop-shaped space station of mass 4.5 × 1 0 5 kg 4.5 \times 10^5 \, \text{kg} and radius 1 km 1 \, \text{km} rotates once every 24 24 hours to maintain artificial gravity equal to that on IPhOO. Dr. Doom plans to mount two thruster rockets, one rocket on opposite sides of the space station, to stop its rotation. Dr. Doom must accomplish his crime within a time 1 hr 1 \, \text{hr} to avoid getting caught. How much force should each rocket deliver in order to despin the Federation's space station in 1 hour? Express your answer in Newtons rounded to 4 significant figures.

Note : A more general version of this problem was proposed by Kimberly Geddes for the IPhOO .


The answer is 4.545.

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1 solution

Tran Duc
Apr 13, 2014

The inital velocity of the space station is ω = 2 π 86400 ( r a d / s ) \omega = \frac {2\pi}{86400} (rad/s)

\Rightarrow The needed deceleration to stop the station is α = 2 π 86400 1 3600 ( r a d / s 2 ) \alpha = \frac {2\pi}{86400} \cdot \frac {1}{3600} (rad/s^2)

The inertia of a hoop is I = m r 2 = 4.5 1 0 11 ( k g m 2 ) I = mr^2 = 4.5 \cdot 10^{11} (kg\cdot m^2)

The torque is τ = I α = F r \tau = I\alpha = Fr

F = I α r = 4.5 1 0 11 2 π 86400 1 3600 1 1 0 3 = 9.0902 \Rightarrow F = \frac {I\alpha}{r} = 4.5 \cdot 10^{11} \cdot \frac {2\pi}{86400} \cdot \frac {1}{3600} \cdot \frac {1}{10^3} = 9.0902

There are 2 rockets, so the force is 9.0902 / 2 = 4.545 9.0902 / 2 = \boxed{4.545}

Not sure if I was right. I think it's a bit "too" straightforward.

Btw, it seems that I can't post another solution once I deleted the first one.

Yep, its right , I did the same way.

Amish Naidu - 7 years, 1 month ago

how is the time period is 86400 seconds? pls explain.

Sudipan Mallick - 6 years, 11 months ago

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1 day = 86400 seconds.

Ahaan Rungta - 6 years, 11 months ago

The general form is much better.I like working with variables unlike the heavy calculations in this problem.

Ayon Ghosh - 3 years, 8 months ago

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