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?

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    $\begingroup$ Do they behave differently? $\endgroup$ – LDC3 Apr 3 '14 at 13:13
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    $\begingroup$ You have to know the weight on the hinge. Then you can deduce the weight on the other side. $\endgroup$ – John Alexiou Apr 3 '14 at 15:27
  • $\begingroup$ @ja72: assuming the lever is masslessl $\endgroup$ – Quora Feans Apr 3 '14 at 16:31
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    $\begingroup$ @QuoraFeans Not weightless, but of known weight. $\endgroup$ – John Alexiou Apr 3 '14 at 17:31
  • $\begingroup$ @ja72 It seems odd to assume that you know the mass of the rod but not its length. $\endgroup$ – David Richerby Apr 3 '14 at 22:47

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.

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    $\begingroup$ Yes, it's a trick question, the answer is to wiggle your lever ;-) $\endgroup$ – user7917 Apr 4 '14 at 9:02

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:

Lever 2

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.

  • $\begingroup$ If the lever was massless and one pivoted the bar so as to have maximum static gravitational torque from the weight on it, wouldn't the rotational acceleration caused by the 3kg weight be three times as fast as that of the 1kg weight? I would think that would be very noticeable. $\endgroup$ – supercat Apr 4 '14 at 0:13
  • $\begingroup$ @supercat: that's covered by Draksis' answer. He got in first so I had to come up with something a bit more extreme. Note that my method is purely static. You just measure a static torque. Draksis/your method is dynamic i.e. you have to measure an acceleration. $\endgroup$ – John Rennie Apr 4 '14 at 5:57
  • $\begingroup$ @JohnRennie: Draskis' answer speaks in terms of response to applied torque. I was thinking in simpler terms: if both weights start 90 degrees out, the 1m arm will have a rotational acceleration of 9.8 radians/sec/sec, while the 3m arm will have a rotational acceleration 1/3 that. No need for any sort of calibrated torque maximum (just finding the angle of maximum torque as an initial condition). $\endgroup$ – supercat Apr 4 '14 at 14:58

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