Does the elevator speed going up effect the free fall time?

[Physicsisfun2005]

I ran across this physics problem in class yesterday, it seems simple but...... (assume simplest case) ok...a screw at the top of an elevator that is traveling upwards comes loose and falls. Does the elevator speed going up effect the free fall time? (like does the elevator move toward the screw?) If the elevator is 2.5m tall and is not moving i figure free fall is .714 sec. (h=.5gt^2), is it the same time say if the elevator it traveling at 5 m/s?

 

[Doc Al]

Assuming the elevator is moving uniformly, its speed won't affect the time the screw takes to fall. If you didn't know the elevator were moving, you would not be able to tell by any mechanical experiment done within the elevator. This is called the principle of Galilean relativity.

 

[Physicsisfun2005]

thank you

 

[HallsofIvy]

Doc Al gave the simplest answer. Here's the more complicated, detailed answer:

If the elevator is moving upward with speed (say) v₀, then, until it actually comes loose and starts to fall the screw is also moving upward at that speed. That will be the initial speed when it starts to fall. Take the time at which the screw starts to fall to be 0. As soon as the screw comes loose, it starts to accelerate (downward) with acceleration -g. At time t after coming loose, it has speed -gt+ v₀. If we take the position of the base of the elevator at t=0 to be 0 and the elevator has height h, then the intial position of the screw is h so the position of the screw at any time t is -(g/2)t²+ v₀t+ h. The position of the bottom of the elevator is v₀t. The screw hits the floor when -(g/2)t²+ v₀t+h= v₀t. The point is that the two "v₀t" terms cancel so the speed of the elevator, v₀, is irrelevant.

 

[pallidin]

Relative to the ground "the bottom of the shaft" the screw will fall at a constant rate regardless of elevator speed.
Relative to the floor of the ascending elavator, however, the situation is much different from an occupants viewers point. The rate of the screws' acceleration is the same, but the time for point of impact with the floor is reduced.

 

[Doc Al]

Originally posted by pallidin
Relative to the ground "the bottom of the shaft" the screw will fall at a constant rate regardless of elevator speed.

I don't understand this statement. If you are talking about the speed of the screw with respect to the ground, its speed is V₀-gt (as Halls explained). V₀ is the speed of the elevator (and thus the initial speed of the screw) so the speed of the screw (with respect to the ground) does depend on the speed of the elevator.

Relative to the floor of the ascending elavator, however, the situation is much different from an occupants viewers point. The rate of the screws' acceleration is the same, but the time for point of impact with the floor is reduced.

Another puzzling statement. An observer on the ground and an observer in the elevator will both agree on the time it takes for the screw to fall to the floor of the elevator.

 

[pallidin]

Doc,

I see I managed to confuse myself. Thanks for your corrections.

Pallidin