Torque on Axle and Wheel I was solving a physics problem when I ran into the following conundrum.
Consider a wheel attached to an axle. The wheel has a radius of $R$, which is twice the radius of the axle, $r$. Assume both wheel and axle have the same mass. We spin the axle (through the horizontal axis) by applying a force on it. Now as a result, the axle starts moving with an angular acceleration of $\alpha$. Additionally, the wheel also starts moving with an angular acceleration of $\alpha$.
Thus, the axle has a torque equal to $I_{axle} \alpha$, while the wheel has a torque equal to $I_{wheel} \alpha$. We can simplify this as follows:
$I_{wheel} = mR^2$
$I_{axle} = mr^2$
Thus, we see that the two torques are indeed different. However, the solution that I was reading stated that the torque on a wheel would always be equal to the torque on its axle. But with the logic we just used, we prove otherwise. What have I missed?
 A: 
Thus, we see that the two torques are indeed different. However, the
  solution that I was reading stated that the torque on a wheel would
  always be equal to the torque on its axle. But with the logic we just
  used, we prove otherwise. What have I missed?

It is correct the torques would be different if you were to consider the axle and the wheel as two separate objects. But they are not separate objects. They comprise one object subjected to a single torque. What is different is that the force on the wheel is less than that on the axle because torque equals the force applied at the radius, times the radius. 
Thus for the wheel (outer) portion we have torque 
$$T_{wheel}=F_{wheel}R$$
And for the axle (inner portion) we have torque
$$T_{axle}=F_{axle}r$$
The two torques are equal so we have
$$F_{wheel}R=F_{axle}r$$
or
$$F_{wheel}=\frac{F_{axle}r}{R}$$
So the force at the maximum radius of the wheel is less than the force at the radius of the axle to achieve the same torque. This tells us we can obtain the same torque by increasing the force tangent to the radius linearly with the radius;
We can come to the same conclusion using formulas for torque in terms of the moment of inertia times angular acceleration. First for the wheel.
$$T_{wheel}=mR^{2}α=F_{wheel}R$$
Therefore
$$F_{wheel}=mRα$$
Likewise for the axle
$$T_{axle}=mr^{2}α=F_{axle}r$$
Therefore
$$F_{axle}=mrα$$
Again since the torques are the same
$$F_{axle}r=F_{wheel}R$$
or 
$$F_{wheel}=\frac{F_{axle}r}{R}$$
as before.
Hope this helps.
A: I think you are confusing a simple idea because you are focussing on torque as an example. If instead you consider force you might find the principle easier to ponder. 
Imagine, therefore, that you are pushing a 20kg shopping trolley that contains a 10kg carton. In this example, the trolley is the equivalent of the axle, and the carton represents the wheel. You can model this in two ways. One is to assume that the trolley and carton are a single entity with a single force acting on them. On the other hand, you can argue that since the trolley and carton are accelerating at the same rate, by F=ma the carton must be subject to half the force that the trolley experiences, because it has has half the mass. You can now persuade yourself that there is a contradiction, because one view says the same force applies to both objects while the other says they must experience different forces. This is exactly what you are doing in your example. There is no contradiction- one view is simply looking at the two objects in aggregate, while the other is distinguishing between components of the overall system at a lower level of detail.
In the case of the trolley and carton, what is 'really' happening is that you are pushing the trolley with three units of force, sufficient to accelerate both objects, the trolley is pushing the carton with one unit of force, and the reaction of the carton against the trolley means that the trolley is subject to two units of force. However, it is much easier to consider the trolley and carton as collectively experiencing the same three units of force- the mathematical answer is identical in either case.
