Does Jumping in a Rotating Space Cylinder Make You Land Ahead?

AI Thread Summary
Jumping inside a rotating space cylinder results in landing ahead of the starting point due to the combination of tangential and radial velocities. When a jumper leaves the ground, they maintain their initial tangential velocity while following a straight path in free fall. The cylinder rotates beneath them, causing the jumper to intersect the edge of the cylinder before the starting point rotates to that location. This phenomenon is explained by conservation of angular momentum, where the jumper's radius decreases during the jump, increasing their tangential speed. Overall, the physics of rotation in a cylinder effectively simulates gravity in space, confirming that the jumper lands ahead.
Null-Set
Messages
2
Reaction score
0
Gravity in space is best simulated by rotation. If I were inside a rotating cylinder facing the direction of rotation (i.e. a window at my feet would show objects coming into view from the top of the window) and jumped straight into the air, would I land ahead of where I started or behind where I started?

I read that you should land behind where you started, which does not make sense to me. I think that you do drift as you jump due to Coriolis acceleration, but I never learned about that so I tried to work it all out in the inertial frame. The jumper and the space station start with a tangential velocity v_{t}. The jumper then jumps and leaves the ground with radial velocity v_{r}, from his perspective. Now that he is in free fall he will travel in a straight path until he intersects the edge of the cylinder again. Meanwhile, the cylinder rotates beneath him. The jumper is following a straight path at a speed \sqrt{v^{2}_{t} + v^{2}_{r}} and the starting point is moving along a curved path at a speed |v_{t}|. The jumper will intersect the cylinder again before the starting point rotates to that point, because the jumper is moving on a shorter path at a greater speed. Thus, the jumper lands ahead of where he started after jumping straight up.

Did I make any mistakes there?
 
Physics news on Phys.org
You're completely right. Another way to see it is conservation of angular momentum/ r v_t is constant for you and for the cylinder. Since your r is smaller while you jump, v_t must be larger, and your angular speed, which is v_t/r is also larger than that of the cylinder.

If you jump on the rotating earth, you will end up a little bit behind, because your r will be larger while you jump.
 
Null-Set said:
Gravity in space is best simulated by rotation. If I were inside a rotating cylinder facing the direction of rotation (i.e. a window at my feet would show objects coming into view from the top of the window) and jumped straight into the air, would I land ahead of where I started or behind where I started?

I read that you should land behind where you started, which does not make sense to me. I think that you do drift as you jump due to Coriolis acceleration, but I never learned about that so I tried to work it all out in the inertial frame. The jumper and the space station start with a tangential velocity v_{t}. The jumper then jumps and leaves the ground with radial velocity v_{r}, from his perspective. Now that he is in free fall he will travel in a straight path until he intersects the edge of the cylinder again. Meanwhile, the cylinder rotates beneath him. The jumper is following a straight path at a speed \sqrt{v^{2}_{t} + v^{2}_{r}} and the starting point is moving along a curved path at a speed |v_{t}|. The jumper will intersect the cylinder again before the starting point rotates to that point, because the jumper is moving on a shorter path at a greater speed. Thus, the jumper lands ahead of where he started after jumping straight up.

You can verify your reasoning visually with the following Java applet that is on my website:
http://www.cleonis.nl/physics/ejs/spacestation_vertical_throw_simulation.php"
Which has exactly the setup you are describing here
The applet allows you to vary several settings, so you can explore a range of cases.

Your reasoning is correct: if the jumper jumps up perpendicular to the local cilinder wall, then his velocity relative to the inertial frame is the vector sum of the tangential velocity of co-rotating with the cilinder, and the radial velocity of the jump. Hence with such a jump you will allways land ahead of the spot where you jumped from.

Cleonis
http://www.cleonis.nl
 
Last edited by a moderator:
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top