Torque and circular rotation question for high school physics?

Question

hi .... i'm hoping someone can help me with a question for my high school physics class ..... it's not homework but it's in the book and i want to figure it out ........ can some please help? i'm not sure what to do...... :-)


A space station consists of a circular tube which is set rotating about its center (like a tubular bicycle inner tube). The circle formed by the tube has a inner diameter of 1.5 km and an outer diameter of 1.6 km.

-- Where will the people inhabit this space station?

-- What must the rotation speed be if an effect equal to Earth’s gravity at the surface is to be felt by its occupants?

-- What torque must be applied to the station if it is to attain this speed from rest in two hours and the mass of the station is 187,016 kg?


can someone help? thanks

:-)

Answer 1

MIND!! PEOPLE WOULD LIVE INSIDE THE WHEEL TUBE, THEIR FEET ON THE OUTER DIAMETER, THEIR HEADS POINTED TO THE WHEEL AXIS!
Centripetal acceleration g=R*w^2, R=1600/2=800 m, angular speed to be found when g=9.8 m/s^2 exactly, hence w=sqrt(g/R)=sqrt(9.8/800)=0,1107 rad/s.
To reach the angular speed w in t=2h=7200s, we must apply angular acceleration w’=w/t=0.1107/7200=1.537*10^-5 rad/s^2. Inertia moment of the wheel is I=m*Rm^2, m=187016kg, mean radius Rm=(1600-1500)/2 m, so I=4.68*10^8 kg*m^2.
Thus torque T=I*w’=7186 Nm – never believe us! Check it!

Answer 2

hmm. where do i start..

the people will be inside this tube, and actually be walking with the inner diameter as the "floor".

gravitational field strength is 9.81N per kg. this is also known as acceleration. we will equate this to the centripetal acceleration of a spinning space station, where 9.81=v²/r. where v is the tangential velocity we want to find, and r is the distance from the centre of the space station to the inner diameter (750m). any object that has a mass will feel a gravitational acceleration of 9.81 when it is resting on the inner diameter.

v=sqrt(750x9.81)=85.8m/s

well now we know the speed, we are given 2hours to accelerate the station to it. this acceleration would be 85.8/7200seconds=0.0119 m/s². using newton's F=ma, we can find the force required to make this acceleration. F=0.0119x187016=2228N.

note that this force is applied at the inner diameter since the gravity calculation and speed calculation were both related to the inner diameter only. torque would be 2228 x 750m=1.67million Nm

have fun on your space station

Answer 3

- they will live standing oposite to the rotation center since the rotation is what is supposed to simulate gravity. That is, head to the rotation center, feet outside - I am mathematically stupid, so I can not calculate that, but I guess you will have to rotate it in a speed that will drive people out with a force equeal to 9.8m/s^2 - Again mathematically stupid. At least I knew 1 answer, didn't I?

Answer 4

Go with kaksi_guy. He's got the right method.