0
$\begingroup$

While answering a question about gears and levers, I found I could not get the work in to equal the work out in this simple gear system illustrated below:

enter image description here

The gears are set up to replicate a lever, by having two gears connected by a third idler gear in the middle, which ensures the two outer gears rotate in the same direction. The size of the middle gear is not important, as it is only there to reverse the directions of the outer gears relative to each other. (We could replace the idler with a bicycle chain around the perimeters of the gears and the result would be the same.)

Pulleys that are the same diameter as their respective gears are attached to the shafts of the outer gears so that string wrapped around the pulleys will raise or lower the weights vertically. A mass of M attached to the larger gear with radius 2R should balance a mass of 2M attached to the gear/pulley with radius R, because the torques are equal, just like a lever with one arm twice as long as the other.

Now consider what happens when we add a small weight of say 0.1M to the mass on the left. The anticlockwise torque due to the weight of 1.1 Mg acting on the large gear now exceeds the clockwise torque due to the weight of 2 Mg acting on the smaller gear and the mass on the right should rise. This is where the paradox arises. The work done by the weights on the left appears to be less than the work output on the right.

For the larger gear the work input is:

$$W_1 = F_1 \times h_1 = (1.1M) g \times \theta (2R) = 2.2Mg\theta R$$

For the smaller gear, the work output is:

$$W_2 = F_2 \times h_2 = (2M) g \times (2\theta R) = 4Mg \theta R$$

This appears to violate conservation of energy. Work out exceeds work in. Obviously that cannot happen, so I must be overlooking or misunderstanding something.

If we look at the changes in gravitational energy due the vertical movement of the weights, there is still a problem. The weight on the left loses Mgh in potential energy and the weight on the right gains 2Mgh. The distance both masses move is the same because the arc lengths are the same. (Arc length on the left equals $\theta (2 R)$ and arc length on the right equals $(2 \theta) R$.) Since the distance the weights move in a given time interval, the velocities of the weights must be equal at any given time. This means if the kinetic energy of the falling weight on the left is $KE_1 = (1/2) 1.1M v^2$ at a given instant, then the kinetic energy of the rising weight on the right would $KE_2 = (1/2) 2M v^2$ and still the energy output appears to be greater than the energy input.

The apparent paradox is this. If the weight on left is 1.1Mg and falls a distance h, the torque on the left exceeds the opposing torque on the right and the gears must rotate and the weight on the right must rise by the same distance h, but if it does, the work out exceeds the work in. How is this paradoxical situation resolved?

$\endgroup$

2 Answers 2

2
$\begingroup$

A mass of M attached to the larger gear with radius 2R should balance a mass of 2M attached to the gear/pulley with radius R, because the torques are equal, just like a lever with one arm twice as long as the other.

Here is the first mistake in your analysis. The way that you have the gearing set up this is just a confusing Atwood machine. The forces do not balance, this is not analogous to a lever. The mass of $2M$ simply goes down and raises the mass of $M$ the same distance, gaining KE for both masses in the process.

To see that it is not balanced, consider the forces acting on the middle gear. The force required to balance the red gear is $Mg$, while the force required to balance the blue gear is $2Mg$. Those two forces are at the same radius for the middle gear, so there is a net torque contradicting the assumption that it is balanced.

Now consider what happens when we add a small weight of say 0.1M to the mass on the left. The anticlockwise torque due to the weight of 1.1 Mg acting on the large gear now exceeds the clockwise torque due to the weight of 2 Mg acting on the smaller gear and the mass on the right should rise.

This is incorrect, the mass on the left still rises. The same qualitative behavior as before occurs, but the KE is slightly less.

$\endgroup$
2
  • $\begingroup$ Your answer is way better than mine. Please take my upvote. In fact, please downvote my answer so that your answer will rise to the top. $\endgroup$ Commented May 19 at 20:07
  • $\begingroup$ @Dale I think you have found the solution. I can now see that it takes a mass of just M suspended on the smaller gear to obtain equilibrium. The opposing torques on each gear wheel are then equal and the work in is equal to the work out. +1 $\endgroup$
    – KDP
    Commented May 19 at 22:55
1
$\begingroup$

You asserted that the gear in the middle doesn't matter, but it does. When the gears are turning, you will have a ridiculous situation, because the turning of the gear in the middle cannot be reconciled with both turning angles of the other two gears. Your system is thus impossible even theoretically.

$\endgroup$
7
  • $\begingroup$ Are you saying the smaller gear on the right, will rotate at the same rate as the larger gear on the left? $\endgroup$
    – KDP
    Commented May 19 at 14:59
  • $\begingroup$ @KDP It is right that middle gear matters because it is a disc and changing its radius change its moment of inertia in the order of $R^2$. $\endgroup$ Commented May 19 at 15:15
  • $\begingroup$ What about if the middle gear is replaced by a chain like on a bicycle? I still get the same outcome. $\endgroup$
    – KDP
    Commented May 19 at 15:17
  • 1
    $\begingroup$ Check out this very cool real time gear simulator. geargenerator.com/… I have set it initially so that the middle gear is the same radius as the gear on the right. The output gear always rotates at twice the angular velocity of the input gear, independently of the radius of the middle gear. You can change the radius of the middle gear by clicking on it and changing the number of teeth and see for yourself. $\endgroup$
    – KDP
    Commented May 19 at 15:22
  • $\begingroup$ Here is gear simulation set up with a chain drive so the issue with the middle idle gear is removed. geargenerator.com/beta/… Clearly my gear set up can exist and function and your answer does not resolve the issue. $\endgroup$
    – KDP
    Commented May 19 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.