The other side of the lever If I have a lever, but I can see only up to the hinge and not the other half, can I know whether the other half is 1 m long with a weight of 3 kg on it, or 3 m long with a weight of 1 kg on it? 
 A: Yes, you can. The moment of inertia of the lever will be different in each of the two situations. Let us assume that the lever is massless and the weight is a point mass. In the first situation, $I = mr^2 = (3)(1)^2 = 3 \text{ N} \cdot \text{m}$. In the second situation, $I = mr^2 = (1)(3)^2 = 9 \text{ N} \cdot \text{m}$.
Because the moment of inertia is greater in the second setup, the lever arm will be slower to respond to applied torques (that is, since $\tau = I \alpha$, a larger moment of inertia corresponds to a smaller angular acceleration for a given torque). To practically determine the difference, you can place a heavy weight on one end of the lever and use a stopwatch to record the amount of time the lever takes to reach a certain angular displacement. The one with a smaller time is setup with a 3 kg mass 1 m away, and the one with a higher time is a 1 kg mass 3 m away.
A: Just for fun let me suggest another rather impractical way to tell the difference.

The diagram shows the far side of the lever. It has a length $L$ that you don't know and there is mass $m$ on the end that you don't know. The torque is equal to $Fd = FL\cos\theta$, and the force is the gravitational force $F = GMm/r^2$, where $M$ is the mass of the Earth and $r$ is the radius of the Earth. So:
$$ T = \frac{GM mL\cos\theta}{r^2} $$
Only this isn't quite right because the distance from the centre of the Earth isn't $r$, but rather $r + L\sin\theta$. So a more accurate calculation of the torque is:
$$ T' = \frac{GM mL\cos\theta}{(r + L\sin\theta)^2} $$
The equation for the torque ignoring the change of gravity, $T$, only contains the product $mL$, so you can't tell the difference between $m = 1, L = 3$ and $m = 3, L = 1$ as in both cases the product $mL = 3$. However the second equation that includes the change in gravity has a separate dependance on just $L$ in the denominator, so you can tell the difference. In fact if you graph the difference between the equations, $T - T'$, you get:

The difference is very small, only around $10^{-5}$ Nm, but it is there. So by very precisely measuring the torque as a function of angle you can tell the difference between the two cases.
