How can one determine if they are not experiencing Earth gravity? Suppose you are kidnapped and you don't know who did it. You have no memory of the trip that brought you to the building where you are now being held. You have reason, however, to suspect that you are not on earth but rather in a large (5km diameter possibly larger) space habitat that uses gravity simulated through rotation. There are many typical objects in the room, possibly even some devices for making measurements if needed. 
So, trapped inside of a room without windows, is there any way to prove you are not on earth using simple found objects and your understanding of physics?
Would the answer to this question change if you are on a non-spinning spaceship that accelerates at 9.8 m/s*s?

 A: If you want to check whether you're rotating this is pretty easy. (1) Measure the Coriolis force. (2) Measure nonuniformity of the centrifugal force (gradients indicating variations in strength and direction). This is how we can prove that the earth is spinning, using a Foucault pendulum.

Would the answer to this question change if you are on a non-spinning spaceship that accelerates at 9.8 m/s*s?

The equivalence principle says that you can't tell the difference between this situation and a uniform gravitational field. However, you could check for the nonuniformities that you expect if you're near a spherical gravitating body.
A: I think it would be worth picking up on Ben's point about measuring non-uniformities, which I assume means measuring the variation in the gravitational acceleration with height.
Assuming you have a ruler and some sort of timing device you can measure the gravitational acceleration by making a pendulum and measuring its period. Since the period is given by:
$$ \tau = 2\pi \sqrt{\frac{\ell}{g}} $$
You can use the pendulum to measure $g$, and you can measure how $g$ changes with height. On the Earth's surface the value of $g$ is given by:
$$ g = \frac{GM}{r^2} $$
with $r$ equal to the radius of the Earth, $r_e$. Differentiating this gives us the equation describing how $g$ varies with height:
$$ \frac{da}{dr} = -\frac{2GM}{r^3} $$
At the Earth's surface this has the value $3.07 \times 10^{-6}$ s$^{-2}$.
In a uniformly accelerating spaceship the variation of $a$ with height would be zero, so that's easily distinguishable.
In the stereotypical rotating torus the acceleration is given by:
$$ a = r\omega^2 $$
with $r$ being the radius of the torus and $\omega$ the angular velocity, so we get:
$$ \frac{da}{dr} = \omega^2 $$
So this time we find $da/dr$ is a constant rather than proportional to $r^{-3}$. We'd also measure a different value. Suppose we have a space station with a radius of $100$m, so the angular velocity has to be about $0.31$ radians per second to get an acceleration equal to $g$. In that case when we measured $dg/dr$ we'd get the value $0.0981$ s$^{-2}$. And again this is easily distinuishable from the value on Earth.
A: General answer without getting into mathematics -
First Part - 


*

*Set a pulley at pretty high place in the room. 

*Put a weight directly below the pulley

*Tie a good quality string to the weight

*Pull the string to lift the weight so it goes significantly higher than its original position.

*Due to conservation of angular momentum, the weight will lean to forward direction of rotation. This way, you can also find direction of rotation. You may use a laser pointer at the bottom of the weight and note the shift in the laser projected on the floor as the weight goes from bottom to the top.
You may also drop something from a height that would act as a projectile and would not land exactly below it.
Second Part is tricky due to equivalence principle so, I will try and wait for others to comment - 
Due to the practicality of imparting a constant force, it can be distinguished in following manner - 
Lift something significantly heavy to the roof. Leave a balance with a weight on it at a distance. The balance should start recording the weight now. Then let the heavy weight fall freely to the floor. Suppose, it does not bounce at all post fall.
I will take four scenarios here - 


*

*The space ship is accelerated by a constant force. In this case, the time taken by the fall will be smaller than the one predicted by the perceived g as the ship will accelerate more during the fall. The balance will record higher weight during the fall, and lower weight on impact and then correct weight. i.e. just one cycle of weight change - higher, lower, correct.

*The room is a blimp hovering in the atmosphere of a planet - In this case, the balance would record multiple cycles of higher and lower weight due to its movement back and forth before it balances again.

*The room is fitted on springs on the surface of a planet - In this case, the balance would record multiple cycles of higher and lower weight due to its movement back and forth before it balances again.

*The room is actually on the surface of a planet - In this case, the balance would record no cycles of higher and lower weight.
I will wait for others to point out if there are any problems with this explanation, or if there are other scenarios that need to be explained.
